Mixing
Non-integral powers of a matrix
$begingroup$
Question
Given a square complex matrix $A$, what ways are there to define and compute $A^p$ for non-integral scalar exponents $pinmathbb R$, and for what matrices do they work?
My thoughts
Integral exponents
Defining $A^k$ for $kinmathbb N$ is easy in terms of repeated multiplication, and works for every matrix. This includes $A^0=I$. Using $A^{-1}$ as the inverse, $A^{-k}=left(A^{-1}right)^k$ is easy to define, but requires the matrix to be invertible. So much for integral exponents.
Rational definition
I guess for a rational exponent, one could define
$$A^{frac pq}=Bquad:Leftrightarrowquad A^p=B^q$$
This will allow for more than one solution, and I'm not sure if the computations I'll describe below will find all solutions satisfying the above equation. So I'm not sure whether that's a reasonable definition. For non-rational exponents, a limit using a convergent series of rational exponents might work.
Diagonalizable computation
If $A$ is diagonalizable, then one has $A=W,D,W^{-1}$ for some diagonal matrix $D$. One can simply raise all the diagonal elements to the $p$-th power, obtaining a matrix which will satisfy the above equation. For each diagonal element, I'd define $lambda^p=e^{(plnlambda)}$, and since $lnlambda$ is only defined up to $2pi imathbb Z$, this allows for multiple possible solutions. If one requires $-pi<operatorname{Im}(lnlambda)lepi$, then the solution should be well defined, and I guess this definition even has a name, although I don't know it.
Non-diagonalizable computation
If $A$ is not diagonalizable, then there is still a Jordan normal form, so instead of raising diagonal elements to a fractional power, one could attempt to do the same with Jordan blocks. Unless I made a mistake, this appears to be possible. At least for my example of a $3times3$ Jordan block, I was able to obtain a $k$-th root.
$$
begin{pmatrix}
lambda^{frac1k} & tfrac1klambda^{frac1k-1} & tfrac{1-k}{2k^2}lambda^{frac1k-2} & \
0 & lambda^{frac1k} & tfrac1klambda^{frac1k-1} \
0 & 0 & lambda^{frac1k}
end{pmatrix}^k
=
begin{pmatrix}
lambda & 1 & 0 \
0 & lambda & 1 \
0 & 0 & lambda
end{pmatrix}
$$
If the eigenvalue $lambda$ of this block is zero, then the root as computed above would be the zero matrix, which doesn't result in a Jordan block. But otherwise it should work.
Conclusion
Edited since this question was first asked.
So it seems that every invertible matrix can be raised to every rational power, as long as uniqueness is not a strong requirement. A non-invertible matrix apparently can be raised to non-negative powers as long as all Jordan blocks for eigenvalue zero have size one.
Is this true? If not, where is my mistake? If it is, is there a good reference for this?
linear-algebra complex-numbers eigenvalues-eigenvectors exponentiation jordan-normal-form
asked Dec 22 '13 at 14:40
MvGMvG
30.9k450104
$endgroup$
add a comment |
$begingroup$
Question
Given a square complex matrix $A$, what ways are there to define and compute $A^p$ for non-integral scalar exponents $pinmathbb R$, and for what matrices do they work?
My thoughts
Integral exponents
Defining $A^k$ for $kinmathbb N$ is easy in terms of repeated multiplication, and works for every matrix. This includes $A^0=I$. Using $A^{-1}$ as the inverse, $A^{-k}=left(A^{-1}right)^k$ is easy to define, but requires the matrix to be invertible. So much for integral exponents.
Rational definition
I guess for a rational exponent, one could define
$$A^{frac pq}=Bquad:Leftrightarrowquad A^p=B^q$$
This will allow for more than one solution, and I'm not sure if the computations I'll describe below will find all solutions satisfying the above equation. So I'm not sure whether that's a reasonable definition. For non-rational exponents, a limit using a convergent series of rational exponents might work.
Diagonalizable computation
If $A$ is diagonalizable, then one has $A=W,D,W^{-1}$ for some diagonal matrix $D$. One can simply raise all the diagonal elements to the $p$-th power, obtaining a matrix which will satisfy the above equation. For each diagonal element, I'd define $lambda^p=e^{(plnlambda)}$, and since $lnlambda$ is only defined up to $2pi imathbb Z$, this allows for multiple possible solutions. If one requires $-pi<operatorname{Im}(lnlambda)lepi$, then the solution should be well defined, and I guess this definition even has a name, although I don't know it.
Non-diagonalizable computation
If $A$ is not diagonalizable, then there is still a Jordan normal form, so instead of raising diagonal elements to a fractional power, one could attempt to do the same with Jordan blocks. Unless I made a mistake, this appears to be possible. At least for my example of a $3times3$ Jordan block, I was able to obtain a $k$-th root.
$$
begin{pmatrix}
lambda^{frac1k} & tfrac1klambda^{frac1k-1} & tfrac{1-k}{2k^2}lambda^{frac1k-2} & \
0 & lambda^{frac1k} & tfrac1klambda^{frac1k-1} \
0 & 0 & lambda^{frac1k}
end{pmatrix}^k
=
begin{pmatrix}
lambda & 1 & 0 \
0 & lambda & 1 \
0 & 0 & lambda
end{pmatrix}
$$
If the eigenvalue $lambda$ of this block is zero, then the root as computed above would be the zero matrix, which doesn't result in a Jordan block. But otherwise it should work.
Conclusion
Edited since this question was first asked.
So it seems that every invertible matrix can be raised to every rational power, as long as uniqueness is not a strong requirement. A non-invertible matrix apparently can be raised to non-negative powers as long as all Jordan blocks for eigenvalue zero have size one.
Is this true? If not, where is my mistake? If it is, is there a good reference for this?
linear-algebra complex-numbers eigenvalues-eigenvectors exponentiation jordan-normal-form
asked Dec 22 '13 at 14:40
MvGMvG
30.9k450104
$endgroup$
1
$begingroup$
As with real numbers you can define power with logarithm. Check wiki for matrix logarithm en.wikipedia.org/wiki/Logarithm_of_a_matrix
$endgroup$
– tom
Dec 22 '13 at 15:09
$begingroup$
@tom: That link states that the logarithm exists iff the matrix is invertible. Now suppose a diagonalizable matrix has one eigenvalue being zero. Then my diagonalization approach should still work, since $0^p=0$ for all $pneq 0$. So it seems that definition doesn't exactly match my thoughts, right? Nevertheless, it looks like a very reasonable definition, and therefore a good answer to my original question. I would like to see this posted as a full answer.
$endgroup$
– MvG
Dec 22 '13 at 15:23
$begingroup$
Yes this is similar to the real number case. You don't define $log{0}$ but you define $0^a=0$ for $aneq 0$.
$endgroup$
– tom
Dec 22 '13 at 15:35
$begingroup$
Since already $x^y$ is not uniquely defined when $y$ is non-integer and $x$ is not a non-negative real (and forcing an extension to these cases necessarily makes usual laws for exponentiation fail) there seems little to be gained (for $y$ non-integer) by allowing $x$ to be a matrix: the same definitional problems reappear. Sometimes they are worse; for instance the matrix square root may take infinitely many values. Note however that when $x$ is a positive real you can define $x^A$ without any problem, as $exp(ln(x)A)$.
$endgroup$
– Marc van Leeuwen
Dec 23 '13 at 7:16
add a comment |
$begingroup$
Question
Given a square complex matrix $A$, what ways are there to define and compute $A^p$ for non-integral scalar exponents $pinmathbb R$, and for what matrices do they work?
My thoughts
Integral exponents
Defining $A^k$ for $kinmathbb N$ is easy in terms of repeated multiplication, and works for every matrix. This includes $A^0=I$. Using $A^{-1}$ as the inverse, $A^{-k}=left(A^{-1}right)^k$ is easy to define, but requires the matrix to be invertible. So much for integral exponents.
Rational definition
I guess for a rational exponent, one could define
$$A^{frac pq}=Bquad:Leftrightarrowquad A^p=B^q$$
This will allow for more than one solution, and I'm not sure if the computations I'll describe below will find all solutions satisfying the above equation. So I'm not sure whether that's a reasonable definition. For non-rational exponents, a limit using a convergent series of rational exponents might work.
Diagonalizable computation
If $A$ is diagonalizable, then one has $A=W,D,W^{-1}$ for some diagonal matrix $D$. One can simply raise all the diagonal elements to the $p$-th power, obtaining a matrix which will satisfy the above equation. For each diagonal element, I'd define $lambda^p=e^{(plnlambda)}$, and since $lnlambda$ is only defined up to $2pi imathbb Z$, this allows for multiple possible solutions. If one requires $-pi<operatorname{Im}(lnlambda)lepi$, then the solution should be well defined, and I guess this definition even has a name, although I don't know it.
Non-diagonalizable computation
If $A$ is not diagonalizable, then there is still a Jordan normal form, so instead of raising diagonal elements to a fractional power, one could attempt to do the same with Jordan blocks. Unless I made a mistake, this appears to be possible. At least for my example of a $3times3$ Jordan block, I was able to obtain a $k$-th root.
$$
begin{pmatrix}
lambda^{frac1k} & tfrac1klambda^{frac1k-1} & tfrac{1-k}{2k^2}lambda^{frac1k-2} & \
0 & lambda^{frac1k} & tfrac1klambda^{frac1k-1} \
0 & 0 & lambda^{frac1k}
end{pmatrix}^k
=
begin{pmatrix}
lambda & 1 & 0 \
0 & lambda & 1 \
0 & 0 & lambda
end{pmatrix}
$$
If the eigenvalue $lambda$ of this block is zero, then the root as computed above would be the zero matrix, which doesn't result in a Jordan block. But otherwise it should work.
Conclusion
Edited since this question was first asked.
So it seems that every invertible matrix can be raised to every rational power, as long as uniqueness is not a strong requirement. A non-invertible matrix apparently can be raised to non-negative powers as long as all Jordan blocks for eigenvalue zero have size one.
Is this true? If not, where is my mistake? If it is, is there a good reference for this?
linear-algebra complex-numbers eigenvalues-eigenvectors exponentiation jordan-normal-form
asked Dec 22 '13 at 14:40
MvGMvG
30.9k450104
$endgroup$
Question
Given a square complex matrix $A$, what ways are there to define and compute $A^p$ for non-integral scalar exponents $pinmathbb R$, and for what matrices do they work?
My thoughts
Integral exponents
Defining $A^k$ for $kinmathbb N$ is easy in terms of repeated multiplication, and works for every matrix. This includes $A^0=I$. Using $A^{-1}$ as the inverse, $A^{-k}=left(A^{-1}right)^k$ is easy to define, but requires the matrix to be invertible. So much for integral exponents.
Rational definition
I guess for a rational exponent, one could define
$$A^{frac pq}=Bquad:Leftrightarrowquad A^p=B^q$$
This will allow for more than one solution, and I'm not sure if the computations I'll describe below will find all solutions satisfying the above equation. So I'm not sure whether that's a reasonable definition. For non-rational exponents, a limit using a convergent series of rational exponents might work.
Diagonalizable computation
If $A$ is diagonalizable, then one has $A=W,D,W^{-1}$ for some diagonal matrix $D$. One can simply raise all the diagonal elements to the $p$-th power, obtaining a matrix which will satisfy the above equation. For each diagonal element, I'd define $lambda^p=e^{(plnlambda)}$, and since $lnlambda$ is only defined up to $2pi imathbb Z$, this allows for multiple possible solutions. If one requires $-pi<operatorname{Im}(lnlambda)lepi$, then the solution should be well defined, and I guess this definition even has a name, although I don't know it.
Non-diagonalizable computation
If $A$ is not diagonalizable, then there is still a Jordan normal form, so instead of raising diagonal elements to a fractional power, one could attempt to do the same with Jordan blocks. Unless I made a mistake, this appears to be possible. At least for my example of a $3times3$ Jordan block, I was able to obtain a $k$-th root.
$$
begin{pmatrix}
lambda^{frac1k} & tfrac1klambda^{frac1k-1} & tfrac{1-k}{2k^2}lambda^{frac1k-2} & \
0 & lambda^{frac1k} & tfrac1klambda^{frac1k-1} \
0 & 0 & lambda^{frac1k}
end{pmatrix}^k
=
begin{pmatrix}
lambda & 1 & 0 \
0 & lambda & 1 \
0 & 0 & lambda
end{pmatrix}
$$
If the eigenvalue $lambda$ of this block is zero, then the root as computed above would be the zero matrix, which doesn't result in a Jordan block. But otherwise it should work.
Conclusion
Edited since this question was first asked.
So it seems that every invertible matrix can be raised to every rational power, as long as uniqueness is not a strong requirement. A non-invertible matrix apparently can be raised to non-negative powers as long as all Jordan blocks for eigenvalue zero have size one.
Is this true? If not, where is my mistake? If it is, is there a good reference for this?
linear-algebra complex-numbers eigenvalues-eigenvectors exponentiation jordan-normal-form
linear-algebra complex-numbers eigenvalues-eigenvectors exponentiation jordan-normal-form
asked Dec 22 '13 at 14:40
MvGMvG
30.9k450104
asked Dec 22 '13 at 14:40
MvGMvG
30.9k450104
edited Dec 23 '13 at 6:20
MvG
asked Dec 22 '13 at 14:40
MvGMvG
30.9k450104
asked Dec 22 '13 at 14:40
MvGMvG
30.9k450104
asked Dec 22 '13 at 14:40
MvGMvG
30.9k450104
30.9k450104
1
$begingroup$
As with real numbers you can define power with logarithm. Check wiki for matrix logarithm en.wikipedia.org/wiki/Logarithm_of_a_matrix
$endgroup$
– tom
Dec 22 '13 at 15:09
$begingroup$
@tom: That link states that the logarithm exists iff the matrix is invertible. Now suppose a diagonalizable matrix has one eigenvalue being zero. Then my diagonalization approach should still work, since $0^p=0$ for all $pneq 0$. So it seems that definition doesn't exactly match my thoughts, right? Nevertheless, it looks like a very reasonable definition, and therefore a good answer to my original question. I would like to see this posted as a full answer.
$endgroup$
– MvG
Dec 22 '13 at 15:23
$begingroup$
Yes this is similar to the real number case. You don't define $log{0}$ but you define $0^a=0$ for $aneq 0$.
$endgroup$
– tom
Dec 22 '13 at 15:35
$begingroup$
Since already $x^y$ is not uniquely defined when $y$ is non-integer and $x$ is not a non-negative real (and forcing an extension to these cases necessarily makes usual laws for exponentiation fail) there seems little to be gained (for $y$ non-integer) by allowing $x$ to be a matrix: the same definitional problems reappear. Sometimes they are worse; for instance the matrix square root may take infinitely many values. Note however that when $x$ is a positive real you can define $x^A$ without any problem, as $exp(ln(x)A)$.
$endgroup$
– Marc van Leeuwen
Dec 23 '13 at 7:16
add a comment |
1
$begingroup$
As with real numbers you can define power with logarithm. Check wiki for matrix logarithm en.wikipedia.org/wiki/Logarithm_of_a_matrix
$endgroup$
– tom
Dec 22 '13 at 15:09
$begingroup$
@tom: That link states that the logarithm exists iff the matrix is invertible. Now suppose a diagonalizable matrix has one eigenvalue being zero. Then my diagonalization approach should still work, since $0^p=0$ for all $pneq 0$. So it seems that definition doesn't exactly match my thoughts, right? Nevertheless, it looks like a very reasonable definition, and therefore a good answer to my original question. I would like to see this posted as a full answer.
$endgroup$
– MvG
Dec 22 '13 at 15:23
$begingroup$
Yes this is similar to the real number case. You don't define $log{0}$ but you define $0^a=0$ for $aneq 0$.
$endgroup$
– tom
Dec 22 '13 at 15:35
$begingroup$
Since already $x^y$ is not uniquely defined when $y$ is non-integer and $x$ is not a non-negative real (and forcing an extension to these cases necessarily makes usual laws for exponentiation fail) there seems little to be gained (for $y$ non-integer) by allowing $x$ to be a matrix: the same definitional problems reappear. Sometimes they are worse; for instance the matrix square root may take infinitely many values. Note however that when $x$ is a positive real you can define $x^A$ without any problem, as $exp(ln(x)A)$.
$endgroup$
– Marc van Leeuwen
Dec 23 '13 at 7:16
1
1
$begingroup$
As with real numbers you can define power with logarithm. Check wiki for matrix logarithm en.wikipedia.org/wiki/Logarithm_of_a_matrix
$endgroup$
– tom
Dec 22 '13 at 15:09
$begingroup$
As with real numbers you can define power with logarithm. Check wiki for matrix logarithm en.wikipedia.org/wiki/Logarithm_of_a_matrix
$endgroup$
– tom
Dec 22 '13 at 15:09
$begingroup$
@tom: That link states that the logarithm exists iff the matrix is invertible. Now suppose a diagonalizable matrix has one eigenvalue being zero. Then my diagonalization approach should still work, since $0^p=0$ for all $pneq 0$. So it seems that definition doesn't exactly match my thoughts, right? Nevertheless, it looks like a very reasonable definition, and therefore a good answer to my original question. I would like to see this posted as a full answer.
$endgroup$
– MvG
Dec 22 '13 at 15:23
$begingroup$
@tom: That link states that the logarithm exists iff the matrix is invertible. Now suppose a diagonalizable matrix has one eigenvalue being zero. Then my diagonalization approach should still work, since $0^p=0$ for all $pneq 0$. So it seems that definition doesn't exactly match my thoughts, right? Nevertheless, it looks like a very reasonable definition, and therefore a good answer to my original question. I would like to see this posted as a full answer.
$endgroup$
– MvG
Dec 22 '13 at 15:23
$begingroup$
Yes this is similar to the real number case. You don't define $log{0}$ but you define $0^a=0$ for $aneq 0$.
$endgroup$
– tom
Dec 22 '13 at 15:35
$begingroup$
Yes this is similar to the real number case. You don't define $log{0}$ but you define $0^a=0$ for $aneq 0$.
$endgroup$
– tom
Dec 22 '13 at 15:35
$begingroup$
Since already $x^y$ is not uniquely defined when $y$ is non-integer and $x$ is not a non-negative real (and forcing an extension to these cases necessarily makes usual laws for exponentiation fail) there seems little to be gained (for $y$ non-integer) by allowing $x$ to be a matrix: the same definitional problems reappear. Sometimes they are worse; for instance the matrix square root may take infinitely many values. Note however that when $x$ is a positive real you can define $x^A$ without any problem, as $exp(ln(x)A)$.
$endgroup$
– Marc van Leeuwen
Dec 23 '13 at 7:16
$begingroup$
Since already $x^y$ is not uniquely defined when $y$ is non-integer and $x$ is not a non-negative real (and forcing an extension to these cases necessarily makes usual laws for exponentiation fail) there seems little to be gained (for $y$ non-integer) by allowing $x$ to be a matrix: the same definitional problems reappear. Sometimes they are worse; for instance the matrix square root may take infinitely many values. Note however that when $x$ is a positive real you can define $x^A$ without any problem, as $exp(ln(x)A)$.
$endgroup$
– Marc van Leeuwen
Dec 23 '13 at 7:16
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As @tom pointed out in a comment, the power of a matrix can be defined in terms of logarithm of a matrix and matrix exponential, using
$$A^p:=expleft(pln Aright)$$
Using the principal logarithm (this is the name for that choice described in the question without giving a name), the above even yields unique results.
The matrix exponential is defined for every matrix, the matrix logarithm only for invertible matrices.
The case of singular matrices mentioned in the question is therefore not covered by this definition.
$endgroup$
add a comment |
$begingroup$
You are correct that your proposed definition for rational exponents can run into issues of uniqueness. Consider just the problem of trying to find the square root of a matrix. If $I$ is the 2x2 identity, then any matrix of the form
begin{pmatrix}
pm1 & a \
0 & mp1 \
end{pmatrix}
satisfies $A^2=I$. Now, there is a case where you can define a unique square root. In particular, your matrix must be positive definite [1]
For a more general discussion, see this
edited Apr 13 '17 at 12:21
Community♦
1
answered Dec 22 '13 at 15:03
Philip HoskinsPhilip Hoskins
1,445715
$endgroup$
$begingroup$
Since talking about positivity in the presence of complex numbers is problematic, so is positive definiteness for complex matrices. So while these considerations are valuable for real matrices, they don't exactly match the setup I've been asking about. The non-uniqueness obervation of course still stands, so this is a valuable answer in any case.
$endgroup$
– MvG
Dec 23 '13 at 6:36
$begingroup$
I'm not sure what you mean by positive definiteness being problematic for complex matrices, but I won't belabor the point since uniqueness isn't a concern for the moment.
$endgroup$
– Philip Hoskins
Dec 23 '13 at 7:41
add a comment |
$begingroup$
I came to the need/wish to extend $A^n$ to $A^z$ from a geometrical perspective (where $n$ is an integer number and $z$ real one), so I hope this can help, by providing a geometrical meaning for "real powers of a matrix".
this is the geometrical setup
The hyperboloid shown on the image comes from the Pells's equation:
$$x^2 - 2y^2 = 1$$
The "vertical planes" shown there (only 3 shown, for $n=-1,0,1$) are identified by their $z$ coordinate and placed at $n$ (integer) values only.
That was the starting point: extending integer values $n$ to all real values $z$, will mean considering a continuous set of "vertical planes": that will involve powers of matrices, as $n$ will show up as a matrix exponent, in the following.
I assume each $z=n$ vertical plane "hosts" the coordinate system $(x_n, y_n)$, just like as these coordinate systems "lived" on those planes.
With this setup, the matrix
begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}
is a representation for this coordinate transform:
$$(x_n, y_n) = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x_0, y_0)$$
which also includes "jumping" from plane $z=0$ to plane $z=n$.
BTW: the Pells's equation above is invariant to this coordinate transform.
Noticeably, the same matrix was used by ancient Greeks "to generate the next integer solution" (to Pells's equation), as $(x, y)_{(0)}=(1,0)$ was known to be "a trivial solution":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
or
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x, y)_{(0)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (1, 0)$$
where
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n$$
means "apply transform $n$ times" and
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^{(-1)}$$
means "apply inverse transform once".
But, considering that matrix as a coordinate transform, we see point $(1,0)$ on plane $n=0$ "jumping FWD" to plane $n=1$ and "rotating CW" to point $(3,2)$, when applying the "forward transform", while we see that same point $(1,0)$ "jumping BWD" to plane $n=-1$ and "rotating CCW" to point $(3,-2)$, when applying the "inverse transform".
Due to the recursivity of the "next solution generator":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
that kind of "jump and rotate" applies to all "integer solutions" to Pells's equation: indeed, as a coordinate transform, all the points in the plane "rotate to" other points.
That would also suggest that "all points on the paraboloids 'rotate' in 3D":
$(1,0,0)$ gets transformed to $(3,2,1)$
$(1,0,0)$ gets inverse-transformed to $(3,-2,-1)$
Having put the "vertical planes" in between "helps visualizing the motion": it's a discrete motion ($n$ steps by $1$, not by a $dz$): could it be "extended" so it's continuous?
With the geometrical interpretation above, "extending" this "motion" to be "continuous" means "extending integer powers of matrices to real powers".
Would you agree?
Cheers,
.k.
answered Jan 9 at 11:19
ccampisanoccampisano
62
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add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As @tom pointed out in a comment, the power of a matrix can be defined in terms of logarithm of a matrix and matrix exponential, using
$$A^p:=expleft(pln Aright)$$
Using the principal logarithm (this is the name for that choice described in the question without giving a name), the above even yields unique results.
The matrix exponential is defined for every matrix, the matrix logarithm only for invertible matrices.
The case of singular matrices mentioned in the question is therefore not covered by this definition.
$endgroup$
add a comment |
$begingroup$
As @tom pointed out in a comment, the power of a matrix can be defined in terms of logarithm of a matrix and matrix exponential, using
$$A^p:=expleft(pln Aright)$$
Using the principal logarithm (this is the name for that choice described in the question without giving a name), the above even yields unique results.
The matrix exponential is defined for every matrix, the matrix logarithm only for invertible matrices.
The case of singular matrices mentioned in the question is therefore not covered by this definition.
$endgroup$
add a comment |
$begingroup$
As @tom pointed out in a comment, the power of a matrix can be defined in terms of logarithm of a matrix and matrix exponential, using
$$A^p:=expleft(pln Aright)$$
Using the principal logarithm (this is the name for that choice described in the question without giving a name), the above even yields unique results.
The matrix exponential is defined for every matrix, the matrix logarithm only for invertible matrices.
The case of singular matrices mentioned in the question is therefore not covered by this definition.
$endgroup$
As @tom pointed out in a comment, the power of a matrix can be defined in terms of logarithm of a matrix and matrix exponential, using
$$A^p:=expleft(pln Aright)$$
Using the principal logarithm (this is the name for that choice described in the question without giving a name), the above even yields unique results.
The matrix exponential is defined for every matrix, the matrix logarithm only for invertible matrices.
The case of singular matrices mentioned in the question is therefore not covered by this definition.
edited Apr 13 '17 at 12:20
community wiki
2 revs
MvG
add a comment |
add a comment |
$begingroup$
You are correct that your proposed definition for rational exponents can run into issues of uniqueness. Consider just the problem of trying to find the square root of a matrix. If $I$ is the 2x2 identity, then any matrix of the form
begin{pmatrix}
pm1 & a \
0 & mp1 \
end{pmatrix}
satisfies $A^2=I$. Now, there is a case where you can define a unique square root. In particular, your matrix must be positive definite [1]
For a more general discussion, see this
edited Apr 13 '17 at 12:21
Community♦
1
answered Dec 22 '13 at 15:03
Philip HoskinsPhilip Hoskins
1,445715
$endgroup$
$begingroup$
Since talking about positivity in the presence of complex numbers is problematic, so is positive definiteness for complex matrices. So while these considerations are valuable for real matrices, they don't exactly match the setup I've been asking about. The non-uniqueness obervation of course still stands, so this is a valuable answer in any case.
$endgroup$
– MvG
Dec 23 '13 at 6:36
$begingroup$
I'm not sure what you mean by positive definiteness being problematic for complex matrices, but I won't belabor the point since uniqueness isn't a concern for the moment.
$endgroup$
– Philip Hoskins
Dec 23 '13 at 7:41
add a comment |
$begingroup$
You are correct that your proposed definition for rational exponents can run into issues of uniqueness. Consider just the problem of trying to find the square root of a matrix. If $I$ is the 2x2 identity, then any matrix of the form
begin{pmatrix}
pm1 & a \
0 & mp1 \
end{pmatrix}
satisfies $A^2=I$. Now, there is a case where you can define a unique square root. In particular, your matrix must be positive definite [1]
For a more general discussion, see this
edited Apr 13 '17 at 12:21
Community♦
1
answered Dec 22 '13 at 15:03
Philip HoskinsPhilip Hoskins
1,445715
$endgroup$
$begingroup$
Since talking about positivity in the presence of complex numbers is problematic, so is positive definiteness for complex matrices. So while these considerations are valuable for real matrices, they don't exactly match the setup I've been asking about. The non-uniqueness obervation of course still stands, so this is a valuable answer in any case.
$endgroup$
– MvG
Dec 23 '13 at 6:36
$begingroup$
I'm not sure what you mean by positive definiteness being problematic for complex matrices, but I won't belabor the point since uniqueness isn't a concern for the moment.
$endgroup$
– Philip Hoskins
Dec 23 '13 at 7:41
add a comment |
$begingroup$
You are correct that your proposed definition for rational exponents can run into issues of uniqueness. Consider just the problem of trying to find the square root of a matrix. If $I$ is the 2x2 identity, then any matrix of the form
begin{pmatrix}
pm1 & a \
0 & mp1 \
end{pmatrix}
satisfies $A^2=I$. Now, there is a case where you can define a unique square root. In particular, your matrix must be positive definite [1]
For a more general discussion, see this
edited Apr 13 '17 at 12:21
Community♦
1
answered Dec 22 '13 at 15:03
Philip HoskinsPhilip Hoskins
1,445715
$endgroup$
You are correct that your proposed definition for rational exponents can run into issues of uniqueness. Consider just the problem of trying to find the square root of a matrix. If $I$ is the 2x2 identity, then any matrix of the form
begin{pmatrix}
pm1 & a \
0 & mp1 \
end{pmatrix}
satisfies $A^2=I$. Now, there is a case where you can define a unique square root. In particular, your matrix must be positive definite [1]
For a more general discussion, see this
edited Apr 13 '17 at 12:21
Community♦
1
answered Dec 22 '13 at 15:03
Philip HoskinsPhilip Hoskins
1,445715
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Dec 22 '13 at 15:03
Philip HoskinsPhilip Hoskins
1,445715
answered Dec 22 '13 at 15:03
Philip HoskinsPhilip Hoskins
1,445715
answered Dec 22 '13 at 15:03
Philip HoskinsPhilip Hoskins
1,445715
1,445715
$begingroup$
Since talking about positivity in the presence of complex numbers is problematic, so is positive definiteness for complex matrices. So while these considerations are valuable for real matrices, they don't exactly match the setup I've been asking about. The non-uniqueness obervation of course still stands, so this is a valuable answer in any case.
$endgroup$
– MvG
Dec 23 '13 at 6:36
$begingroup$
I'm not sure what you mean by positive definiteness being problematic for complex matrices, but I won't belabor the point since uniqueness isn't a concern for the moment.
$endgroup$
– Philip Hoskins
Dec 23 '13 at 7:41
add a comment |
$begingroup$
Since talking about positivity in the presence of complex numbers is problematic, so is positive definiteness for complex matrices. So while these considerations are valuable for real matrices, they don't exactly match the setup I've been asking about. The non-uniqueness obervation of course still stands, so this is a valuable answer in any case.
$endgroup$
– MvG
Dec 23 '13 at 6:36
$begingroup$
I'm not sure what you mean by positive definiteness being problematic for complex matrices, but I won't belabor the point since uniqueness isn't a concern for the moment.
$endgroup$
– Philip Hoskins
Dec 23 '13 at 7:41
$begingroup$
Since talking about positivity in the presence of complex numbers is problematic, so is positive definiteness for complex matrices. So while these considerations are valuable for real matrices, they don't exactly match the setup I've been asking about. The non-uniqueness obervation of course still stands, so this is a valuable answer in any case.
$endgroup$
– MvG
Dec 23 '13 at 6:36
$begingroup$
Since talking about positivity in the presence of complex numbers is problematic, so is positive definiteness for complex matrices. So while these considerations are valuable for real matrices, they don't exactly match the setup I've been asking about. The non-uniqueness obervation of course still stands, so this is a valuable answer in any case.
$endgroup$
– MvG
Dec 23 '13 at 6:36
$begingroup$
I'm not sure what you mean by positive definiteness being problematic for complex matrices, but I won't belabor the point since uniqueness isn't a concern for the moment.
$endgroup$
– Philip Hoskins
Dec 23 '13 at 7:41
$begingroup$
I'm not sure what you mean by positive definiteness being problematic for complex matrices, but I won't belabor the point since uniqueness isn't a concern for the moment.
$endgroup$
– Philip Hoskins
Dec 23 '13 at 7:41
add a comment |
$begingroup$
I came to the need/wish to extend $A^n$ to $A^z$ from a geometrical perspective (where $n$ is an integer number and $z$ real one), so I hope this can help, by providing a geometrical meaning for "real powers of a matrix".
this is the geometrical setup
The hyperboloid shown on the image comes from the Pells's equation:
$$x^2 - 2y^2 = 1$$
The "vertical planes" shown there (only 3 shown, for $n=-1,0,1$) are identified by their $z$ coordinate and placed at $n$ (integer) values only.
That was the starting point: extending integer values $n$ to all real values $z$, will mean considering a continuous set of "vertical planes": that will involve powers of matrices, as $n$ will show up as a matrix exponent, in the following.
I assume each $z=n$ vertical plane "hosts" the coordinate system $(x_n, y_n)$, just like as these coordinate systems "lived" on those planes.
With this setup, the matrix
begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}
is a representation for this coordinate transform:
$$(x_n, y_n) = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x_0, y_0)$$
which also includes "jumping" from plane $z=0$ to plane $z=n$.
BTW: the Pells's equation above is invariant to this coordinate transform.
Noticeably, the same matrix was used by ancient Greeks "to generate the next integer solution" (to Pells's equation), as $(x, y)_{(0)}=(1,0)$ was known to be "a trivial solution":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
or
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x, y)_{(0)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (1, 0)$$
where
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n$$
means "apply transform $n$ times" and
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^{(-1)}$$
means "apply inverse transform once".
But, considering that matrix as a coordinate transform, we see point $(1,0)$ on plane $n=0$ "jumping FWD" to plane $n=1$ and "rotating CW" to point $(3,2)$, when applying the "forward transform", while we see that same point $(1,0)$ "jumping BWD" to plane $n=-1$ and "rotating CCW" to point $(3,-2)$, when applying the "inverse transform".
Due to the recursivity of the "next solution generator":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
that kind of "jump and rotate" applies to all "integer solutions" to Pells's equation: indeed, as a coordinate transform, all the points in the plane "rotate to" other points.
That would also suggest that "all points on the paraboloids 'rotate' in 3D":
$(1,0,0)$ gets transformed to $(3,2,1)$
$(1,0,0)$ gets inverse-transformed to $(3,-2,-1)$
Having put the "vertical planes" in between "helps visualizing the motion": it's a discrete motion ($n$ steps by $1$, not by a $dz$): could it be "extended" so it's continuous?
With the geometrical interpretation above, "extending" this "motion" to be "continuous" means "extending integer powers of matrices to real powers".
Would you agree?
Cheers,
.k.
answered Jan 9 at 11:19
ccampisanoccampisano
62
$endgroup$
add a comment |
$begingroup$
I came to the need/wish to extend $A^n$ to $A^z$ from a geometrical perspective (where $n$ is an integer number and $z$ real one), so I hope this can help, by providing a geometrical meaning for "real powers of a matrix".
this is the geometrical setup
The hyperboloid shown on the image comes from the Pells's equation:
$$x^2 - 2y^2 = 1$$
The "vertical planes" shown there (only 3 shown, for $n=-1,0,1$) are identified by their $z$ coordinate and placed at $n$ (integer) values only.
That was the starting point: extending integer values $n$ to all real values $z$, will mean considering a continuous set of "vertical planes": that will involve powers of matrices, as $n$ will show up as a matrix exponent, in the following.
I assume each $z=n$ vertical plane "hosts" the coordinate system $(x_n, y_n)$, just like as these coordinate systems "lived" on those planes.
With this setup, the matrix
begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}
is a representation for this coordinate transform:
$$(x_n, y_n) = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x_0, y_0)$$
which also includes "jumping" from plane $z=0$ to plane $z=n$.
BTW: the Pells's equation above is invariant to this coordinate transform.
Noticeably, the same matrix was used by ancient Greeks "to generate the next integer solution" (to Pells's equation), as $(x, y)_{(0)}=(1,0)$ was known to be "a trivial solution":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
or
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x, y)_{(0)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (1, 0)$$
where
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n$$
means "apply transform $n$ times" and
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^{(-1)}$$
means "apply inverse transform once".
But, considering that matrix as a coordinate transform, we see point $(1,0)$ on plane $n=0$ "jumping FWD" to plane $n=1$ and "rotating CW" to point $(3,2)$, when applying the "forward transform", while we see that same point $(1,0)$ "jumping BWD" to plane $n=-1$ and "rotating CCW" to point $(3,-2)$, when applying the "inverse transform".
Due to the recursivity of the "next solution generator":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
that kind of "jump and rotate" applies to all "integer solutions" to Pells's equation: indeed, as a coordinate transform, all the points in the plane "rotate to" other points.
That would also suggest that "all points on the paraboloids 'rotate' in 3D":
$(1,0,0)$ gets transformed to $(3,2,1)$
$(1,0,0)$ gets inverse-transformed to $(3,-2,-1)$
Having put the "vertical planes" in between "helps visualizing the motion": it's a discrete motion ($n$ steps by $1$, not by a $dz$): could it be "extended" so it's continuous?
With the geometrical interpretation above, "extending" this "motion" to be "continuous" means "extending integer powers of matrices to real powers".
Would you agree?
Cheers,
.k.
answered Jan 9 at 11:19
ccampisanoccampisano
62
$endgroup$
add a comment |
$begingroup$
I came to the need/wish to extend $A^n$ to $A^z$ from a geometrical perspective (where $n$ is an integer number and $z$ real one), so I hope this can help, by providing a geometrical meaning for "real powers of a matrix".
this is the geometrical setup
The hyperboloid shown on the image comes from the Pells's equation:
$$x^2 - 2y^2 = 1$$
The "vertical planes" shown there (only 3 shown, for $n=-1,0,1$) are identified by their $z$ coordinate and placed at $n$ (integer) values only.
That was the starting point: extending integer values $n$ to all real values $z$, will mean considering a continuous set of "vertical planes": that will involve powers of matrices, as $n$ will show up as a matrix exponent, in the following.
I assume each $z=n$ vertical plane "hosts" the coordinate system $(x_n, y_n)$, just like as these coordinate systems "lived" on those planes.
With this setup, the matrix
begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}
is a representation for this coordinate transform:
$$(x_n, y_n) = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x_0, y_0)$$
which also includes "jumping" from plane $z=0$ to plane $z=n$.
BTW: the Pells's equation above is invariant to this coordinate transform.
Noticeably, the same matrix was used by ancient Greeks "to generate the next integer solution" (to Pells's equation), as $(x, y)_{(0)}=(1,0)$ was known to be "a trivial solution":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
or
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x, y)_{(0)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (1, 0)$$
where
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n$$
means "apply transform $n$ times" and
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^{(-1)}$$
means "apply inverse transform once".
But, considering that matrix as a coordinate transform, we see point $(1,0)$ on plane $n=0$ "jumping FWD" to plane $n=1$ and "rotating CW" to point $(3,2)$, when applying the "forward transform", while we see that same point $(1,0)$ "jumping BWD" to plane $n=-1$ and "rotating CCW" to point $(3,-2)$, when applying the "inverse transform".
Due to the recursivity of the "next solution generator":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
that kind of "jump and rotate" applies to all "integer solutions" to Pells's equation: indeed, as a coordinate transform, all the points in the plane "rotate to" other points.
That would also suggest that "all points on the paraboloids 'rotate' in 3D":
$(1,0,0)$ gets transformed to $(3,2,1)$
$(1,0,0)$ gets inverse-transformed to $(3,-2,-1)$
Having put the "vertical planes" in between "helps visualizing the motion": it's a discrete motion ($n$ steps by $1$, not by a $dz$): could it be "extended" so it's continuous?
With the geometrical interpretation above, "extending" this "motion" to be "continuous" means "extending integer powers of matrices to real powers".
Would you agree?
Cheers,
.k.
answered Jan 9 at 11:19
ccampisanoccampisano
62
$endgroup$
I came to the need/wish to extend $A^n$ to $A^z$ from a geometrical perspective (where $n$ is an integer number and $z$ real one), so I hope this can help, by providing a geometrical meaning for "real powers of a matrix".
this is the geometrical setup
The hyperboloid shown on the image comes from the Pells's equation:
$$x^2 - 2y^2 = 1$$
The "vertical planes" shown there (only 3 shown, for $n=-1,0,1$) are identified by their $z$ coordinate and placed at $n$ (integer) values only.
That was the starting point: extending integer values $n$ to all real values $z$, will mean considering a continuous set of "vertical planes": that will involve powers of matrices, as $n$ will show up as a matrix exponent, in the following.
I assume each $z=n$ vertical plane "hosts" the coordinate system $(x_n, y_n)$, just like as these coordinate systems "lived" on those planes.
With this setup, the matrix
begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}
is a representation for this coordinate transform:
$$(x_n, y_n) = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x_0, y_0)$$
which also includes "jumping" from plane $z=0$ to plane $z=n$.
BTW: the Pells's equation above is invariant to this coordinate transform.
Noticeably, the same matrix was used by ancient Greeks "to generate the next integer solution" (to Pells's equation), as $(x, y)_{(0)}=(1,0)$ was known to be "a trivial solution":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
or
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (x, y)_{(0)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n (1, 0)$$
where
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^n$$
means "apply transform $n$ times" and
$$begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix}^{(-1)}$$
means "apply inverse transform once".
But, considering that matrix as a coordinate transform, we see point $(1,0)$ on plane $n=0$ "jumping FWD" to plane $n=1$ and "rotating CW" to point $(3,2)$, when applying the "forward transform", while we see that same point $(1,0)$ "jumping BWD" to plane $n=-1$ and "rotating CCW" to point $(3,-2)$, when applying the "inverse transform".
Due to the recursivity of the "next solution generator":
$$(x, y)_{(n+1)} = begin{pmatrix}
3 & 4 \
2 & 3 \
end{pmatrix} (x, y)_{(n)}$$
that kind of "jump and rotate" applies to all "integer solutions" to Pells's equation: indeed, as a coordinate transform, all the points in the plane "rotate to" other points.
That would also suggest that "all points on the paraboloids 'rotate' in 3D":
$(1,0,0)$ gets transformed to $(3,2,1)$
$(1,0,0)$ gets inverse-transformed to $(3,-2,-1)$
Having put the "vertical planes" in between "helps visualizing the motion": it's a discrete motion ($n$ steps by $1$, not by a $dz$): could it be "extended" so it's continuous?
With the geometrical interpretation above, "extending" this "motion" to be "continuous" means "extending integer powers of matrices to real powers".
Would you agree?
Cheers,
.k.
answered Jan 9 at 11:19
ccampisanoccampisano
62
answered Jan 9 at 11:19
ccampisanoccampisano
62
answered Jan 9 at 11:19
ccampisanoccampisano
62
answered Jan 9 at 11:19
ccampisanoccampisano
62
62
add a comment |
add a comment |
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1
$begingroup$
As with real numbers you can define power with logarithm. Check wiki for matrix logarithm en.wikipedia.org/wiki/Logarithm_of_a_matrix
$endgroup$
– tom
Dec 22 '13 at 15:09
$begingroup$
@tom: That link states that the logarithm exists iff the matrix is invertible. Now suppose a diagonalizable matrix has one eigenvalue being zero. Then my diagonalization approach should still work, since $0^p=0$ for all $pneq 0$. So it seems that definition doesn't exactly match my thoughts, right? Nevertheless, it looks like a very reasonable definition, and therefore a good answer to my original question. I would like to see this posted as a full answer.
$endgroup$
– MvG
Dec 22 '13 at 15:23
$begingroup$
Yes this is similar to the real number case. You don't define $log{0}$ but you define $0^a=0$ for $aneq 0$.
$endgroup$
– tom
Dec 22 '13 at 15:35
$begingroup$
Since already $x^y$ is not uniquely defined when $y$ is non-integer and $x$ is not a non-negative real (and forcing an extension to these cases necessarily makes usual laws for exponentiation fail) there seems little to be gained (for $y$ non-integer) by allowing $x$ to be a matrix: the same definitional problems reappear. Sometimes they are worse; for instance the matrix square root may take infinitely many values. Note however that when $x$ is a positive real you can define $x^A$ without any problem, as $exp(ln(x)A)$.
$endgroup$
– Marc van Leeuwen
Dec 23 '13 at 7:16