Mixing
implied eigenvalue equations for an arbitrary, symmetric and positive definite matrix
$begingroup$
I have a matrix $M$, where $M$ = $begin{bmatrix}a & b\c & dend{bmatrix}$. It is known that $M$ is symmetric and positive definite. Also, it is known that $x^TMy$ is a valid dot product in $R^2$. I know that to characterize $M$ by its eigenvectors, it can be shown as
$Mv_i$ = λ$v_i$.
I was told in my class that based on this context, we can also gather the following information regarding the eigenvalues of this matrix:
$a - c = λ_1$ , $c - d = -λ_1$ , $a + c = λ_2$ and $c + d= λ_2$.
Why is this true? Is this always the case?
eigenvalues-eigenvectors symmetric-matrices
asked Jan 15 at 0:59
kkunkkun
51
$endgroup$
add a comment |
$begingroup$
I have a matrix $M$, where $M$ = $begin{bmatrix}a & b\c & dend{bmatrix}$. It is known that $M$ is symmetric and positive definite. Also, it is known that $x^TMy$ is a valid dot product in $R^2$. I know that to characterize $M$ by its eigenvectors, it can be shown as
$Mv_i$ = λ$v_i$.
I was told in my class that based on this context, we can also gather the following information regarding the eigenvalues of this matrix:
$a - c = λ_1$ , $c - d = -λ_1$ , $a + c = λ_2$ and $c + d= λ_2$.
Why is this true? Is this always the case?
eigenvalues-eigenvectors symmetric-matrices
asked Jan 15 at 0:59
kkunkkun
51
$endgroup$
$begingroup$
this is completely wrong.
$endgroup$
– Will Jagy
Jan 15 at 1:06
$begingroup$
If $a=d$ then it is true.
$endgroup$
– Sota Antonino
Jan 15 at 1:16
$begingroup$
is it true if $b = c$? How can you prove this is true?
$endgroup$
– kkun
Jan 15 at 1:29
add a comment |
$begingroup$
I have a matrix $M$, where $M$ = $begin{bmatrix}a & b\c & dend{bmatrix}$. It is known that $M$ is symmetric and positive definite. Also, it is known that $x^TMy$ is a valid dot product in $R^2$. I know that to characterize $M$ by its eigenvectors, it can be shown as
$Mv_i$ = λ$v_i$.
I was told in my class that based on this context, we can also gather the following information regarding the eigenvalues of this matrix:
$a - c = λ_1$ , $c - d = -λ_1$ , $a + c = λ_2$ and $c + d= λ_2$.
Why is this true? Is this always the case?
eigenvalues-eigenvectors symmetric-matrices
asked Jan 15 at 0:59
kkunkkun
51
$endgroup$
I have a matrix $M$, where $M$ = $begin{bmatrix}a & b\c & dend{bmatrix}$. It is known that $M$ is symmetric and positive definite. Also, it is known that $x^TMy$ is a valid dot product in $R^2$. I know that to characterize $M$ by its eigenvectors, it can be shown as
$Mv_i$ = λ$v_i$.
I was told in my class that based on this context, we can also gather the following information regarding the eigenvalues of this matrix:
$a - c = λ_1$ , $c - d = -λ_1$ , $a + c = λ_2$ and $c + d= λ_2$.
Why is this true? Is this always the case?
eigenvalues-eigenvectors symmetric-matrices
eigenvalues-eigenvectors symmetric-matrices
asked Jan 15 at 0:59
kkunkkun
51
asked Jan 15 at 0:59
kkunkkun
51
asked Jan 15 at 0:59
kkunkkun
51
asked Jan 15 at 0:59
kkunkkun
51
asked Jan 15 at 0:59
kkunkkun
51
51
$begingroup$
this is completely wrong.
$endgroup$
– Will Jagy
Jan 15 at 1:06
$begingroup$
If $a=d$ then it is true.
$endgroup$
– Sota Antonino
Jan 15 at 1:16
$begingroup$
is it true if $b = c$? How can you prove this is true?
$endgroup$
– kkun
Jan 15 at 1:29
add a comment |
$begingroup$
this is completely wrong.
$endgroup$
– Will Jagy
Jan 15 at 1:06
$begingroup$
If $a=d$ then it is true.
$endgroup$
– Sota Antonino
Jan 15 at 1:16
$begingroup$
is it true if $b = c$? How can you prove this is true?
$endgroup$
– kkun
Jan 15 at 1:29
$begingroup$
this is completely wrong.
$endgroup$
– Will Jagy
Jan 15 at 1:06
$begingroup$
this is completely wrong.
$endgroup$
– Will Jagy
Jan 15 at 1:06
$begingroup$
If $a=d$ then it is true.
$endgroup$
– Sota Antonino
Jan 15 at 1:16
$begingroup$
If $a=d$ then it is true.
$endgroup$
– Sota Antonino
Jan 15 at 1:16
$begingroup$
is it true if $b = c$? How can you prove this is true?
$endgroup$
– kkun
Jan 15 at 1:29
$begingroup$
is it true if $b = c$? How can you prove this is true?
$endgroup$
– kkun
Jan 15 at 1:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $M=begin{bmatrix} a & c \ c & dend{bmatrix}$ be a symmetric matrix. The eigenvalues of this matrix are given by $$(a-lambda)(d-lambda)-c^2=0$$ $$lambda^2-(a+d)lambda +ad-c^2$$ Using the quadratic formula we obtain $$frac{(a+d)pm sqrt{(a+d)^2-4(ad-c^2)}}{2}$$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad+4c^2}}{2}$$ Note that for the symmetric and positive definite matrix $M=begin{bmatrix}2 & 1 \ 2 & 3 end{bmatrix}$ the quadratic formula will give us two irrational values, while the stated formula only gives whole values. So the stated formulae are not true in general but in just a few cases.
Case 1: $c=0$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad}}{2}$$ $$frac{(a+d)pm (a-d)}{2}$$ $$lambda = a,d$$
Case 2: $a=d$ $$frac{2apm sqrt{4c^2}}{2}$$ $$apm c$$ $$lambda = a-c, a+c$$
answered Jan 15 at 1:41
Ryan GreylingRyan Greyling
3229
$endgroup$
add a comment |
$begingroup$
This is patently untrue for a general positive-definite matrix: Combining the first and third equations produces $lambda_1+lambda_2 = operatorname{tr}(A)=2a$, while the second and fourth give $operatorname{tr}(A)=2d$. Any diagonal matrix with different positive entires on the diagonal is a counterexample.
Note, too, that the statements that $A$ is symmetric and positive-definite, and that $x^TAy$ is a valid inner (not “dot”) product are equivalent.
answered Jan 15 at 2:33
amdamd
30.4k21050
$endgroup$
$begingroup$
Thanks for the comment. Regarding your note about dot and inner products-- is it true that a dot product an example of an inner product?
$endgroup$
– kkun
Jan 15 at 3:00
$begingroup$
@kkun Yes. I’m careful to distinguish the dot product—the matrix product of a pair of coordinate tuples—from a scalar product, which is a more general concept. An inner product is a positive-definite scalar product. Many sources use the terms interchangeably when working in $mathbb R^n$ with the Euclidean inner product and the standard basis because the formula for this scalar product is then given by the dot product. If you change to a non-orthonormal basis, though, the value of the inner product of two vectors doesn’t change, but its formula in terms of coordinates does.
$endgroup$
– amd
Jan 15 at 3:13
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $M=begin{bmatrix} a & c \ c & dend{bmatrix}$ be a symmetric matrix. The eigenvalues of this matrix are given by $$(a-lambda)(d-lambda)-c^2=0$$ $$lambda^2-(a+d)lambda +ad-c^2$$ Using the quadratic formula we obtain $$frac{(a+d)pm sqrt{(a+d)^2-4(ad-c^2)}}{2}$$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad+4c^2}}{2}$$ Note that for the symmetric and positive definite matrix $M=begin{bmatrix}2 & 1 \ 2 & 3 end{bmatrix}$ the quadratic formula will give us two irrational values, while the stated formula only gives whole values. So the stated formulae are not true in general but in just a few cases.
Case 1: $c=0$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad}}{2}$$ $$frac{(a+d)pm (a-d)}{2}$$ $$lambda = a,d$$
Case 2: $a=d$ $$frac{2apm sqrt{4c^2}}{2}$$ $$apm c$$ $$lambda = a-c, a+c$$
answered Jan 15 at 1:41
Ryan GreylingRyan Greyling
3229
$endgroup$
add a comment |
$begingroup$
Let $M=begin{bmatrix} a & c \ c & dend{bmatrix}$ be a symmetric matrix. The eigenvalues of this matrix are given by $$(a-lambda)(d-lambda)-c^2=0$$ $$lambda^2-(a+d)lambda +ad-c^2$$ Using the quadratic formula we obtain $$frac{(a+d)pm sqrt{(a+d)^2-4(ad-c^2)}}{2}$$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad+4c^2}}{2}$$ Note that for the symmetric and positive definite matrix $M=begin{bmatrix}2 & 1 \ 2 & 3 end{bmatrix}$ the quadratic formula will give us two irrational values, while the stated formula only gives whole values. So the stated formulae are not true in general but in just a few cases.
Case 1: $c=0$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad}}{2}$$ $$frac{(a+d)pm (a-d)}{2}$$ $$lambda = a,d$$
Case 2: $a=d$ $$frac{2apm sqrt{4c^2}}{2}$$ $$apm c$$ $$lambda = a-c, a+c$$
answered Jan 15 at 1:41
Ryan GreylingRyan Greyling
3229
$endgroup$
add a comment |
$begingroup$
Let $M=begin{bmatrix} a & c \ c & dend{bmatrix}$ be a symmetric matrix. The eigenvalues of this matrix are given by $$(a-lambda)(d-lambda)-c^2=0$$ $$lambda^2-(a+d)lambda +ad-c^2$$ Using the quadratic formula we obtain $$frac{(a+d)pm sqrt{(a+d)^2-4(ad-c^2)}}{2}$$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad+4c^2}}{2}$$ Note that for the symmetric and positive definite matrix $M=begin{bmatrix}2 & 1 \ 2 & 3 end{bmatrix}$ the quadratic formula will give us two irrational values, while the stated formula only gives whole values. So the stated formulae are not true in general but in just a few cases.
Case 1: $c=0$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad}}{2}$$ $$frac{(a+d)pm (a-d)}{2}$$ $$lambda = a,d$$
Case 2: $a=d$ $$frac{2apm sqrt{4c^2}}{2}$$ $$apm c$$ $$lambda = a-c, a+c$$
answered Jan 15 at 1:41
Ryan GreylingRyan Greyling
3229
$endgroup$
Let $M=begin{bmatrix} a & c \ c & dend{bmatrix}$ be a symmetric matrix. The eigenvalues of this matrix are given by $$(a-lambda)(d-lambda)-c^2=0$$ $$lambda^2-(a+d)lambda +ad-c^2$$ Using the quadratic formula we obtain $$frac{(a+d)pm sqrt{(a+d)^2-4(ad-c^2)}}{2}$$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad+4c^2}}{2}$$ Note that for the symmetric and positive definite matrix $M=begin{bmatrix}2 & 1 \ 2 & 3 end{bmatrix}$ the quadratic formula will give us two irrational values, while the stated formula only gives whole values. So the stated formulae are not true in general but in just a few cases.
Case 1: $c=0$ $$frac{(a+d)pm sqrt{a^2+d^2-2ad}}{2}$$ $$frac{(a+d)pm (a-d)}{2}$$ $$lambda = a,d$$
Case 2: $a=d$ $$frac{2apm sqrt{4c^2}}{2}$$ $$apm c$$ $$lambda = a-c, a+c$$
answered Jan 15 at 1:41
Ryan GreylingRyan Greyling
3229
answered Jan 15 at 1:41
Ryan GreylingRyan Greyling
3229
answered Jan 15 at 1:41
Ryan GreylingRyan Greyling
3229
answered Jan 15 at 1:41
Ryan GreylingRyan Greyling
3229
3229
add a comment |
add a comment |
$begingroup$
This is patently untrue for a general positive-definite matrix: Combining the first and third equations produces $lambda_1+lambda_2 = operatorname{tr}(A)=2a$, while the second and fourth give $operatorname{tr}(A)=2d$. Any diagonal matrix with different positive entires on the diagonal is a counterexample.
Note, too, that the statements that $A$ is symmetric and positive-definite, and that $x^TAy$ is a valid inner (not “dot”) product are equivalent.
answered Jan 15 at 2:33
amdamd
30.4k21050
$endgroup$
$begingroup$
Thanks for the comment. Regarding your note about dot and inner products-- is it true that a dot product an example of an inner product?
$endgroup$
– kkun
Jan 15 at 3:00
$begingroup$
@kkun Yes. I’m careful to distinguish the dot product—the matrix product of a pair of coordinate tuples—from a scalar product, which is a more general concept. An inner product is a positive-definite scalar product. Many sources use the terms interchangeably when working in $mathbb R^n$ with the Euclidean inner product and the standard basis because the formula for this scalar product is then given by the dot product. If you change to a non-orthonormal basis, though, the value of the inner product of two vectors doesn’t change, but its formula in terms of coordinates does.
$endgroup$
– amd
Jan 15 at 3:13
add a comment |
$begingroup$
This is patently untrue for a general positive-definite matrix: Combining the first and third equations produces $lambda_1+lambda_2 = operatorname{tr}(A)=2a$, while the second and fourth give $operatorname{tr}(A)=2d$. Any diagonal matrix with different positive entires on the diagonal is a counterexample.
Note, too, that the statements that $A$ is symmetric and positive-definite, and that $x^TAy$ is a valid inner (not “dot”) product are equivalent.
answered Jan 15 at 2:33
amdamd
30.4k21050
$endgroup$
$begingroup$
Thanks for the comment. Regarding your note about dot and inner products-- is it true that a dot product an example of an inner product?
$endgroup$
– kkun
Jan 15 at 3:00
$begingroup$
@kkun Yes. I’m careful to distinguish the dot product—the matrix product of a pair of coordinate tuples—from a scalar product, which is a more general concept. An inner product is a positive-definite scalar product. Many sources use the terms interchangeably when working in $mathbb R^n$ with the Euclidean inner product and the standard basis because the formula for this scalar product is then given by the dot product. If you change to a non-orthonormal basis, though, the value of the inner product of two vectors doesn’t change, but its formula in terms of coordinates does.
$endgroup$
– amd
Jan 15 at 3:13
add a comment |
$begingroup$
This is patently untrue for a general positive-definite matrix: Combining the first and third equations produces $lambda_1+lambda_2 = operatorname{tr}(A)=2a$, while the second and fourth give $operatorname{tr}(A)=2d$. Any diagonal matrix with different positive entires on the diagonal is a counterexample.
Note, too, that the statements that $A$ is symmetric and positive-definite, and that $x^TAy$ is a valid inner (not “dot”) product are equivalent.
answered Jan 15 at 2:33
amdamd
30.4k21050
$endgroup$
This is patently untrue for a general positive-definite matrix: Combining the first and third equations produces $lambda_1+lambda_2 = operatorname{tr}(A)=2a$, while the second and fourth give $operatorname{tr}(A)=2d$. Any diagonal matrix with different positive entires on the diagonal is a counterexample.
Note, too, that the statements that $A$ is symmetric and positive-definite, and that $x^TAy$ is a valid inner (not “dot”) product are equivalent.
answered Jan 15 at 2:33
amdamd
30.4k21050
answered Jan 15 at 2:33
amdamd
30.4k21050
answered Jan 15 at 2:33
amdamd
30.4k21050
answered Jan 15 at 2:33
amdamd
30.4k21050
30.4k21050
$begingroup$
Thanks for the comment. Regarding your note about dot and inner products-- is it true that a dot product an example of an inner product?
$endgroup$
– kkun
Jan 15 at 3:00
$begingroup$
@kkun Yes. I’m careful to distinguish the dot product—the matrix product of a pair of coordinate tuples—from a scalar product, which is a more general concept. An inner product is a positive-definite scalar product. Many sources use the terms interchangeably when working in $mathbb R^n$ with the Euclidean inner product and the standard basis because the formula for this scalar product is then given by the dot product. If you change to a non-orthonormal basis, though, the value of the inner product of two vectors doesn’t change, but its formula in terms of coordinates does.
$endgroup$
– amd
Jan 15 at 3:13
add a comment |
$begingroup$
Thanks for the comment. Regarding your note about dot and inner products-- is it true that a dot product an example of an inner product?
$endgroup$
– kkun
Jan 15 at 3:00
$begingroup$
@kkun Yes. I’m careful to distinguish the dot product—the matrix product of a pair of coordinate tuples—from a scalar product, which is a more general concept. An inner product is a positive-definite scalar product. Many sources use the terms interchangeably when working in $mathbb R^n$ with the Euclidean inner product and the standard basis because the formula for this scalar product is then given by the dot product. If you change to a non-orthonormal basis, though, the value of the inner product of two vectors doesn’t change, but its formula in terms of coordinates does.
$endgroup$
– amd
Jan 15 at 3:13
$begingroup$
Thanks for the comment. Regarding your note about dot and inner products-- is it true that a dot product an example of an inner product?
$endgroup$
– kkun
Jan 15 at 3:00
$begingroup$
Thanks for the comment. Regarding your note about dot and inner products-- is it true that a dot product an example of an inner product?
$endgroup$
– kkun
Jan 15 at 3:00
$begingroup$
@kkun Yes. I’m careful to distinguish the dot product—the matrix product of a pair of coordinate tuples—from a scalar product, which is a more general concept. An inner product is a positive-definite scalar product. Many sources use the terms interchangeably when working in $mathbb R^n$ with the Euclidean inner product and the standard basis because the formula for this scalar product is then given by the dot product. If you change to a non-orthonormal basis, though, the value of the inner product of two vectors doesn’t change, but its formula in terms of coordinates does.
$endgroup$
– amd
Jan 15 at 3:13
$begingroup$
@kkun Yes. I’m careful to distinguish the dot product—the matrix product of a pair of coordinate tuples—from a scalar product, which is a more general concept. An inner product is a positive-definite scalar product. Many sources use the terms interchangeably when working in $mathbb R^n$ with the Euclidean inner product and the standard basis because the formula for this scalar product is then given by the dot product. If you change to a non-orthonormal basis, though, the value of the inner product of two vectors doesn’t change, but its formula in terms of coordinates does.
$endgroup$
– amd
Jan 15 at 3:13
add a comment |
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$begingroup$
this is completely wrong.
$endgroup$
– Will Jagy
Jan 15 at 1:06
$begingroup$
If $a=d$ then it is true.
$endgroup$
– Sota Antonino
Jan 15 at 1:16
$begingroup$
is it true if $b = c$? How can you prove this is true?
$endgroup$
– kkun
Jan 15 at 1:29