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Help with this generalised eigenvector

I was wondering if anyone could explain, from the following information that we put z=0, when I have always, when given the information that follows thought we could let z equal anything and so let z = 1.

We have, with:

\lambda = 1

A = \begin{bmatrix} 0 & 0 & 0 \\1 & 2 & 0 \\0 & 1 & 0 \end{bmatrix}

v_1 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

(A - \lambda I)v_2 = v_1

\begin{bmatrix} 0 & 0 & 0 \\1 & 2 & 0 \\0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

So we can infer:

y = 1

x + 2y= 0 \Rightarrow x= -2

We also have:

(A - \lambda I)^2v_2 = 0

\begin{bmatrix} 0 & 0 & 0 \\2 & 4 & 0 \\1 & 2 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

So we still have the same equations as before.

0x + 0y + 0z = 0

x + 2y + 0z = 0

0x + y + 0z = 0

I thought in a case like this that z could take any value so we let z = 1. But z = 0, can anyone explain why? I think I should point out that v_1 and v_2 need to be linearly independent, so if putting z = 0 makes them linearly independent, could someone explain why.

v_2 = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}

Thanks guys.

(edited 13 years ago)

Reply 1

Mark13

Original post by TheBhramaBull

Any value of z will give you the property you want for

v_2

. However, letting z=0 makes

v_2

orthogonal to

v_1

, so it may be preferable to let z=0.

In reference to your linear independence bit at the end - do you know the definition of linear independence? If you can visualise how a set of linearly independent vectors in

\mathbb{R}^3

looks, then it should be fairly easy to see when

v_1

and

v_2

are linearly independent.

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