# Eigenvalues homework

1. Apr 28, 2006

### UrbanXrisis

matrix $$A = \left(\begin{array}{ccc}3&0&0 \\ 0&3&0 \\3&0&0 \end{array}\right)$$

has two real eigenvalues lambda_1=3 of multiplicity 2, and lambda_2=0 of multiplicity 1. find the eigenspace.

$$A = \left(\begin{array}{ccc}3-3 &0&0 \\ 0&3-3&0 \\3&0&0-3 \end{array}\right)$$

$$A = \left(\begin{array}{ccc}0 &0&0 \\ 0&0&0 \\3&0&-3 \end{array}\right)$$

$$A = \left(\begin{array}{ccc}1&0&-1 \\ 0&0&0 \\0&0&0 \end{array}\right)$$

that means the eigenspace for lambda_1 is the column vector [0 1 0] and [1 0 1]

for lambda_2, there will be two different eigenspaces such that...

$$A = \left(\begin{array}{ccc}1&0&0 \\ 0&1&0 \\0&0&0 \end{array}\right)$$

I know the answer has two vectors, how do i get two vectors when I have 4 that defines the eigenspace?

2. Apr 28, 2006

### Gokul43201

Staff Emeritus
Looks correct up to here ...
How do you get that ?

3. Apr 28, 2006

### J77

Is your eigenvector for $$\lambda_3$$ not $$v_3=(0, 0, k)^T,\;k\in\mathbb{R}$$, and for $$\lambda_{1,2}$$ not $$v_{1,2}=(k_1,k_2,k_1)^T,\;k_{1,2}\in\matbb{R}$$?

4. Apr 28, 2006

### HallsofIvy

Staff Emeritus
If
$$A = \left(\begin{array}{ccc}1&0&0 \\ 0&1&0 \\0&0&0 \end{array}\right)\left(\begin{array}{c}x \\y\\ z\end{array}\right)= \left(\begin{array}{c}0 \\ 0 \\0 \end{array}\right)$$
Which says x= 0, y= 0. Looks to me like you have just a one dimensional eigen space, spanned by (0, 0, 1)T.

5. Apr 28, 2006

### UrbanXrisis

oh right, so three vectors then... [0, 0, 1], [0 1 0] and [1 0 1]

the solution gives 2 vectors, how would I find the eigenspace from 3 vectors?

6. Apr 28, 2006

### HallsofIvy

Staff Emeritus
WHAT solution gives 2 vectors and what eigenspace are you talking about?

Not all texts use that terminology but I think by "eigenspace" you mean the sub-space consisting of eigenvectors haveing the same eigenvalue. In this case, you have two distinct "eigenspaces", one having dimension 2 and the other dimension 1.

Last edited: Apr 28, 2006
7. Apr 28, 2006

### J77

Depending on the definition, I think you could also use the general eigenvectors that I gave and form your basis by sticking in ones and zeroes.

8. Apr 28, 2006

### yeahright

yeah right.

If you start with a three dimensional square hermitian matrix, you'll get a 3D eigenspace. period. If you have degenerate eigenvalues - like you do, then you get one unambiguous eigenvector (for the eigenvalue 0) and a 2D eigenspace (corresponding to the degenerate eigenvalue 1) from which you have the freedom of choosing any two linearly independent eigenvectors.

9. Apr 28, 2006

### HallsofIvy

Staff Emeritus
That's not the question I was responding to.
(And, the matrix we are talking about is not Hermitian.)

Last edited: Apr 28, 2006
10. Apr 28, 2006

### UrbanXrisis

so... there are three vectors the span the space... this is because:

$$\lambda_1=3$$

$$A = \left(\begin{array}{ccc}1&0&-1 \\ 0&0&0 \\0&0&0 \end{array}\right)$$

$$\left(\begin{array}{c}a \\b \\c \end{array}\right) = s \left(\begin{array}{c}0 \\1 \\0 \end{array}\right) + t \left(\begin{array}{c}1 \\0 \\1 \end{array}\right)$$

$$\lambda_1=0$$

$$A = \left(\begin{array}{ccc}1&0&0 \\ 0&1&0 \\0&0&0 \end{array}\right)$$

$$\left(\begin{array}{c}a \\b \\c \end{array}\right) = u \left(\begin{array}{c}0 \\0 \\1 \end{array}\right)$$

so, there are three linearly independent vetors that span the eigenspace

they are: $$\left(\begin{array}{c}0 \\0 \\1 \end{array}\right) , \left(\begin{array}{c}0 \\1 \\0 \end{array}\right) , \left(\begin{array}{c}1 \\0 \\1 \end{array}\right)$$

however, my solution is telling me that I need two vectors instead of three such that:
$$\left(\begin{array}{c}a \\b \\c \end{array}\right) , \left(\begin{array}{c}d \\e \\f \end{array}\right)$$

Last edited: Apr 28, 2006
11. Apr 28, 2006

### HallsofIvy

Staff Emeritus
Then what was the question? Yes, the three vectors span the space- it is, after all, 3 dimensional.

"my solution is telling me that I need two vectors". Your solution? What solution are you talking about? Your matrix has two distinct eigenvalues, 3 and 0. The eigenspace corresponding to 3 is two dimensional and is spanned by
$$\left(\begin{array}{c}1 \\0 \\1 \end{array}\right)$$
and
$$\left(\begin{array}{c}0 \\1 \\0 \end{array}\right)$$
The eigenspace corresponding to 0 is one dimensional and is spanned by
$$\left(\begin{array}{c}0 \\0 \\1 \end{array}\right)$$

Those three vectors, of course, span all of R3.

12. Apr 28, 2006

### UrbanXrisis

oh whoops, I was to find the eigenspace for only $$\lambda_1=3$$

so I didnt have to take into account $$\lambda_2$$