# Thread: proof of similar matrices

1. ## proof of similar matrices

A is a 2x2 matrices
$\displaystyle A^2=-I$
and there is a transformation for which $\displaystyle T^2=-I$
there is a basis B={v,T(v)} for which
$\displaystyle [T]_B$=
(0 1)
(-1 0)

prove that A is similar to
(0 1)
(-1 0)
matrices.

how i tried:

i got T being represented by base B
$\displaystyle [T]_B$=
(0 1)
(-1 0)

i got T being represented by the stansdart base which is
A matrices.
what so A is similar to
(0 1)
(-1 0)
??

bacause of what?
is it correct?

2. ## Re: proof of similar matrices

Originally Posted by transgalactic
A is a 2x2 matrices
$\displaystyle A^2=-I$
and there is a transformation for which $\displaystyle T^2=-I$
there is a basis B={v,T(v)} for which
$\displaystyle [T]_B$=
(0 1)
(-1 0)

prove that A is similar to
(0 1)
(-1 0)
matrices.

how i tried:

i got T being represented by base B
$\displaystyle [T]_B$=
(0 1)
(-1 0)

i got T being represented by the stansdart base which is
A matrices.
what so A is similar to
(0 1)
(-1 0)
??

bacause of what?
is it correct?
Sadly, this question makes no sense. Is there some way you can reword it?

3. ## Re: proof of similar matrices

Originally Posted by Drexel28
Sadly, this question makes no sense. Is there some way you can reword it?
I agree. If $\displaystyle B=\{v,T(v)\}$ is a basis of $\displaystyle \mathbb{R}^{2\times 2}$ and the coordinates on $\displaystyle B$ are expressed by rows, then $\displaystyle [T]_B=\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ . So, it seems we are saying that $\displaystyle A$ is similar to $\displaystyle A$ .

4. ## Re: proof of similar matrices

Originally Posted by FernandoRevilla
I agree. If $\displaystyle B=\{v,T(v)\}$ is a basis of $\displaystyle \mathbb{R}^{2\times 2}$ and the coordinates on $\displaystyle B$ are expressed by rows, then $\displaystyle [T]_B=\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ . So, it seems we are saying that $\displaystyle A$ is similar to $\displaystyle A$ .
this question is the third part of some bigger question in which i solved the first two parts.

V is lenear space dimV=n n>1
T:V->V is a transformation for wheach $\displaystyle T^2=-I$
i prooved that {T(v),v} basis $\displaystyle [T]_B=$
(0 1)
(-1 0)

in the third part i was asked:
A is 2x2 matrices
$\displaystyle A^2=-I$
prove that A is similar to
(0 1)
(-1 0)

i am used to prove that if $\displaystyle B=P^-1AP$ then they are similar

but here in the book they some thing liketwo representation in diffeerent basis
so they are similar
i cant understand this thing
why is that?

5. ## Re: proof of similar matrices

Originally Posted by transgalactic
in the third part i was asked:
A is 2x2 matrices $\displaystyle A^2=-I$ prove that A is similar to
(0 1)
(-1 0)
That is false. Choose for example $\displaystyle A=\begin{bmatrix}{i}&{0}\\{0}&{i}\end{bmatrix}$ , we have $\displaystyle A^2=-I$. However $\displaystyle \det A=-1$ and $\displaystyle \det \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}=1$ so, the matrices are not similar.

6. ## Re: proof of similar matrices

Originally Posted by FernandoRevilla
That is false. Choose for example $\displaystyle A=\begin{bmatrix}{i}&{0}\\{0}&{i}\end{bmatrix}$ , we have $\displaystyle A^2=-I$. However $\displaystyle \det A=-1$ and $\displaystyle \det \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}=1$ so, the matrices are not similar.
But the result is true for matrices with real entries. Let T be the linear transformation whose matrix (with respect to the standard basis) is A. Then $\displaystyle T^2=-I$, so T has no real eigenvalues and hence no (real) eigenvectors. It follows that if x is any nonzero vector then the vectors x and –Tx are linearly independent. If B is the basis consisting of those two vectors then the matrix of T with respect to B is $\displaystyle \bigl[{\scriptstyle{0\atop-1}\:{1\atop0}}\bigr].$ So that matrix is similar to A.

7. ## Re: proof of similar matrices

why if $\displaystyle A^2=-I$ then $\displaystyle T^2=-I$
?

8. ## Re: proof of similar matrices

Originally Posted by Opalg
But the result is true for matrices with real entries. Let T be the linear transformation whose matrix (with respect to the standard basis) is A. Then $\displaystyle T^2=-I$, so T has no real eigenvalues and hence no (real) eigenvectors. It follows that if x is any nonzero vector then the vectors x and –Tx are linearly independent. If B is the basis consisting of those two vectors then the matrix of T with respect to B is $\displaystyle \bigl[{\scriptstyle{0\atop-1}\:{1\atop0}}\bigr].$ So that matrix is similar to A.
Of course. These things happen when the formulation of the problem is not clear. For example in the answer #2 we already commented that for a basis $\displaystyle B$ satisfying $\displaystyle B=\{v,T(v)\}$ the matrix of $\displaystyle T$ is $\displaystyle \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ (row representation) or equivalently, for $\displaystyle B'=\{v,-T(v)\}$ is $\displaystyle \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ (column representation). Now in the third part of the problem it seems we have a $\displaystyle 2\times 2$ matrix without a reference to the field $\displaystyle \mathbb{K}$ .

9. ## Re: proof of similar matrices

Originally Posted by transgalactic
why if $\displaystyle A^2=-I$ then $\displaystyle T^2=-I$ ?
If $\displaystyle A$ is the matrix of $\displaystyle T$ with respect to a determined basis $\displaystyle B$ and we write the coordinates of the vectors in columns then, by a well known result, $\displaystyle Y=AX$ where $\displaystyle X$ are the coordinates of $\displaystyle x\in V$ with respect to $\displaystyle B$ and $\displaystyle Y$ the coordinates of $\displaystyle T(x)\in V$ with respect to $\displaystyle B$ . Conclude.