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The matrix $A = \begin{bmatrix}9&-1\\1&7\end{bmatrix}$ has one eigenvalue of multiplicity 2. Find this eigenvalue and the dimension of the eigenspace.

So I found the eigenvalue by doing $A - \lambda I$ to get:

$\lambda = 8$

But how exactly do I find the dimension of the eigenspace?

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  • $\begingroup$ Just FYI for a quicker determination... $detA=64$. The eigenvalues have this product. If the eigenvalues are equal, then they must be $\pm \sqrt{64}$ $\endgroup$
    – David P
    Dec 3, 2021 at 23:27

2 Answers 2

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The dimension of the eigenspace is given by the dimension of the nullspace of $A - 8I = \left(\begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right)$, which one can row reduce to $\left(\begin{matrix} 1 & -1 \\ 0 & 0 \end{matrix} \right)$, so the dimension is $1$.

Note that the number of pivots in this matrix counts the rank of $A-8I$. Thinking of $A-8I$ as a linear operator from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$, the dimension of the nullspace of $A-8I$ is given by $\dim(\mathbb{R}^{2}) - \mathrm{rank}(A-8I) = 2 - 1 = 1$ by the so-called rank-nullity theorem. Of course, one can be more explicit: it is straightforward to see that the nullspace of $A - 8I$ is spanned by the vector $(1, 1)$, and hence has dimension $1$.

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  • $\begingroup$ I understand the RREF, and how you got the matrix that tells you the dimension, but how exactly do you determine that the dimension is 1? Is it just because there is only one pivot column? $\endgroup$
    – Shammy
    Jul 15, 2016 at 0:03
  • $\begingroup$ @Shammy: I've updated my answer. Does this help? $\endgroup$ Jul 15, 2016 at 0:19
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By definition, an eigenvector $v$ with eigenvalue $\lambda$ satisfies $Av = \lambda v$, so we have $Av-\lambda v =Av - \lambda I v = 0$, where $I$ is the identity matrix. Thus, $$(A-\lambda I)v = 0,$$ and $v$ is in the nullspace of $A-\lambda I$.

Since the eigenvalue in your example is $\lambda = 8$, to find the eigenspace related to this eigenvalue we need to find the nullspace of $A - 8I$, which is the matrix $$\left[\begin{array}{cc}1 & -1 \\ 1 & -1 \\ \end{array} \right].$$ We can row-reduce it to obtain $$\left[\begin{array}{cc} 1 & -1 \\ 0 & 0 \\ \end{array} \right].$$ This corresponds to the equation $$x-y = 0,$$ so $x = y$ for every eigenvector associated to the eigenvalue $\lambda = 8$. Therefore, if $(x,y)$ is an eigenvector, then $(x,y) = (x,x) = x(1,1)$, meaning that the eigenspace is $$W=[(1,1)],$$ and its dimension is $1$.

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