Question

Let A and B be 2x2 matrices. Determine the matrix of the operator T: M->AMB on the space \(F^{2x2}\) of 2x2 matrices, with respect to the basis \((e_{11}, e_{12}, e_{21}, e_{22})\) of \(F^{2x2}\).

Answer 1

This problem seems easier than I'm making it, I'm just getting mixed up. I know B transpose is involved, but I don't think \(B^tA\) is right.

Answer 2

\[e_{11}=\left[\begin{matrix}1&0\\0&0\end{matrix}\right]\]\[e_{12}=\left[\begin{matrix}0&1\\0&0\end{matrix}\right]\]etc.?

Answer 3

Yes, those are definitely the basis matrices, but I'm still lost as to figuring out the matrix of the operator.

Answer 4

I only know what I was taught in school, sorry.

Answer 5

I'm so confused as to why this problem is hard. It seems simple. And it's question 1 of the section of the book >.<

Answer 6

of which book?

Answer 7

Title, Author, and Edition Please

Answer 8

Artin's Algebra, second edition. Question 4.2.1

Answer 9

not sure I understand the question. Is it asking for the result of this multiplication?\[\left[\begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{matrix}\right]\left[\begin{matrix}e_{11}&e_{12}\\e_{21}&e_{22}\end{matrix}\right]\left[\begin{matrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{matrix}\right]\]

Answer 10

Some background for the question. We're talking about linear operators, which are linear transformations from a vector space back to itself. It is a theorem that any linear operator can be represented as left multiplication by a matrix. The question is to find the matrix for this given operator.

Answer 11

Can't find an online version of the book anywhere

Answer 12

but the operator just says take some matrix M and transform it by applying AMB - correct?

Answer 13

Nevermind, I found it:

http://www.coursesmart.com/algebra-second-edition/michael-artin/dp/9780321746580

Answer 14

asna: Correct. But we need to find a matrix for that operator that transforms it by left multiplication only. So, we need to find what matrix, call it P, such that PM=AMB. Now that I've worded it that way... maybe I can just manipulate that equivalence to figure it out? Will try it out.

Answer 15

ok - so yes - if you carry out the matrix multiplications then PM and AMB will give you 2x2 matrices. so you then just have to compare corresponding elements to work out what P should be.

Answer 16

I tried that, but it gets really messy and I can't figure it out =/

Answer 17

In AMB you end up with, for example, the 1,1 entry has all four bases in it. So I don't know how you make one matrix that will lead to that.

Answer 18

If we view B as a column operation matrix, then we can say that B transpose would be an equivalent row operation matrix, which is why I was thinking that B transpose was involved. But, I'm not sure quite how.

Answer 19

one second, if we write this out as matrix manipulations, then we get:
\[\begin{align}
PM &= AMB\\
P^{-1}PM&=P^{-1}AMB\\
M&=P^{-1}AMB\\
\end{align}\]which implies:\[P=A^{-1}\]and\[B=I\text{ (the identity matrix)}\]

Answer 20

sorry I meant P=A

Answer 21

I don't think that it implies that. There are other ways for the equivalence to hold.

Answer 22

ok - let me try the messy algebraic way...

Answer 23

Proof that that implication doesn't hold: Look at conjugation by an elementary matrix \(E\). We can have \(A=E^{-1}AE\) without it being the case that \(E=I\). For example, any permutation matrix.

Answer 24

true

Answer 25

hmmm... algebraic method is not leading anywhere sensible - there must be something obvious we are missing in the question.

Answer 26

Right?? That's what I keep thinking, but I don't see it.

Answer 27

let me mull over for a while...