Find $P$ where $P^{-1}AP$ for a given matrix $A$

Question

I am doing a past paper and I have been given a matrix A:

\begin{bmatrix} 4 & -1 & -3 & 2 \\ 4 & -2 & -4 & 4 \\ -4 & 4 & 6 & -4 \\ -6 & 5 & 7 & -4 \\ \end{bmatrix}

and I need to find a matrix $P$ such that $P^{-1}AP$ =

\begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & -2 \\ \end{bmatrix}

I'm really not sure how to begin with this question, would anyone be able to help out?

Answer 1

This corresponds to the so-called Jordan decomposition.

Theorem:

If $A \in \mathbb{C}^{n\times n}$, then there exists a nonsingular $P \in \mathbb{C}^{n\times n}$ such that: $$ P^{-1}AP=\text{diag}(J_1, \dots, J_k) $$ where $$ J_i= \begin{bmatrix} \lambda_i & 1 & \dots & 0 \\ 0 & \lambda_i & \ddots & \vdots\\ \vdots & & \ddots & 1\\ 0 & \dots & 0 & \lambda_i \end{bmatrix} \in \mathbb{C}^{n_i\times n_i} $$ and $n_1 + \dots + n_k = n$.

In your example, you have $$ P^{-1} A P = J = \text{diag}(J_1, J_2, J_3) $$ $n_1 = n_3 = 1, n_2 = 2$ and $\lambda_1 = \lambda_2 = 2, \lambda_3 = -2$. That is, $J_1 = 2, J_3 = -2$ and $J_2 = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix}. $ To find $P$, you can write the decomposition as $AP = PJ$. Assuming $P = [p_1, p_2, p_3, p_4]$, with $p_i$ columns of $P$, you have $$ A [p_1, p_2, p_3, p_4] = [p_1, p_2, p_3, p_4]J $$ leading to $$ A [p_1, p_2, p_3, p_4] = [2p_1, 2p_2, p_2+2p_3, -2p_4] $$ which gives (column by column) $$ A p_1 = 2p_1 \qquad \text{(1)}\\ A p_2 = 2p_2 \qquad \text{(2)}\\ A p_3 = p_2 + 2p_3 \qquad \text{(3)}\\ A p_4 = -2p_4 \qquad \text{(4)} $$ Now, (1), (2) indicate that $p_1$ and $p_2$ are the eigenvectors of $A$ associated with the same eigenvalue $\lambda = 2$. Similarly, (4) indicates that $p_4$ is the eigenvector of $A$ associated with $\lambda = -2$. Moreover (3) can be written as $$ (A - 2I) p_3 = p_2 $$ multiplying both sides by (A - 2I) we get $$ (A - 2I)^2 p_3 = (A - 2I) p_2 = 0 $$ since $(A - 2I) p_2 = 0$ comes from (2). This indicates that $p_3$ is the generalized eigenvector of rank 2 of $A$ (see here) corresponding to the (generalized) eigenvalue $\lambda = 2$.

Answer 2

From the given information, we know that matrix A had eigenvalues 2, -2, and 2 is an eigenvalue of multiplicity.

We want to find eigenvectors that are associated with each of these eigenvalues.

find $v$ such that $(A-\lambda I) v = 0$

$\lambda = 2$

$\begin{bmatrix} 2 & -1 & -3 & 2 \\ 4 & -4 & -4 & 4 \\ -4 & 4 & 4 & -4 \\ -6 & 5 & 7 & -6 \\ \end{bmatrix}v = 0$

With a little bit of trial and error we find $\begin{bmatrix} 1\\1\\1\\1 \end{bmatrix}, \begin{bmatrix} -1\\0\\0\\1 \end{bmatrix} $

You might have found different eigenvectors, any two independent vectors in the eigenspace will do.

for the remaining vector we are looking for some vector $w$ such that

$Aw = 2w + v$ where $v$ is one of the the eigenvectors that is associated with the eigenvalue 2.

if such a vector exists then

$(A-2I) w = v\\ (A-2I)^2 w = (A-2I)v = 0$

$(A-2I)^2 = \begin{bmatrix} 0&0&0&0\\-16&16&16&-16\\16&-16&-16&16\\16&-16&-16&16\end{bmatrix}$

Which means that $w$ must be some linear combination of

$\begin{bmatrix} 1\\1\\0\\0\end{bmatrix},\begin{bmatrix} 1\\0\\1\\0\end{bmatrix},\begin{bmatrix} 1\\0\\0\\1\end{bmatrix}$

find some combination of these such that $(A-2I)w$ equals one of the eigenvectors you already have.

$(A-2I)\begin{bmatrix} 1\\0\\1\\0\end{bmatrix} =\begin{bmatrix} -1\\0\\0\\1 \end{bmatrix}$

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

And I will leave it to you to find the remaining eigenvector and complete the matrix.