Mixing
symmetric and regular matrix
$begingroup$
A is symmetric, regular matrix. I need to prove that A ^ 17 is symmetric too.
There is some sentence about symmetric matrix I can use here?
linear-algebra matrices
asked Jan 27 at 15:25
user3523226user3523226
347
$endgroup$
add a comment |
$begingroup$
A is symmetric, regular matrix. I need to prove that A ^ 17 is symmetric too.
There is some sentence about symmetric matrix I can use here?
linear-algebra matrices
asked Jan 27 at 15:25
user3523226user3523226
347
$endgroup$
$begingroup$
What do you mean by "sentence about symmetric matrix"?
$endgroup$
– user635162
Jan 27 at 15:27
$begingroup$
Do you know that $(AB)' = (B') (A')$ , for matrices $A$ and $B$ in general?
$endgroup$
– kimchi lover
Jan 27 at 15:41
$begingroup$
@kimchilover I don't understand how it helps..
$endgroup$
– user3523226
Jan 27 at 17:13
$begingroup$
So what have you tried already? What do you know about this problem. Or about symmetric matrices? Can you answer the (presumably simpler) problem about $A^2$?
$endgroup$
– kimchi lover
Jan 27 at 17:19
add a comment |
$begingroup$
A is symmetric, regular matrix. I need to prove that A ^ 17 is symmetric too.
There is some sentence about symmetric matrix I can use here?
linear-algebra matrices
asked Jan 27 at 15:25
user3523226user3523226
347
$endgroup$
A is symmetric, regular matrix. I need to prove that A ^ 17 is symmetric too.
There is some sentence about symmetric matrix I can use here?
linear-algebra matrices
linear-algebra matrices
asked Jan 27 at 15:25
user3523226user3523226
347
asked Jan 27 at 15:25
user3523226user3523226
347
asked Jan 27 at 15:25
user3523226user3523226
347
asked Jan 27 at 15:25
user3523226user3523226
347
asked Jan 27 at 15:25
user3523226user3523226
347
347
$begingroup$
What do you mean by "sentence about symmetric matrix"?
$endgroup$
– user635162
Jan 27 at 15:27
$begingroup$
Do you know that $(AB)' = (B') (A')$ , for matrices $A$ and $B$ in general?
$endgroup$
– kimchi lover
Jan 27 at 15:41
$begingroup$
@kimchilover I don't understand how it helps..
$endgroup$
– user3523226
Jan 27 at 17:13
$begingroup$
So what have you tried already? What do you know about this problem. Or about symmetric matrices? Can you answer the (presumably simpler) problem about $A^2$?
$endgroup$
– kimchi lover
Jan 27 at 17:19
add a comment |
$begingroup$
What do you mean by "sentence about symmetric matrix"?
$endgroup$
– user635162
Jan 27 at 15:27
$begingroup$
Do you know that $(AB)' = (B') (A')$ , for matrices $A$ and $B$ in general?
$endgroup$
– kimchi lover
Jan 27 at 15:41
$begingroup$
@kimchilover I don't understand how it helps..
$endgroup$
– user3523226
Jan 27 at 17:13
$begingroup$
So what have you tried already? What do you know about this problem. Or about symmetric matrices? Can you answer the (presumably simpler) problem about $A^2$?
$endgroup$
– kimchi lover
Jan 27 at 17:19
$begingroup$
What do you mean by "sentence about symmetric matrix"?
$endgroup$
– user635162
Jan 27 at 15:27
$begingroup$
What do you mean by "sentence about symmetric matrix"?
$endgroup$
– user635162
Jan 27 at 15:27
$begingroup$
Do you know that $(AB)' = (B') (A')$ , for matrices $A$ and $B$ in general?
$endgroup$
– kimchi lover
Jan 27 at 15:41
$begingroup$
Do you know that $(AB)' = (B') (A')$ , for matrices $A$ and $B$ in general?
$endgroup$
– kimchi lover
Jan 27 at 15:41
$begingroup$
@kimchilover I don't understand how it helps..
$endgroup$
– user3523226
Jan 27 at 17:13
$begingroup$
@kimchilover I don't understand how it helps..
$endgroup$
– user3523226
Jan 27 at 17:13
$begingroup$
So what have you tried already? What do you know about this problem. Or about symmetric matrices? Can you answer the (presumably simpler) problem about $A^2$?
$endgroup$
– kimchi lover
Jan 27 at 17:19
$begingroup$
So what have you tried already? What do you know about this problem. Or about symmetric matrices? Can you answer the (presumably simpler) problem about $A^2$?
$endgroup$
– kimchi lover
Jan 27 at 17:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Well, here says that for integer $n space,A^{n}$ is symmetric if $A$ is symmetric. But if you want really to prove that, you can use formula which says that matrix is diagonalizable if and only if has n different linear independent eigenvectors. It means that matrix is symmilar to some diagonal matrix D. And matrix S is matrix that has eigenvectors from matrix A as colums. $Rightarrow A=Scdot D cdot S^{-1}$
$$A^{n}=underbrace{Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}...}\ _{n times}$$ $$A^{n}=Scdot D^{n}cdot S^{-1}$$ I think it is the easiest way to prove that matrix $space A^{n}$ is symmetric if $A$ is symmetric for some integer $n$.
answered Jan 27 at 19:09
FiggaroFiggaro
234
$endgroup$
$begingroup$
You can use the fact that $(AB)'=B'A'$ to calculate $(A^n)' = (A')^n=A^n$, knowing nothing about diagonalizability.
$endgroup$
– kimchi lover
Jan 27 at 22:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, here says that for integer $n space,A^{n}$ is symmetric if $A$ is symmetric. But if you want really to prove that, you can use formula which says that matrix is diagonalizable if and only if has n different linear independent eigenvectors. It means that matrix is symmilar to some diagonal matrix D. And matrix S is matrix that has eigenvectors from matrix A as colums. $Rightarrow A=Scdot D cdot S^{-1}$
$$A^{n}=underbrace{Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}...}\ _{n times}$$ $$A^{n}=Scdot D^{n}cdot S^{-1}$$ I think it is the easiest way to prove that matrix $space A^{n}$ is symmetric if $A$ is symmetric for some integer $n$.
answered Jan 27 at 19:09
FiggaroFiggaro
234
$endgroup$
$begingroup$
You can use the fact that $(AB)'=B'A'$ to calculate $(A^n)' = (A')^n=A^n$, knowing nothing about diagonalizability.
$endgroup$
– kimchi lover
Jan 27 at 22:17
add a comment |
$begingroup$
Well, here says that for integer $n space,A^{n}$ is symmetric if $A$ is symmetric. But if you want really to prove that, you can use formula which says that matrix is diagonalizable if and only if has n different linear independent eigenvectors. It means that matrix is symmilar to some diagonal matrix D. And matrix S is matrix that has eigenvectors from matrix A as colums. $Rightarrow A=Scdot D cdot S^{-1}$
$$A^{n}=underbrace{Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}...}\ _{n times}$$ $$A^{n}=Scdot D^{n}cdot S^{-1}$$ I think it is the easiest way to prove that matrix $space A^{n}$ is symmetric if $A$ is symmetric for some integer $n$.
answered Jan 27 at 19:09
FiggaroFiggaro
234
$endgroup$
$begingroup$
You can use the fact that $(AB)'=B'A'$ to calculate $(A^n)' = (A')^n=A^n$, knowing nothing about diagonalizability.
$endgroup$
– kimchi lover
Jan 27 at 22:17
add a comment |
$begingroup$
Well, here says that for integer $n space,A^{n}$ is symmetric if $A$ is symmetric. But if you want really to prove that, you can use formula which says that matrix is diagonalizable if and only if has n different linear independent eigenvectors. It means that matrix is symmilar to some diagonal matrix D. And matrix S is matrix that has eigenvectors from matrix A as colums. $Rightarrow A=Scdot D cdot S^{-1}$
$$A^{n}=underbrace{Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}...}\ _{n times}$$ $$A^{n}=Scdot D^{n}cdot S^{-1}$$ I think it is the easiest way to prove that matrix $space A^{n}$ is symmetric if $A$ is symmetric for some integer $n$.
answered Jan 27 at 19:09
FiggaroFiggaro
234
$endgroup$
Well, here says that for integer $n space,A^{n}$ is symmetric if $A$ is symmetric. But if you want really to prove that, you can use formula which says that matrix is diagonalizable if and only if has n different linear independent eigenvectors. It means that matrix is symmilar to some diagonal matrix D. And matrix S is matrix that has eigenvectors from matrix A as colums. $Rightarrow A=Scdot D cdot S^{-1}$
$$A^{n}=underbrace{Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}cdot Scdot D cdot S^{-1}...}\ _{n times}$$ $$A^{n}=Scdot D^{n}cdot S^{-1}$$ I think it is the easiest way to prove that matrix $space A^{n}$ is symmetric if $A$ is symmetric for some integer $n$.
answered Jan 27 at 19:09
FiggaroFiggaro
234
answered Jan 27 at 19:09
FiggaroFiggaro
234
answered Jan 27 at 19:09
FiggaroFiggaro
234
answered Jan 27 at 19:09
FiggaroFiggaro
234
234
$begingroup$
You can use the fact that $(AB)'=B'A'$ to calculate $(A^n)' = (A')^n=A^n$, knowing nothing about diagonalizability.
$endgroup$
– kimchi lover
Jan 27 at 22:17
add a comment |
$begingroup$
You can use the fact that $(AB)'=B'A'$ to calculate $(A^n)' = (A')^n=A^n$, knowing nothing about diagonalizability.
$endgroup$
– kimchi lover
Jan 27 at 22:17
$begingroup$
You can use the fact that $(AB)'=B'A'$ to calculate $(A^n)' = (A')^n=A^n$, knowing nothing about diagonalizability.
$endgroup$
– kimchi lover
Jan 27 at 22:17
$begingroup$
You can use the fact that $(AB)'=B'A'$ to calculate $(A^n)' = (A')^n=A^n$, knowing nothing about diagonalizability.
$endgroup$
– kimchi lover
Jan 27 at 22:17
add a comment |
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$begingroup$
What do you mean by "sentence about symmetric matrix"?
$endgroup$
– user635162
Jan 27 at 15:27
$begingroup$
Do you know that $(AB)' = (B') (A')$ , for matrices $A$ and $B$ in general?
$endgroup$
– kimchi lover
Jan 27 at 15:41
$begingroup$
@kimchilover I don't understand how it helps..
$endgroup$
– user3523226
Jan 27 at 17:13
$begingroup$
So what have you tried already? What do you know about this problem. Or about symmetric matrices? Can you answer the (presumably simpler) problem about $A^2$?
$endgroup$
– kimchi lover
Jan 27 at 17:19