Finding a Jordan canonical form $J$ and an invertible matrix $Q$ such that $J = Q^{-1}AQ$

Question

$$ A = \begin{bmatrix} 0 & 1 & -1 \\ -4 & 4 & -2 \\ -2 & 1 & 1 \end{bmatrix} $$

The characteristic polynomial can be found to be: $p(t)= -(t-1)(t-2)^2$.

I found $E_{\lambda_1} = N(A - I) = (1, 2, 1)$.

I'm having problems finding $E_{\lambda_2}$: $$ (A - 2I) = \begin{bmatrix} -2 & 1 & -1 \\ -4 & 2 & -2 \\ -2 & 1 & -1 \end{bmatrix} $$ and $$ (A - 2I)^2 = \begin{bmatrix} 2 & -1 & 1 \\ 4 & -2 & 2 \\ 2 & -1 & 1 \end{bmatrix} $$

But there are numerous $\beta$'s that can be found for $(A - 2I)$, such as $\beta = (1, 1, -1)$ or $\beta = (1, 0, -2)$, etc., etc.

Then for $(A - 2I)^2$, these same $\beta$'s can be used, so I already know that I have messed up somewhere. Can anyone offer some suggestions?

Answer 1

We asked to find the Jordan Normal Form of the matrix: $$ A = \begin{bmatrix} 0 & 1 & -1 \\ -4 & 4 & -2 \\ -2 & 1 & 1 \end{bmatrix} $$

We start out by solving $|A - \lambda I| = 0$ to find the eigenvalues, which gives:

$$-(\lambda - 2)^2(\lambda-1) = 0 \rightarrow \lambda_1 = 1,~ \lambda_2 = 2, ~\lambda_3 = 2$$

We have one eigenvalue that has algebraic multiplicity one and a second one that has multiplicity two.

Next, we want to find the three linearly independent eigenvectors $[A - \lambda_i I]v_i = 0$.

For $\lambda_1 = 1$, we get:

$$\begin{bmatrix} -1 & 1 & -1 \\ -4 & 3 & -2 \\ -2 & 1 & 0 \end{bmatrix}v_1 = 0$$

The RREF yields:

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}v_1 = 0$$

This leads to the eigenvector $v_1 = (1,2,1)$.

For $\lambda_{2,3} = 2$, we get:

$$\begin{bmatrix} -2 & 1 & -1 \\ -4 & 2 & -2 \\ -2 & 1 & 1 \end{bmatrix}v_{2,3} = 0$$

The RREF yields:

$$\begin{bmatrix} 1 &-\dfrac{1}{2} & \dfrac{1}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}v_{2,3} = 0$$

This fortunately leads to two independent eigenvectors (not always the case):

• $v_2 = (-1,0,2)$
• $v_3 = (1,2,0)$

We are now able to find the diagonal matrix as:

$$J = Q^{-1} \cdot A \cdot Q$$

$Q$ is made up of the column eigenvectors as:

$$Q = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}$$

This should yield:

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

Now, you can also do this the way you were trying and should get an identical result (can be different as eigenvectors are not unique).