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 $2$.
Two Matrices with the Same Characteristic Polynomial. Diagonalize if Possible.
Let
\[A=\begin{bmatrix}
1 & 3 & 3 \\
-3 &-5 &-3 \\
3 & 3 & 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
2 & 4 & 3 \\
-4 &-6 &-3 \\
3 & 3 & 1
\end{bmatrix}.\]
For this problem, you may use the fact that both matrices have the same characteristic […]
How to Diagonalize a Matrix. Step by Step Explanation.
In this post, we explain how to diagonalize a matrix if it is diagonalizable.
As an example, we solve the following problem.
Diagonalize the matrix
\[A=\begin{bmatrix}
4 & -3 & -3 \\
3 &-2 &-3 \\
-1 & 1 & 2
\end{bmatrix}\]
by finding a nonsingular […]
Maximize the Dimension of the Null Space of $A-aI$
Let
\[ A=\begin{bmatrix}
5 & 2 & -1 \\
2 &2 &2 \\
-1 & 2 & 5
\end{bmatrix}.\]
Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.
Your score of this problem is equal to that […]
Determine the Values of $a$ such that the 2 by 2 Matrix is Diagonalizable
Let
\[A=\begin{bmatrix}
1-a & a\\
-a& 1+a
\end{bmatrix}\]
be a $2\times 2$ matrix, where $a$ is a complex number.
Determine the values of $a$ such that the matrix $A$ is diagonalizable.
(Nagoya University, Linear Algebra Exam Problem)
Proof.
To […]