# Linear Transformation of Matrix

1. Apr 17, 2012

### h4v0k

1. The problem statement, all variables and given/known data

Let A$_{2x2}$ have all entries=1 and let T: M$_{2x2}$$\rightarrow$M$_{2x2}$ be the linear transformation defined by T(B)=AB for all B$\in$M$_{2x2}$

Find the matrix C=[T]s,s, where S is the standard basis for M$_{2x2}$

My solution:

Standard basis for M$_{2x2}$={(1,0),(0,1)}
T(1,0)=(1,1)
T(0,1)=(1,1)
[T]s,s=(1,1;1,1)

I'm not sure how correct this is. Any advice would be appreciated.

2. Apr 17, 2012

### h4v0k

Still confused on this one

3. Apr 17, 2012

### Dick

The standard basis for M_2x2 is four matrices [[1,0],[0,0]], [[0,1],[0,0]], [[0,0],[1,0]] and [[0,0],[0,1]]. You can express any other matrix as a linear sum of those. Now take a look at the problem again.

4. Apr 17, 2012

### h4v0k

Doesn't this still produce the same vector?

5. Apr 17, 2012

### Dick

What vector? What is A times the first basis matrix?

6. Apr 17, 2012

### h4v0k

Sorry, the same matrix

A times first basis matrix is

[[1,0][0,0]]
then
[[0,0][1,0]], [[0,1][0,0]], and [[0,0][0,1]]

7. Apr 17, 2012

### Dick

[[1,1],[1,1]]*[[1,0],[0,0]] isn't equal to [[1,0],[0,0]].

8. Apr 17, 2012

### h4v0k

It appears it isn't

[[1,0][1,0]]

then

[[0,1][0,1]], [[1,0][1,0]], and [[0,1][0,1]]

9. Apr 17, 2012

### Dick

Ok, so work on what the matrix C should be. It should be 4x4 since you have four basis elements.

10. Apr 17, 2012

### h4v0k

Would this be

[[1,0,0,1],[1,0,0,1],[1,0,0,1],[1,0,0,1]]?

11. Apr 17, 2012

### Dick

It would depend on which column represents which basis element. You should spell that out. But no I don't think that's it. How did you conclude that?