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?
Answer
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 $A$ is given by $\dim(\mathbb{R}^{2}) - \mathrm{rank}(A) = 2 - 1 = 1$ 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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