Thread: 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?

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?

Originally Posted by Drexel28

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

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 .

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?

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.

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.

why if then
?

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 , 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.

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 satisfying the matrix of is (row representation) or equivalently, for is (column representation). Now in the third part of the problem it seems we have a matrix without a reference to the field .

Originally Posted by transgalactic

why if then ?

If is the matrix of with respect to a determined basis and we write the coordinates of the vectors in columns then, by a well known result, where are the coordinates of with respect to and the coordinates of with respect to . Conclude.