# Determine Dimensions of Eigenspaces From Characteristic Polynomial of Diagonalizable Matrix

## Problem 384

Let $A$ be an $n\times n$ matrix with the characteristic polynomial
$p(t)=t^3(t-1)^2(t-2)^5(t+2)^4.$ Assume that the matrix $A$ is diagonalizable.

(a) Find the size of the matrix $A$.

(b) Find the dimension of the eigenspace $E_2$ corresponding to the eigenvalue $\lambda=2$.

(c) Find the nullity of $A$.

(The Ohio State University, Linear Algebra Final Exam Problem)

## Hint/Definition.

• Recall that when a matrix is diagonalizable, the algebraic multiplicity of each eigenvalue is the same as the geometric multiplicity.
• The geometric multiplicity of an eigenvalue $\lambda$ is the dimension of the eigenspace $E_{\lambda}=\calN(A-\lambda I)$ corresponding to $\lambda$.
• The nullity of $A$ is the dimension of the null space $\calN(A)$ of $A$.

## Solution.

### (a) Find the size of the matrix $A$.

In general, if $A$ is an $n\times n$ matrix, then its characteristic polynomials has degree $n$.
Since the degree of $p(t)$ is $14$, the size of $A$ is $14 \times 14$.

### (b) Find the dimension of the eigenspace $E_2$ corresponding to the eigenvalue $\lambda=2$.

Note that the dimension of the eigenspace $E_2$ is the geometric multiplicity of the eigenvalue $\lambda=2$ by definition.

From the characteristic polynomial $p(t)$, we see that $\lambda=2$ is an eigenvalue of $A$ with algebraic multiplicity $5$.
Since $A$ is diagonalizable, the algebraic multiplicity of each eigenvalue is the same as the geometric multiplicity.

It follows that the geometric multiplicity of $\lambda=2$ is $5$, hence the dimension of the eigenspace $E_2$ is $5$.

### (c) Find the nullity of $A$.

We first observe that $\lambda=0$ is an eigenvalue of $A$ with algebraic multiplicity $3$ from the characteristic polynomial.

By definition, the nullity of $A$ is the dimension of the null space $\calN(A)$, and furthermore the null space $\calN(A)$ is the eigenspace $E_0$.
Thus, the nullity of $A$ is the same as the geometric multiplicity of the eigenvalue $\lambda=0$.

Since $A$ is diagonalizable, the algebraic and geometric multiplicities are the same. Hence the nullity of $A$ is $3$.

Let $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 &0 &1 \\ 0 & 0 & 1 \end{bmatrix}$ be a...