Finding eigenvectors (and JCF) of a $3 \times 3 $matrix

Question

Let

$$A = \begin{pmatrix}-3&17&-4\\ -2&9&-2 \\ -2&8&-1\end{pmatrix}.$$

I can't find the Jordan canonical form of matrix $A$, as I get confused due to the complex eigenvalues and equations with variables.

I know that the eigenvalues are $1$, $2+i$, and $2-i$. So, from here I need to find $v_1$ such that $(A-I)v_1 = 0$ and I get three equations.Two of them are same, so I'm not sure how to solve this (it should be simple but I'm rusty at the moment).

Also, how would I deal with the complex eigenvalues? Once I find the three eigenvectors I would use them as column vectors to make the matrix $P$ such that $J = P^{-1}AP$. But I cannot seem to get the eigenvectors.

Answer 1

You find the eigenvectors, one at a time, for each eigenvalue.

So, for $\lambda = 1$, we have $[A- \lambda I]v = 0$ as:

$$\begin{bmatrix}-4&17&-4\\ -2&8&-2 \\ -2&8&-2 \end{bmatrix}v_1 = 0$$

Using Gaussian Elimination, the RREF is:

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

This gives us an eigenvector $v_1 = (-1,0,1)$.

For one of the imaginary eigenvalues $\lambda = 2 + i$, we get $[A- \lambda I]v_2 = 0$ of:

$$\begin{bmatrix}-3-(2+i)&17&-4\\ -2&9-(2+i)&-2 \\ -2&8&-1-(2+i) \end{bmatrix}v_2 = 0$$

Using Gaussian Elimination, the RREF is:

$$\begin{bmatrix}1 & 0 & -\dfrac{5}{2} + \dfrac {i}{2}\\ 0& 1 & - 1 \\ 0 & 0 & 0 \end{bmatrix}v_2 = 0$$

This leads to an eigenvector of $v_2 = \left(\dfrac{5}{2} - \dfrac {i}{2}, 1, 1\right)$.

Of course, since we have complex conjugate eigenvalues, the other eigenvector is the complex conjugate of this one.

Can you now write out P? You already know $J$ since we have three unique eigenvalues.

Answer 2

I believe you're learning Jordan Canonical forms. Here are some questions you might want to ask yourself? I am answering them comprehensively here.

(1) How do I find eigenvalues, dimension and basis of the eigenspace corresponding to an eigenvalue?

(2) What is this question of finding $P$ really?

Answer to (1)

To find the eigenvalues, generally, the most useful tool is the characteristic polynomial. A scalar $\lambda$ is an eigenvalue if and only if $|\lambda I - A| = 0$.

The polynomial in $x$ given by $|xI - A| = 0$ is called the characteristic polynomial. If $\lambda$ is a root of this polynomial with multiplicity $a_{\lambda}$, then, $\lambda$ is an eigenvalue of algebraic multiplicity $a_{\lambda}$.

We say that a non-zero vector $v$ is an eigenvector with eigenvalue $\lambda$, if $Av = \lambda v$. Put alternatively, $(A - \lambda) v = 0$, that is, $v \in \ker(A-\lambda I)$. This motivates:

The kernel of $A - \lambda I$ is called the eigenspace corresponding to $\lambda$. Its dimension $k_{\lambda}$ is called the geometric multiplicity of $\lambda$.

Answer to (2)

We are looking for a $P$ such that $P^{-1}AP = J$ where $J$ is the Jordan Canonical Form of $A$. Abstractly, we have an operator on a vector space $V$ over an algebraically closed field. Let $T : V \to V$ be the operator affording this matrix representation with respect to the standard basis.

The point to realise is that $P$ has dual property: $P$ changes the standard basis $\mathfrak{B}$ to another basis $\mathfrak{B}'$ such that the matrix representation of $T$ with respect to $\mathfrak{B}'$ is the matrix $J$. That is, if we pick as basis $\mathfrak{B}'$, the columns of $P$, then, the operator $T$ behaves like $J$. (See the image at the end summarising change of basis matrices!)

Once this idea is clear, then, there is a natural way to find out what this $P$ must be! Suppose, for simplicity, that $J$ is a $2 \times 2$ Jordan block corresponding to $\lambda$. If $v$ is an eigenvector of eigenvalue $\lambda$, then, if we pick $v$ as first vector of our basis $\mathfrak{B}'$, then, $Tv = \lambda v$, or in basis notation, $$[T]_{\mathfrak{B}'}[e_1]_{\mathfrak{B}'} = \lambda [e_1]_{\mathfrak{B}'}$$ Put in yet another form, the first column of $[T]_{\mathfrak{B}'}$ is $$\begin{pmatrix}\lambda \\ 0\end{pmatrix}.$$ So, you may pick as $v$, the eigen vector corresponding to $\lambda$.

Now, Jordan's theorem tells us that, there is a vector $w$ such that $Tw = \lambda w + v$. Now, picking $w$ as the second vector for $\mathfrak{B}'$, we have, $$[T]_{\mathfrak{B}'}[e_2]_{\mathfrak{B}'} = \lambda [e_2]_{\mathfrak{B}'} + [e_1]_{\mathfrak{B}'}.$$

Put in terms of matrix $[T]_{\mathfrak{B}'}$, we get that, its second column is $$\begin{pmatrix}1 \\ \lambda\end{pmatrix}.$$

Hurrah! You've got a Jordan basis, viz. $\mathfrak{B}'$! (The basis $\mathfrak{B}'$ is called Jordan Basis, because, the matrix of $T$ is: $$\begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix},$$ a Jordan block in this basis.)

In order to determine $P$, or equivalently a Jordan basis, determine the eigenvalues $\{\lambda\}$, its algebraic multiplicity $\{a_\lambda\}$ and dimension of eigenspaces$\{k_{\lambda}\}$. Now, $P$ is the block diagonal matrix consisting of column vectors of Jordan chains corresponding to distinct eigenvalues: a Jordan chain corresponding to eigenvalue $\lambda$ is a sequence of vectors $v_1, \dots, v_{a_{\lambda}}$ such that $$\begin{align*} Tv_j &= \lambda v_j &\text{for } &&1 \leqslant j\leqslant k_{\lambda}\\ Tv_l &= \lambda v_l + v_{l-1} &\text{for } &&k_{\lambda}+1 \leqslant l \leqslant a_{\lambda} \end{align*}$$

Summary of Base Change