proof of similar matrices

Printable View

June 29th 2011, 11:00 AM
transgalactic

proof of similar matrices

A is a 2x2 matrices
$A^2=-I$
and there is a transformation for which $T^2=-I$
there is a basis B={v,T(v)} for which
$[T]_B$ =
(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
$[T]_B$ =
(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?
June 29th 2011, 11:39 AM
Drexel28

Re: proof of similar matrices

Quote:

Originally Posted by transgalactic

A is a 2x2 matrices
$A^2=-I$
and there is a transformation for which $T^2=-I$
there is a basis B={v,T(v)} for which
$[T]_B$ =
(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
$[T]_B$ =
(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?
June 29th 2011, 12:41 PM
FernandoRevilla

Re: proof of similar matrices

Quote:

Originally Posted by Drexel28

Sadly, this question makes no sense. Is there some way you can reword it?

I agree. If $B=\{v,T(v)\}$ is a basis of $\mathbb{R}^{2\times 2}$ and the coordinates on $B$ are expressed by rows, then $[T]_B=\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ . So, it seems we are saying that $A$ is similar to $A$ .
June 29th 2011, 02:16 PM
transgalactic

Re: proof of similar matrices

Quote:

Originally Posted by FernandoRevilla

I agree. If $B=\{v,T(v)\}$ is a basis of $\mathbb{R}^{2\times 2}$ and the coordinates on $B$ are expressed by rows, then $[T]_B=\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ . So, it seems we are saying that $A$ is similar to $A$ .

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 $T^2=-I$
i prooved that {T(v),v} basis $[T]_B=$
(0 1)
(-1 0)

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

i am used to prove that if $B=P^-1AP$ 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?
June 30th 2011, 03:24 AM
FernandoRevilla

Re: proof of similar matrices

Quote:

Originally Posted by transgalactic

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

That is false. Choose for example $A=\begin{bmatrix}{i}&{0}\\{0}&{i}\end{bmatrix}$ , we have $A^2=-I$ . However $\det A=-1$ and $\det \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}=1$ so, the matrices are not similar.
June 30th 2011, 12:26 PM
Opalg

Re: proof of similar matrices

Quote:

Originally Posted by FernandoRevilla

That is false. Choose for example $A=\begin{bmatrix}{i}&{0}\\{0}&{i}\end{bmatrix}$ , we have $A^2=-I$ . However $\det A=-1$ and $\det \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}=1$ 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 $T^2=-I$ , 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 $\bigl[{\scriptstyle{0\atop-1}\:{1\atop0}}\bigr].$ So that matrix is similar to A.
June 30th 2011, 01:30 PM
transgalactic

Re: proof of similar matrices

why if $A^2=-I$ then $T^2=-I$
?
June 30th 2011, 07:44 PM
FernandoRevilla

Re: proof of similar matrices

Quote:

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 $T^2=-I$ , 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 $\bigl[{\scriptstyle{0\atop-1}\:{1\atop0}}\bigr].$ 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 $B$ satisfying $B=\{v,T(v)\}$ the matrix of $T$ is $\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ (row representation) or equivalently, for $B'=\{v,-T(v)\}$ is $\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ (column representation). Now in the third part of the problem it seems we have a $2\times 2$ matrix without a reference to the field $\mathbb{K}$ .
June 30th 2011, 07:52 PM
FernandoRevilla

Re: proof of similar matrices

Quote:

Originally Posted by transgalactic

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

If $A$ is the matrix of $T$ with respect to a determined basis $B$ and we write the coordinates of the vectors in columns then, by a well known result, $Y=AX$ where $X$ are the coordinates of $x\in V$ with respect to $B$ and $Y$ the coordinates of $T(x)\in V$ with respect to $B$ . Conclude.