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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?
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    $\begingroup$ You can only diagonalize if the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity. If you only found one eigenvector for a repeated eigenvalue, you won't be able to diagonalize. $\endgroup$ – Adrian Keister Sep 23 '18 at 0:38
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Here's some helpful things to remember:

1.)An $n \times n$ matrix will be diagonalizable if and only if it has $n$ linearly independent eigenvectors.

2.) Because eigenvectors corresponding to distinct eigenvalues are always linearly independent, a matrix with $n$ distinct eigenvalues will always be diagonalizable. This is not a necessary condition though, as sometimes an eigenvalue will have more than one linearly independent eigenvector associated to it.

3.) In response to your question regarding geometric multiplicity, the geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors associated to it. Equivalently, it the dimension of the eigenspace of that eigenvector.

4.) $x_1$ is free above $($for the system $(A-\lambda I)x = 0)$. The first row effectively says $0x_1 + 1x_2 =0$

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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.

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