Mixing
Let $A= begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}$. Find the entry in the first row and second column...
$begingroup$
I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.
So that $A^{2014}=PD^{2014}P^{-1}$
But i just want to know whether there is an alternate method.
linear-algebra eigenvalues-eigenvectors diagonalization
edited Jan 6 at 0:18
vadim123
75.9k897189
asked Jan 6 at 0:16
DD90DD90
2628
$endgroup$
add a comment |
$begingroup$
I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.
So that $A^{2014}=PD^{2014}P^{-1}$
But i just want to know whether there is an alternate method.
linear-algebra eigenvalues-eigenvectors diagonalization
edited Jan 6 at 0:18
vadim123
75.9k897189
asked Jan 6 at 0:16
DD90DD90
2628
$endgroup$
1
$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19
1
$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19
add a comment |
$begingroup$
I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.
So that $A^{2014}=PD^{2014}P^{-1}$
But i just want to know whether there is an alternate method.
linear-algebra eigenvalues-eigenvectors diagonalization
edited Jan 6 at 0:18
vadim123
75.9k897189
asked Jan 6 at 0:16
DD90DD90
2628
$endgroup$
I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.
So that $A^{2014}=PD^{2014}P^{-1}$
But i just want to know whether there is an alternate method.
linear-algebra eigenvalues-eigenvectors diagonalization
linear-algebra eigenvalues-eigenvectors diagonalization
edited Jan 6 at 0:18
vadim123
75.9k897189
asked Jan 6 at 0:16
DD90DD90
2628
edited Jan 6 at 0:18
vadim123
75.9k897189
asked Jan 6 at 0:16
DD90DD90
2628
edited Jan 6 at 0:18
vadim123
75.9k897189
edited Jan 6 at 0:18
vadim123
75.9k897189
edited Jan 6 at 0:18
vadim123
75.9k897189
75.9k897189
asked Jan 6 at 0:16
DD90DD90
2628
asked Jan 6 at 0:16
DD90DD90
2628
asked Jan 6 at 0:16
DD90DD90
2628
2628
1
$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19
1
$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19
add a comment |
1
$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19
1
$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19
1
1
$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19
$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19
1
1
$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19
$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint Show that
$$A^2=6A$$
Note: You can show the stronger statement:
$$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
but this is overkill.
answered Jan 6 at 0:23
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
$endgroup$
– Mindlack
Jan 6 at 0:25
$begingroup$
Very nice indeed!
$endgroup$
– Mike
Jan 6 at 0:33
$begingroup$
Can you please explain it a little further how to proceed with this
$endgroup$
– DD90
Jan 6 at 0:35
1
$begingroup$
If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
$endgroup$
– Mindlack
Jan 6 at 0:52
$begingroup$
@DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
$endgroup$
– N. S.
Jan 6 at 3:34
add a comment |
$begingroup$
By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.
Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.
answered Jan 6 at 1:02
amdamd
29.7k21050
$endgroup$
add a comment |
$begingroup$
Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
$$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$
Which now immediately gives you
$$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$
answered Jan 6 at 3:26
obscuransobscurans
1,027311
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint Show that
$$A^2=6A$$
Note: You can show the stronger statement:
$$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
but this is overkill.
answered Jan 6 at 0:23
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
$endgroup$
– Mindlack
Jan 6 at 0:25
$begingroup$
Very nice indeed!
$endgroup$
– Mike
Jan 6 at 0:33
$begingroup$
Can you please explain it a little further how to proceed with this
$endgroup$
– DD90
Jan 6 at 0:35
1
$begingroup$
If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
$endgroup$
– Mindlack
Jan 6 at 0:52
$begingroup$
@DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
$endgroup$
– N. S.
Jan 6 at 3:34
add a comment |
$begingroup$
Hint Show that
$$A^2=6A$$
Note: You can show the stronger statement:
$$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
but this is overkill.
answered Jan 6 at 0:23
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
$endgroup$
– Mindlack
Jan 6 at 0:25
$begingroup$
Very nice indeed!
$endgroup$
– Mike
Jan 6 at 0:33
$begingroup$
Can you please explain it a little further how to proceed with this
$endgroup$
– DD90
Jan 6 at 0:35
1
$begingroup$
If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
$endgroup$
– Mindlack
Jan 6 at 0:52
$begingroup$
@DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
$endgroup$
– N. S.
Jan 6 at 3:34
add a comment |
$begingroup$
Hint Show that
$$A^2=6A$$
Note: You can show the stronger statement:
$$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
but this is overkill.
answered Jan 6 at 0:23
N. S.N. S.
103k6111208
$endgroup$
Hint Show that
$$A^2=6A$$
Note: You can show the stronger statement:
$$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
but this is overkill.
answered Jan 6 at 0:23
N. S.N. S.
103k6111208
answered Jan 6 at 0:23
N. S.N. S.
103k6111208
answered Jan 6 at 0:23
N. S.N. S.
103k6111208
answered Jan 6 at 0:23
N. S.N. S.
103k6111208
103k6111208
$begingroup$
This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
$endgroup$
– Mindlack
Jan 6 at 0:25
$begingroup$
Very nice indeed!
$endgroup$
– Mike
Jan 6 at 0:33
$begingroup$
Can you please explain it a little further how to proceed with this
$endgroup$
– DD90
Jan 6 at 0:35
1
$begingroup$
If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
$endgroup$
– Mindlack
Jan 6 at 0:52
$begingroup$
@DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
$endgroup$
– N. S.
Jan 6 at 3:34
add a comment |
$begingroup$
This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
$endgroup$
– Mindlack
Jan 6 at 0:25
$begingroup$
Very nice indeed!
$endgroup$
– Mike
Jan 6 at 0:33
$begingroup$
Can you please explain it a little further how to proceed with this
$endgroup$
– DD90
Jan 6 at 0:35
1
$begingroup$
If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
$endgroup$
– Mindlack
Jan 6 at 0:52
$begingroup$
@DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
$endgroup$
– N. S.
Jan 6 at 3:34
$begingroup$
This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
$endgroup$
– Mindlack
Jan 6 at 0:25
$begingroup$
This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
$endgroup$
– Mindlack
Jan 6 at 0:25
$begingroup$
Very nice indeed!
$endgroup$
– Mike
Jan 6 at 0:33
$begingroup$
Very nice indeed!
$endgroup$
– Mike
Jan 6 at 0:33
$begingroup$
Can you please explain it a little further how to proceed with this
$endgroup$
– DD90
Jan 6 at 0:35
$begingroup$
Can you please explain it a little further how to proceed with this
$endgroup$
– DD90
Jan 6 at 0:35
1
1
$begingroup$
If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
$endgroup$
– Mindlack
Jan 6 at 0:52
$begingroup$
If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
$endgroup$
– Mindlack
Jan 6 at 0:52
$begingroup$
@DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
$endgroup$
– N. S.
Jan 6 at 3:34
$begingroup$
@DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
$endgroup$
– N. S.
Jan 6 at 3:34
add a comment |
$begingroup$
By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.
Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.
answered Jan 6 at 1:02
amdamd
29.7k21050
$endgroup$
add a comment |
$begingroup$
By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.
Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.
answered Jan 6 at 1:02
amdamd
29.7k21050
$endgroup$
add a comment |
$begingroup$
By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.
Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.
answered Jan 6 at 1:02
amdamd
29.7k21050
$endgroup$
By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.
Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.
answered Jan 6 at 1:02
amdamd
29.7k21050
edited Jan 6 at 1:08
answered Jan 6 at 1:02
amdamd
29.7k21050
answered Jan 6 at 1:02
amdamd
29.7k21050
answered Jan 6 at 1:02
amdamd
29.7k21050
29.7k21050
add a comment |
add a comment |
$begingroup$
Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
$$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$
Which now immediately gives you
$$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$
answered Jan 6 at 3:26
obscuransobscurans
1,027311
$endgroup$
add a comment |
$begingroup$
Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
$$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$
Which now immediately gives you
$$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$
answered Jan 6 at 3:26
obscuransobscurans
1,027311
$endgroup$
add a comment |
$begingroup$
Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
$$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$
Which now immediately gives you
$$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$
answered Jan 6 at 3:26
obscuransobscurans
1,027311
$endgroup$
Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
$$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$
Which now immediately gives you
$$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$
answered Jan 6 at 3:26
obscuransobscurans
1,027311
answered Jan 6 at 3:26
obscuransobscurans
1,027311
answered Jan 6 at 3:26
obscuransobscurans
1,027311
answered Jan 6 at 3:26
obscuransobscurans
1,027311
1,027311
add a comment |
add a comment |
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1
$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19
1
$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
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– José Alejandro Aburto Araneda
Jan 6 at 0:19