Finding the third eigenvector of a diagonalizing matrix.

by philip   Last Updated August 10, 2018 11:20 AM

Prove that matrix $A$ is diagonalizable, find the bases for the eigenspaces, the diagonalizing matrix $P$, and compute $P^{-1} A P$ where $A= \left(\begin{array}{ccc} 2 & 0 & 3 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right).$

Then finding the eigenvectors, we look at the null space of $A-I\lambda$ for the two eigenvalues.

For $\lambda=2$, we find the nullspace of $\begin{bmatrix}0&0&3\\0&1&0\\0&0&1 \end{bmatrix}$, which eyeballing we can see is the vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$

or $\lambda=3$, we find the nullspace of $\begin{bmatrix}-1&0&3\\0&0&0\\0&0&0 \end{bmatrix}$ which we can eyeball one easily as $\begin{bmatrix}0\\1\\0\end{bmatrix}$.

Then how to find the third eigenvector?



Answers 2


By eyeballing, we can see that the nullspace of $\begin{bmatrix}-1&0&3\\0&0&0\\0&0&0 \end{bmatrix}$ also contains $\begin{bmatrix}3\\0\\1\end{bmatrix}$.

Kenny Lau
Kenny Lau
August 10, 2018 10:56 AM

Your eigenspace for $\lambda=3$ is two dimensional: Solving(not eyeballing)

$$\begin{pmatrix}-1 &0 &3\\0 &0 &0\\0 &0 &0\end{pmatrix}x=\mathbf0$$

yields the following parametric solution:

$$x=c(3,0,1)^\top+c'(0,1,0)^\top$$

Thus, you have two linearly independent eigenvectors(spanning the corresponding eigenspace) corresponding to the eigenvalue $\lambda=3$ and one eigenvector(spanning the corresponding eigenspace) corresponding to $\lambda=2$. These now form a basis s.t. your matrix is diagonalizable.

zzuussee
zzuussee
August 10, 2018 10:58 AM

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