# Math Help - Linear operator and standard matrix representation

1. ## Linear operator and standard matrix representation

Let L: $R^n$ -> $R^n$ be a linear operator. If A is the standard matrix representation of L, then an nxn matrix B is also a matrix representation of L.

I understand that if an operator goes from two spaces that are equal (As in this case, with 2-d going to 2-d), then a matrix representation A such that L(x) = Ax for any x in R^n must be nxn. But how do I know if all nxn matrices can be standard matrix representations?

This is the way the question was written, btw.

2. Originally Posted by Lord Darkin
Let L: $R^n$ -> $R^n$ be a linear operator. If A is the standard matrix representation of L, then an nxn matrix B is also a matrix representation of L.

The above doesn't make sense at all: it seems to be saying that ANY nxn matrix B is a (matrix, of course) representation of some given and fixed

linear map L, which of course is grossly false. Besides this, what is the relation between A and B??

What the question could have been, perhaps, is: prove that any nxn real matrix is the representation of some linear map from R^n to itself, but this is too

remote from the language used in the OP...

Tonio

I understand that if an operator goes from two spaces that are equal (As in this case, with 2-d going to 2-d), then a matrix representation A such that L(x) = Ax for any x in R^n must be nxn. But how do I know if all nxn matrices can be standard matrix representations?

This is the way the question was written, btw.
.

3. I agree, tonio. The wording in the question, which I have written verbatim here, confused me.

Please disregard the question, then. It must be some typo.

4. the only situation in which this can possibly be true, is if rank(A) = rank(B).

but, in general, ANY nxn matrix is a matrix representation of a linear transformation from R^n to R^n, if P is an nxn matrix, we have the

linear transformation P:R^n-->R^n which take v to P(v).

5. Originally Posted by Deveno
the only situation in which this can possibly be true, is if rank(A) = rank(B).

No, not even then. It is false in general that two matrices with the same rank represent the same linear map. For example, take two

matrices (of the same rank) with different eigenvalues or, even easier, with different char. polynomials.

Tonio

but, in general, ANY nxn matrix is a matrix representation of a linear transformation from R^n to R^n, if P is an nxn matrix, we have the

linear transformation P:R^n-->R^n which take v to P(v).
.

6. quite correct, tonio. i did not mean to imply rank(A) = rank(B) --> A,B represent the same linear transformation, but rather the other way around.

(was thinking along the lines of counter-examples).

7. Originally Posted by Lord Darkin
Let L: $R^n$ -> $R^n$ be a linear operator. If A is the standard matrix representation of L, then an nxn matrix B is also a matrix representation of L.
Two $n \times n$ matrices $A$ and $B$ represent the same linear operator under two different bases, if and only if they are similar matrices.
This means that $B = P^{-1} A P$ where $P$ is the matrix of change of basis.