proof of similar matrices

A is a 2x2 matrices

and there is a transformation for which

there is a basis B={v,T(v)} for which

=

(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

=

(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?

Re: proof of similar matrices

Quote:

Originally Posted by

**transgalactic** A is a 2x2 matrices

and there is a transformation for which

there is a basis B={v,T(v)} for which

=

(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

=

(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?

Re: proof of similar matrices

Re: proof of similar matrices

Quote:

Originally Posted by

**FernandoRevilla** I agree. If

is a basis of

and the coordinates on

are expressed by rows, then

. So, it seems we are saying that

is similar to

.

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

i prooved that {T(v),v} basis

(0 1)

(-1 0)

**in the third part i was asked:**

A is 2x2 matrices

prove that A is similar to

(0 1)

(-1 0)

i am used to prove that if 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?

Re: proof of similar matrices

Quote:

Originally Posted by

**transgalactic** **in the third part i was asked:**
A is 2x2 matrices

prove that A is similar to

(0 1)

(-1 0)

That is false. Choose for example , we have . However and so, the matrices are not similar.

Re: proof of similar matrices

Quote:

Originally Posted by

**FernandoRevilla** That is false. Choose for example

, we have

. However

and

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 , 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 So that matrix is similar to A.

Re: proof of similar matrices

why if then

?

Re: proof of similar matrices

Re: proof of similar matrices