# Find the bases for the eigenspaces of the matrix

1. May 9, 2012

### DODGEVIPER13

1. The problem statement, all variables and given/known data
Find the bases for the eigen spaces? I handwrote the question and my work so it would be easier to read and attached it so see attachment.

2. Relevant equations

3. The attempt at a solution
My work is on the attachment. My question is this what do I do when I have 3 eigenvalues like in this case -1, 6, 6 and two are the same like how do I find for a third base it makes no sense to me?

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2. May 10, 2012

### Staff: Mentor

You have a mistake in your calculation for the eigenvectors for λ = 6. For this eigenvalue, there are two eigenvectors.

From your equations,
x1 = (3/4)x2
x2 = .......x2
x3 = ............x3

Both x2 and x3 are arbitrary. You should be able to get two eigenvectors from the above.

3. May 10, 2012

### DODGEVIPER13

Thanks man I got it but can you tell me how abritary values are chosen I mean what if I set x1 to arbitrary?

4. May 10, 2012

### Staff: Mentor

You can do that, but I prefer to use non-leading entries in each pivot row as the arbitrary values, together with any variables whose coefficients are zero.

In the completely row-reduced matrix you end up with
$$\begin{bmatrix} 1 & -3/4 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

This matrix represents this equation:
x1 - (3/4)x2 = 0

Since we have one equation but three variables, we're going to have two arbitrary variables.

The leading entry (1) is the coefficient of x1. The non-leading entry is -3/4, so I choose x2 to be arbitrary. The coefficient of x3 is zero, so I choose it to be arbitrary, too.

From the equation above, we have
x1 = (3/4)x2
x2 = x2 (arb.)
x3 = x3 (arb.)

If you line them up as I did in my previous post you can pick off the eigenvectors by sight.

$$\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = x_2\begin{bmatrix} 3/4 \\ 1 \\ 0\end{bmatrix} + x_3\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$$

5. May 10, 2012

### DODGEVIPER13

Thanks man that cleared it up for me

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