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# Diagonalizable or not

## Problem

Let

$$
A = \begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 1 & 0 \\
0 & 0 & 0 & 2 & 1 \\
0 & 0 & 0 & 0 & 2 \\
\end{bmatrix}.
$$

For each distinct eigenvalue of $A$, find its algebraic multiplicity and its geometric multiplicity. Then determine if $A$ is diagonalizable or not.

## Thought

For any $n\times n$ matrix, its characteristic polynomial can be factored as

$$
p_A(x) = (-1)^n(x - \lambda_1)^{m_1}(x - \lambda_2)^{m_2}\cdots(x - \lambda_q)^{m_q}.
$$

Recall that for the eigenvalue $\lambda_i$, $\am_A(\lambda_i) = m_i$ is the algebraic multiplicity and $\gm_A(\lambda_i) = \nul(A - \lambda_iI)$ is the geometric multiplicity. Intuitively, $\am_A(\lambda_i)$ is the number of eigenvalues that equal to $\lambda_i$, while $\gm_A(\lambda_i)$ is the number of independent eigenvectors that we can find.

It is known that $\gm_A(\lambda_i) \leq \am_A(\lambda_i)$ for any eigenvalue $\lambda_i$. If $\gm_A(\lambda_i) < \am_A(\lambda_i)$ for some $\lambda_i$, then it means we cannot find enough independent eigenvectors for $\lambda_i$, indicating that $A$ not diagonalizable.

## Sample answer

By direct computation, the eigenvalues of $A$ are $\{1,1,2,2,2\}$. Thus, we have $\am_A(1) = 2$ and $\am_A(2) = 3$.

Again by direct computation, we have

$$
A - I = \begin{bmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 \\
\end{bmatrix} \text{ and }
A - 2I = \begin{bmatrix}
-1 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}.
$$

Through row operations, we know $\gm_A(1) = \nul(A - I) = 2$ and $\gm_A(2) = \nul(A - 2I) = 1$.

Since $1 = \gm_A(2) < \am_A(2) = 3$, the matrix $A$ does not have enough independent eigenvectors for the eigenvalue $2$, indicating that $A$ is not diagonalizable.

*This note can be found at Course website > Learning resources.*