# How can I find the dimension of the eigenspace?

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?

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

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 $$1$$.
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 $1$.