# Eigentheory of Transformations between Matrix Spaces

1. Feb 28, 2012

### TranscendArcu

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

My instructor wants me to only solve for the case m=2.

3. The attempt at a solution

So I thought I should discover what T does to the standard basis for matrices of size 2x2:

$T \left| \begin{array}{cc} 1 &0 \\ 0&0 \end{array} \right| = \left| \begin{array}{cc} 1 &0 \\ 0&0 \end{array} \right|$

$T \left| \begin{array}{cc} 0 &1 \\ 0&0 \end{array} \right| =\left| \begin{array}{cc} 0 &0 \\ 1&0 \end{array} \right|$

$T \left| \begin{array}{cc} 0 &0 \\ 1&0 \end{array} \right| =\left| \begin{array}{cc} 0 &1 \\ 0&0 \end{array} \right|$

$T \left| \begin{array}{cc} 0 &0 \\ 0&1 \end{array} \right| =\left| \begin{array}{cc} 0 &0 \\ 0&1 \end{array} \right|$

At this point do I have to create a matrix consisting of these transformed matrices in order to find $Δ_T (t) = det(M - tI)$? I'm kind of confused about my next step.

2. Feb 28, 2012

### micromass

Staff Emeritus
Yes, try to find the matrix of T with respect to your basis. This will be a 4x4 matrix.

3. Feb 28, 2012

### TranscendArcu

But if the matrix is 4x4, how will I apply it to a 2x2 matrix? The inner matrix dimensions won't agree.

4. Feb 28, 2012

### micromass

Staff Emeritus
The 2x2-matrices won't have anything to do with the matrix of T.

The entries of the matrix of T are coordinates.

For example, if we put on $M_{2,2}(\mathbb{R})$ (the 2x2-matrices) the basis you mention, then the 4x4-matrix will contain the coordinates of the image of the basis.

For example, you have to write

$$T\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right)$$

as a linear combination of the basis. So we have

$$T\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right) = 1 \left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right) + 0\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right) + 0\left(\begin{array}{cc} 0 & 0\\ 1 & 0\end{array}\right) +0\left(\begin{array}{cc} 0 & 0\\ 0 & 1\end{array}\right)$$

So the coordinates of

$$T\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right)$$

are (1,0,0,0). This will be the first column of the 4x4-matrix.

5. Feb 28, 2012

### TranscendArcu

Okay, then I have for my matrix:
$$\left| \begin{array}{cccc} 1 &0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{array} \right| = Q$$
$$det(Q- tI) = det\left| \begin{array}{cccc} 1-t &0&0&0 \\ 0&-t&1&0 \\ 0&1&-t&0 \\ 0&0&0&1-t \end{array} \right| = (1-t)(1-t)[(t)(t) - 1] = (t-1)^3 (t+1)$$

That is the characteristic polynomial, and its roots are the eigenvalues. So we have eigenvalues 1,-1. Look about right?

6. Feb 28, 2012

### micromass

Staff Emeritus
That sounds correct. Now you need to find the eigenspaces.