Find the bases for the eigenspaces of the matrix

[DODGEVIPER13]

Homework Statement

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.

Homework Equations

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?

 

[Mark44]

Homework Statement

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.

Homework Equations

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?

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.

 

[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?

 

[Mark44]

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.

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

 

[DODGEVIPER13]

Thanks man that cleared it up for me