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?
Best Answer
For diagonalization you need two linearly independent eigenvectors. In your problem you only have one eigenvector thus your matrix is not diagonalizable.
The free component in the eigenvector is common because any non zero constant multiple of an eigenvector is an eigenvector with the same eigenvalue.