Confused about finding non-brute force way to solve for matrix to the 2019th power The Next CEO of Stack OverflowGiven a matrix $A$ find a matrix $C$ such that $C^3$=$A$$XX^t=A$, $X=?$. Where $X in 0,1^n times m$Cholesky decomposition for sparse matrixIs there a quick way to generate the characteristic polynomial of a Vandermonde matrix?Quick way of finding the eigenvalues and eigenvectors of the matrix $A=operatornametridiag_n(-1,alpha,-1)$Calculate $P(X_16=2|X_0=0)$How can one understand if a system is consistent from RREF?Is there a clever/non-brute force way to compute the kernel of a matrix?Force Eigenvalues of a Matrix to be RationalNeed a matrix to seventh power which is the identity but the original is not or the negative identity
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Confused about finding non-brute force way to solve for matrix to the 2019th power
The Next CEO of Stack OverflowGiven a matrix $A$ find a matrix $C$ such that $C^3$=$A$$XX^t=A$, $X=?$. Where $X in 0,1^n times m$Cholesky decomposition for sparse matrixIs there a quick way to generate the characteristic polynomial of a Vandermonde matrix?Quick way of finding the eigenvalues and eigenvectors of the matrix $A=operatornametridiag_n(-1,alpha,-1)$Calculate $P(X_16=2|X_0=0)$How can one understand if a system is consistent from RREF?Is there a clever/non-brute force way to compute the kernel of a matrix?Force Eigenvalues of a Matrix to be RationalNeed a matrix to seventh power which is the identity but the original is not or the negative identity
$begingroup$
I am attempting to solve this problem, it has four parts. I solved part a (a trivial matrix problem), but the next three parts appear to be a bit confusing to me. I just would like some help getting started so I can see and observe this matrix and come up with a solution.
The Questions Note:
The idea here is NOT to use brute force computation to get $A^2019$ matrix, instead use some observations which can significantly reduce computational work and will also give you an insight into such problems.
Obviously this is some huge numbered matrix, but I do not understand what observation will reduce this? My first thought was just to use a calculator and calculate $A^2019$ and then just multiply that by each vector. But that appears to be not the point of the question.
$$
Let A = left[beginarrayrrr
-4 & -6 & -12 \
-2 & -1 & -4 \
2 & 3 & 6
endarrayright]
$$
And Let $u$ = [6 5 -3], $v$ = [-2 0 1], and $w$ = [-2 -1 1].
b). Compute $A^2019mathbb v$
linear-algebra matrices
asked Mar 19 at 20:38
icodericoder
52
$endgroup$
|
show 3 more comments
$begingroup$
I am attempting to solve this problem, it has four parts. I solved part a (a trivial matrix problem), but the next three parts appear to be a bit confusing to me. I just would like some help getting started so I can see and observe this matrix and come up with a solution.
The Questions Note:
The idea here is NOT to use brute force computation to get $A^2019$ matrix, instead use some observations which can significantly reduce computational work and will also give you an insight into such problems.
Obviously this is some huge numbered matrix, but I do not understand what observation will reduce this? My first thought was just to use a calculator and calculate $A^2019$ and then just multiply that by each vector. But that appears to be not the point of the question.
$$
Let A = left[beginarrayrrr
-4 & -6 & -12 \
-2 & -1 & -4 \
2 & 3 & 6
endarrayright]
$$
And Let $u$ = [6 5 -3], $v$ = [-2 0 1], and $w$ = [-2 -1 1].
b). Compute $A^2019mathbb v$
linear-algebra matrices
asked Mar 19 at 20:38
icodericoder
52
$endgroup$
$begingroup$
If $p(x)$ is the characteristic polynomial of $A$, then $p(A)=0$. If you divide $x^2019=p(x)q(x)+r(x)$, with $r$ of smaller degree than $p$, then $A^2019=p(A)q(A)+r(A)=r(A)$.
$endgroup$
– user647486
Mar 19 at 20:40
$begingroup$
Take into account that the quotient $q$ doesn't need to be computed. Only the remainder is necessary. The characteristic polynomial in your case is $-x^3 + x^2 + 2 x$. The remainder can be found by indeterminate coefficients. For example, evaluating the equation $x^2019=p(x)q(x)+r(x)$ at the roots of $p$. In your case $0,-1,2$. The remainder becomes, I think, $frac2^2019-4cdot (-1)^20196x^2+frac2^2019+2cdot(-1)^20195x$.
$endgroup$
– user647486
Mar 19 at 20:47
$begingroup$
So, you only need to compute $A$ and $A^2$ and plug them in there.
$endgroup$
– user647486
Mar 19 at 20:49
1
$begingroup$
Hint: Is there a simple relationship between $mathbf v$ and $Amathbf v$?
$endgroup$
– amd
Mar 19 at 20:50
$begingroup$
What is $x$ ? Also you can decompose your matrix like this : $A=SJS^-1$ where $J$ is a diagonal matrix and then $A^2019=A J^2019 S^-1$.
$endgroup$
– Alain
Mar 19 at 20:51
|
show 3 more comments
$begingroup$
I am attempting to solve this problem, it has four parts. I solved part a (a trivial matrix problem), but the next three parts appear to be a bit confusing to me. I just would like some help getting started so I can see and observe this matrix and come up with a solution.
The Questions Note:
The idea here is NOT to use brute force computation to get $A^2019$ matrix, instead use some observations which can significantly reduce computational work and will also give you an insight into such problems.
Obviously this is some huge numbered matrix, but I do not understand what observation will reduce this? My first thought was just to use a calculator and calculate $A^2019$ and then just multiply that by each vector. But that appears to be not the point of the question.
$$
Let A = left[beginarrayrrr
-4 & -6 & -12 \
-2 & -1 & -4 \
2 & 3 & 6
endarrayright]
$$
And Let $u$ = [6 5 -3], $v$ = [-2 0 1], and $w$ = [-2 -1 1].
b). Compute $A^2019mathbb v$
linear-algebra matrices
asked Mar 19 at 20:38
icodericoder
52
$endgroup$
I am attempting to solve this problem, it has four parts. I solved part a (a trivial matrix problem), but the next three parts appear to be a bit confusing to me. I just would like some help getting started so I can see and observe this matrix and come up with a solution.
The Questions Note:
The idea here is NOT to use brute force computation to get $A^2019$ matrix, instead use some observations which can significantly reduce computational work and will also give you an insight into such problems.
Obviously this is some huge numbered matrix, but I do not understand what observation will reduce this? My first thought was just to use a calculator and calculate $A^2019$ and then just multiply that by each vector. But that appears to be not the point of the question.
$$
Let A = left[beginarrayrrr
-4 & -6 & -12 \
-2 & -1 & -4 \
2 & 3 & 6
endarrayright]
$$
And Let $u$ = [6 5 -3], $v$ = [-2 0 1], and $w$ = [-2 -1 1].
b). Compute $A^2019mathbb v$
linear-algebra matrices
linear-algebra matrices
asked Mar 19 at 20:38
icodericoder
52
asked Mar 19 at 20:38
icodericoder
52
edited Mar 19 at 21:00
icoder
asked Mar 19 at 20:38
icodericoder
52
asked Mar 19 at 20:38
icodericoder
52
asked Mar 19 at 20:38
icodericoder
52
52
$begingroup$
If $p(x)$ is the characteristic polynomial of $A$, then $p(A)=0$. If you divide $x^2019=p(x)q(x)+r(x)$, with $r$ of smaller degree than $p$, then $A^2019=p(A)q(A)+r(A)=r(A)$.
$endgroup$
– user647486
Mar 19 at 20:40
$begingroup$
Take into account that the quotient $q$ doesn't need to be computed. Only the remainder is necessary. The characteristic polynomial in your case is $-x^3 + x^2 + 2 x$. The remainder can be found by indeterminate coefficients. For example, evaluating the equation $x^2019=p(x)q(x)+r(x)$ at the roots of $p$. In your case $0,-1,2$. The remainder becomes, I think, $frac2^2019-4cdot (-1)^20196x^2+frac2^2019+2cdot(-1)^20195x$.
$endgroup$
– user647486
Mar 19 at 20:47
$begingroup$
So, you only need to compute $A$ and $A^2$ and plug them in there.
$endgroup$
– user647486
Mar 19 at 20:49
1
$begingroup$
Hint: Is there a simple relationship between $mathbf v$ and $Amathbf v$?
$endgroup$
– amd
Mar 19 at 20:50
$begingroup$
What is $x$ ? Also you can decompose your matrix like this : $A=SJS^-1$ where $J$ is a diagonal matrix and then $A^2019=A J^2019 S^-1$.
$endgroup$
– Alain
Mar 19 at 20:51
|
show 3 more comments
$begingroup$
If $p(x)$ is the characteristic polynomial of $A$, then $p(A)=0$. If you divide $x^2019=p(x)q(x)+r(x)$, with $r$ of smaller degree than $p$, then $A^2019=p(A)q(A)+r(A)=r(A)$.
$endgroup$
– user647486
Mar 19 at 20:40
$begingroup$
Take into account that the quotient $q$ doesn't need to be computed. Only the remainder is necessary. The characteristic polynomial in your case is $-x^3 + x^2 + 2 x$. The remainder can be found by indeterminate coefficients. For example, evaluating the equation $x^2019=p(x)q(x)+r(x)$ at the roots of $p$. In your case $0,-1,2$. The remainder becomes, I think, $frac2^2019-4cdot (-1)^20196x^2+frac2^2019+2cdot(-1)^20195x$.
$endgroup$
– user647486
Mar 19 at 20:47
$begingroup$
So, you only need to compute $A$ and $A^2$ and plug them in there.
$endgroup$
– user647486
Mar 19 at 20:49
1
$begingroup$
Hint: Is there a simple relationship between $mathbf v$ and $Amathbf v$?
$endgroup$
– amd
Mar 19 at 20:50
$begingroup$
What is $x$ ? Also you can decompose your matrix like this : $A=SJS^-1$ where $J$ is a diagonal matrix and then $A^2019=A J^2019 S^-1$.
$endgroup$
– Alain
Mar 19 at 20:51
$begingroup$
If $p(x)$ is the characteristic polynomial of $A$, then $p(A)=0$. If you divide $x^2019=p(x)q(x)+r(x)$, with $r$ of smaller degree than $p$, then $A^2019=p(A)q(A)+r(A)=r(A)$.
$endgroup$
– user647486
Mar 19 at 20:40
$begingroup$
If $p(x)$ is the characteristic polynomial of $A$, then $p(A)=0$. If you divide $x^2019=p(x)q(x)+r(x)$, with $r$ of smaller degree than $p$, then $A^2019=p(A)q(A)+r(A)=r(A)$.
$endgroup$
– user647486
Mar 19 at 20:40
$begingroup$
Take into account that the quotient $q$ doesn't need to be computed. Only the remainder is necessary. The characteristic polynomial in your case is $-x^3 + x^2 + 2 x$. The remainder can be found by indeterminate coefficients. For example, evaluating the equation $x^2019=p(x)q(x)+r(x)$ at the roots of $p$. In your case $0,-1,2$. The remainder becomes, I think, $frac2^2019-4cdot (-1)^20196x^2+frac2^2019+2cdot(-1)^20195x$.
$endgroup$
– user647486
Mar 19 at 20:47
$begingroup$
Take into account that the quotient $q$ doesn't need to be computed. Only the remainder is necessary. The characteristic polynomial in your case is $-x^3 + x^2 + 2 x$. The remainder can be found by indeterminate coefficients. For example, evaluating the equation $x^2019=p(x)q(x)+r(x)$ at the roots of $p$. In your case $0,-1,2$. The remainder becomes, I think, $frac2^2019-4cdot (-1)^20196x^2+frac2^2019+2cdot(-1)^20195x$.
$endgroup$
– user647486
Mar 19 at 20:47
$begingroup$
So, you only need to compute $A$ and $A^2$ and plug them in there.
$endgroup$
– user647486
Mar 19 at 20:49
$begingroup$
So, you only need to compute $A$ and $A^2$ and plug them in there.
$endgroup$
– user647486
Mar 19 at 20:49
1
1
$begingroup$
Hint: Is there a simple relationship between $mathbf v$ and $Amathbf v$?
$endgroup$
– amd
Mar 19 at 20:50
$begingroup$
Hint: Is there a simple relationship between $mathbf v$ and $Amathbf v$?
$endgroup$
– amd
Mar 19 at 20:50
$begingroup$
What is $x$ ? Also you can decompose your matrix like this : $A=SJS^-1$ where $J$ is a diagonal matrix and then $A^2019=A J^2019 S^-1$.
$endgroup$
– Alain
Mar 19 at 20:51
$begingroup$
What is $x$ ? Also you can decompose your matrix like this : $A=SJS^-1$ where $J$ is a diagonal matrix and then $A^2019=A J^2019 S^-1$.
$endgroup$
– Alain
Mar 19 at 20:51
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I assume that you're asking about $A^2019v$.
To that end, observe that $Av = 2v$ (I'll leave it to you to verify that this is true).
It follows that
$$
A^2v = AAv = A(2v) = 2Av = 4v.
$$
Similarly, $A^3 v = 8v$, and in general we have $A^k v = 2^k v$ for any positive integer $k$.
answered Mar 19 at 20:54
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Ah, that is an interesting relationship! There are two more parts to this question, but I assume that these will be similarly discovered! Thank you so much
$endgroup$
– icoder
Mar 19 at 21:06
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
I assume that you're asking about $A^2019v$.
To that end, observe that $Av = 2v$ (I'll leave it to you to verify that this is true).
It follows that
$$
A^2v = AAv = A(2v) = 2Av = 4v.
$$
Similarly, $A^3 v = 8v$, and in general we have $A^k v = 2^k v$ for any positive integer $k$.
answered Mar 19 at 20:54
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Ah, that is an interesting relationship! There are two more parts to this question, but I assume that these will be similarly discovered! Thank you so much
$endgroup$
– icoder
Mar 19 at 21:06
add a comment |
$begingroup$
I assume that you're asking about $A^2019v$.
To that end, observe that $Av = 2v$ (I'll leave it to you to verify that this is true).
It follows that
$$
A^2v = AAv = A(2v) = 2Av = 4v.
$$
Similarly, $A^3 v = 8v$, and in general we have $A^k v = 2^k v$ for any positive integer $k$.
answered Mar 19 at 20:54
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Ah, that is an interesting relationship! There are two more parts to this question, but I assume that these will be similarly discovered! Thank you so much
$endgroup$
– icoder
Mar 19 at 21:06
add a comment |
$begingroup$
I assume that you're asking about $A^2019v$.
To that end, observe that $Av = 2v$ (I'll leave it to you to verify that this is true).
It follows that
$$
A^2v = AAv = A(2v) = 2Av = 4v.
$$
Similarly, $A^3 v = 8v$, and in general we have $A^k v = 2^k v$ for any positive integer $k$.
answered Mar 19 at 20:54
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
I assume that you're asking about $A^2019v$.
To that end, observe that $Av = 2v$ (I'll leave it to you to verify that this is true).
It follows that
$$
A^2v = AAv = A(2v) = 2Av = 4v.
$$
Similarly, $A^3 v = 8v$, and in general we have $A^k v = 2^k v$ for any positive integer $k$.
answered Mar 19 at 20:54
OmnomnomnomOmnomnomnom
129k792186
answered Mar 19 at 20:54
OmnomnomnomOmnomnomnom
129k792186
answered Mar 19 at 20:54
OmnomnomnomOmnomnomnom
129k792186
answered Mar 19 at 20:54
OmnomnomnomOmnomnomnom
129k792186
129k792186
$begingroup$
Ah, that is an interesting relationship! There are two more parts to this question, but I assume that these will be similarly discovered! Thank you so much
$endgroup$
– icoder
Mar 19 at 21:06
add a comment |
$begingroup$
Ah, that is an interesting relationship! There are two more parts to this question, but I assume that these will be similarly discovered! Thank you so much
$endgroup$
– icoder
Mar 19 at 21:06
$begingroup$
Ah, that is an interesting relationship! There are two more parts to this question, but I assume that these will be similarly discovered! Thank you so much
$endgroup$
– icoder
Mar 19 at 21:06
$begingroup$
Ah, that is an interesting relationship! There are two more parts to this question, but I assume that these will be similarly discovered! Thank you so much
$endgroup$
– icoder
Mar 19 at 21:06
add a comment |
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$begingroup$
If $p(x)$ is the characteristic polynomial of $A$, then $p(A)=0$. If you divide $x^2019=p(x)q(x)+r(x)$, with $r$ of smaller degree than $p$, then $A^2019=p(A)q(A)+r(A)=r(A)$.
$endgroup$
– user647486
Mar 19 at 20:40
$begingroup$
Take into account that the quotient $q$ doesn't need to be computed. Only the remainder is necessary. The characteristic polynomial in your case is $-x^3 + x^2 + 2 x$. The remainder can be found by indeterminate coefficients. For example, evaluating the equation $x^2019=p(x)q(x)+r(x)$ at the roots of $p$. In your case $0,-1,2$. The remainder becomes, I think, $frac2^2019-4cdot (-1)^20196x^2+frac2^2019+2cdot(-1)^20195x$.
$endgroup$
– user647486
Mar 19 at 20:47
$begingroup$
So, you only need to compute $A$ and $A^2$ and plug them in there.
$endgroup$
– user647486
Mar 19 at 20:49
1
$begingroup$
Hint: Is there a simple relationship between $mathbf v$ and $Amathbf v$?
$endgroup$
– amd
Mar 19 at 20:50
$begingroup$
What is $x$ ? Also you can decompose your matrix like this : $A=SJS^-1$ where $J$ is a diagonal matrix and then $A^2019=A J^2019 S^-1$.
$endgroup$
– Alain
Mar 19 at 20:51