Given matrix $A = \begin{bmatrix} -3 & 2 \\ 0 & -3 \end{bmatrix}$ The eigenvalues of $A$ are $\lambda_{1,2} = -3$ with algebraic multiplicity 2. Now to find the eigenvector, we use Gaussian elimination as follows
$$ \quad (A-\lambda I | 0) \Rightarrow \left( \begin{array}{cc|c} 0 & 2 & 0 \\ 0 & 0 & 0 \end{array} \right)\quad \overset{\implies}{R_2/2} \quad \left(\begin{array}{cc|c} 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right) $$ Then we know $x_2=0$, and the eigenvector corresponding to $\lambda_{1,2}$ is $v_1=[ 1 \quad 0]^{T}$.
My questions are:
- Is $x_1$ what we call a free variable? Since it has no specific value. I'm confused because there isn't a leading $1$ in the first row.
- What is the geometric multiplicity? Is it one?
- Can I diagonalize $A$? I have two eigenvalues and one eigenvector but can I say $\lambda_{1} = -3$ with $v_1=[ 1 \quad 0]^{T}$ and $\lambda_{2} = -3$ with $v_2=[ 1 \quad 0]^{T}$? or I have to have two different eigenvectors for the repeated eigenvector?