# How can I find the dimension of the eigenspace? The matrix \[A= \b

How can I find the dimension of the eigenspace?
The matrix $A=\left[\begin{array}{cc}9& -1\\ 1& 7\end{array}\right]$ has one eigenvalue of multiplicity 2. Find this eigenvalue and the dimension of the eigenspace.
So we found the eigenvalue by doing $A-\lambda I$ to get:
$\lambda =8$
But how exactly do we find the dimension of the eigenspace?
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sonSnubsreose6v
The dimension of the eigenspace is given by the dimension of the nullspace of $A-8I=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)$ , which one can row reduce to $\left(\begin{array}{cc}1& -1\\ 0& 0\end{array}\right)$, 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 ${R}^{2}\to {R}^{2}$, the dimension of the nullspace of A is given by dim$\left({R}^{2}\right)$ −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.
###### Not exactly what you’re looking for?
Wendy Boykin
By definition, an eigenvector v with eigenvalue $\lambda$ satisfies $Av=\lambda v$, so we have $Av-\lambda v=Av-\lambda Iv=0$, where I is the identity matrix. Thus, $\left(A-\lambda I\right)v=0$, 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.