Given the Characteristic Polynomial of a Diagonalizable Matrix, Find the Size of the Matrix, Dimension of Eigenspace

Stanford University Linear Algebra Exam Problems and Solutions

Problem 39

Suppose that $A$ is a diagonalizable matrix with characteristic polynomial
\[f_A(\lambda)=\lambda^2(\lambda-3)(\lambda+2)^3(\lambda-4)^3.\]

(a) Find the size of the matrix $A$.

(b) Find the dimension of $E_4$, the eigenspace corresponding to the eigenvalue $\lambda=4$.

(c) Find the dimension of the kernel(nullspace) of $A$.

(Stanford University Linear Algebra Exam)

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Hint.

If a matrix is diagonalizable, then the algebraic multiplicity of an eigenvalue is the same as the geometric multiplicity of the eigenvalue.

Recall that the geometric multiplicity of an eigenvalue is the dimension of the eigenspace of the eigenvalue.

Solution.

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

If $A$ is an $n\times n$ matrix, then its characteristic polynomial is of degree $n$. Since the degree of $f_A$ is $9$, the size of $A$ is $9 \times 9$.

(b) Find the dimension of $E_4$, the eigenspace corresponding to the eigenvalue $\lambda=4$.

The algebraic multiplicity of the eigenvalue $\lambda =4$ is $3$. (This is the number of times that the factor $\lambda -4$ appears in $f_A$.)
Since $A$ is diagonalizable, the algebraic multiplicity is the same as the geometric multiplicity for each eigenvalue.
The geometric multiplicity is the dimension of the eigenspace by definition. Thus the dimension of $E_4$ is $3$.

(c) Find the dimension of the kernel(nullspace) of $A$.

Note that $\ker(A)=\{x\in \R^9 \mid Ax=0\}=\{x\in \R^9 \mid (A-0I_9)x=0\}=E_0$. Thus the kernel of $A$ is the same as the eigenspace $E_0$ corresponding to the eigenvalue $\lambda=0$.
Since $A$ is diagonalizable, the geometric multiplicity of $\lambda=0$, which is the dimension of the eigenspace $E_0=\ker(A)$, is the same as the algebraic multiplicity of $\lambda=0$.
Thus the dimension of $\ker(A)$ is $2$.


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