# Eigenvalues, eigenvectors, and eigenspaces

1. Feb 27, 2012

### TranscendArcu

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

3. The attempt at a solution
T(1,0,0) = (3,-1,0)
T(0,1,0) = (0,1,0)
T(0,0,1) = (-1,2,4)

Thus, we have the matrix,

$\left| \begin{array}{ccc} 3 &0&-1 \\ -1&1&2 \\ 0&0&4 \end{array} \right|$

$Δ_T (t) = det( \left| \begin{array}{ccc} 3 &0&-1 \\ -1&1&2 \\ 0&0&4 \end{array} \right| - tI)$

I have this equaling: -(t-4)(t-3)(t-1), which is the characteristic polynomial. The roots are the eigenvalues, which are 4,3,1.

To compute the eigenvectors:

When t=4, we have,
-x-z=0
-x-3y+2z=0
0z=0

Which implies that eigenvectors are multiples of (-1,1,1).

When t=3, we have,
-z=0
-x-2y=0
z=0

Which implies that eigenvectors are multiples of (-1,2,0)

When t=1, we have,
2x-z=0
-x+2z=0
3z=0

Which implies that eigenvectors are multiples of (0,1,0).

T is diagonizable because (-1,1,1),(-1,2,0),(0,1,0) are lin. indep.

First of all, are these correct? Also, how does one determine eigenspaces? That's the part I feel that I really don't know.

2. Feb 27, 2012

### lanedance

to see if they're correct, test by multiplying your eignevectors by the matrix - do they behave as expected?

3. Feb 27, 2012

### lanedance

as for the eigenspaces, they will be the span of all eignevectors corresponding to a single eigenvalue

what does the span of a single vector look like?

4. Feb 27, 2012

### fauboca

Eigenspace are the eigenvectors. I obtained a different eigenvector for you second one. I don't believe I made a mistake but I could have.

5. Feb 27, 2012

### TranscendArcu

Okay. I have the equations

-z=0
-x-2y=0
z=0

So clearly z=0 and x=-2y. Thus, is the eigenvector multiples of (1,1/-2,0)? Or, equivalently, multiples of (-2,1,0)?

So the span of a single vector is just a line, no? So, for example, considering the eigenvalue of 4, is the eigenspace just: span(-1,1,1)?

For eigenvalue of 1: span(0,1,0)?

If by behave as expected you mean give back the zero vector, then yes (after I made an adjustment to the second eigenvector calculation). If this is the expected result, why do I expect this matrix multiplication to give back a column of all zeros?

6. Feb 27, 2012

### lanedance

The line is correct, you coud parameterise to describe it if need be

Are you multiplying by $T$ or $(T - \mathbb{I} \lambda)$?

Say your eignevector is $u$ multply by $T$ you should get the original vector multiplied by the corresponding eigenvalue $Tu =\lambda_u u$