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I'm studying for my linear exam and would appreciate any help for this practise question:

You are given that λ = 1 is an eigenvalue of A. What is the dimension of the corresponding eigenspace?

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

Then with my knowing that λ = 1, I got:

$\begin{bmatrix} 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}$

Which I assume right off the bat means my dimension is 0. Is that correct? If not how should I do it?

If we had a different matrix, how would I go ahead to properly find the dimension? In layman terms I think that it would be whichever value is linearly independent?

Thanks for the help.

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  • $\begingroup$ Can you recall the definition of an eigenspace? Then just apply that to your situation (and solve the equation). $\endgroup$ – Marc van Leeuwen Apr 16 '13 at 5:25
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    $\begingroup$ No, the dimension of the eigenspace is the dimension of the null space of the matrix $A - \lambda I$ (the second matrix you mentioned). Note that you have two free variables, $x_2$ and $x_3$, and so the dimension is two. $\endgroup$ – Suugaku Apr 16 '13 at 5:32
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The dimension is two. Note that the vectors $ u=\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ \end{array} \right] $ and $v= \left[ \begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ \end{array} \right] $ are in the null space of $A-I_4=\begin{bmatrix} 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}$, i.e. $$Au=u\mbox{ and } Av=v.$$ So $u$ and $v$ are eigenvectors corresponding to the eigenvalue $1$. In fact, the form a basis for the null space of $A-I_4$. Therefore, the eigenspace for $1$ is spanned by $u$ and $v$, and its dimension is two.

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  • $\begingroup$ Thank you for the explanation. $\endgroup$ – Abushawish Apr 16 '13 at 16:38
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In general, the eigenspace of an eigenvalue $\lambda$ is the set of all vectors $v$ such that $Av=\lambda v$. This also means $Av-\lambda v=0$, or $(A-\lambda I)v=0$. Hence, you can just calculate the kernel of $A-\lambda I$ to find the eigenspace of $\lambda$.

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