After browsing through similar posts, I was wondering if I am understanding the meaning of "$n$ distinct eigenvalues" for the following theorem.
If the $n\times n$ matrix $A$ has $n$ distinct eigenvalues, then the corresponding eigenvectors are linearly independent and $A$ is diagonalizable.
$A = \begin{bmatrix}
3 & 2 & 1 \\
0 & 0 & 2 \\
0 & 2 & 0 \\
\end{bmatrix}$
$\lambda I – A = 0 =
\begin{bmatrix}
\lambda – 3 & -2 & -1 \\
0 & \lambda – 0 & -2 \\
0 & -2 & \lambda – 0 \\
\end{bmatrix} =
(\lambda – 3)\begin{vmatrix}
\lambda & -2 \\
-2 & \lambda \\
\end{vmatrix}
=
(\lambda – 3) (\lambda ^ 2 – 4 )
$
I get $\lambda = 3, 2, -2$. I tried to find an eigenvector with $\lambda = 3$ and got a weird looking matrix that doesn't look linearly independent. Despite this weird matrix, I rearranged the rows via column operations and managed to find an eigenvector. It appears the matrix is diagonalizable? I am not sure I handled $\lambda = 3$ correctly though.
$3I – A
=
\begin{bmatrix}
0 & -2 & -1 \\
0 & 3 & -2 \\
0 & -2 & 3 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
=
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
0x_2 + 0x_3 \\
0x_3 \\
x_3 \\
\end{bmatrix}
=
x_3 \begin{bmatrix}
0 \\
0 \\
1 \\
\end{bmatrix}$
$2I – A
=
\begin{bmatrix}
-1 & -2 & -1 \\
0 & 2 & -2 \\
0 & -2 & 2 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 0 & 3 \\
0 & 1 & -1 \\
0 & 0 & 0 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
3_x3 \\
-x_3 \\
x_3 \\
\end{bmatrix}
=
x_3 \begin{bmatrix}
-3 \\
1 \\
1 \\
\end{bmatrix}$
$-2I – A
=
\begin{bmatrix}
-5 & -2 & -1 \\
0 & -2 & -2 \\
0 & -2 & -2 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 0 & -.2 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}
=
\begin{bmatrix}
-.2x_3 \\
x_3 \\
x_3 \\
\end{bmatrix}
=
x_3 \begin{bmatrix}
.2 \\
-1 \\
1 \\
\end{bmatrix}$
$P =
\begin{bmatrix}
0 & -3 & .2 \\
0 & 1 & -1 \\
1 & 1 & 1 \\
\end{bmatrix}$,
$P^{-1}AP =
\begin{bmatrix}
3 & 0 & 1.4e^{-13} \\
-.5 & 2 & 0 \\
-2.5 & 0 & -2 \\
\end{bmatrix}$
Best Answer
You have three eigenvalues; $\lambda^2-4$ has roots of both $2$ and $-2$.
Additionally, it is not the matrices that will be independent, but the eigenvectors. That "weird" matrix is going to tell you what the eigenvector associated with $\lambda=3$ is, as soon as you find its nullspace.