Generalized Eigenvectors and Jordan Form

As we saw previously, not every matrix is diagonalizable, even when allowing complex eigenvalues/eigenvectors. The matrix below was given as an example of a non-diagonalizable matrix.

$\begin{align*} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \end{align*}$


However, notice that there’s a pattern in the powers of this matrix.

$\begin{align*} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^2 &= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \\ \text{ } \\ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^3 &= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \\ \text{ } \\ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^4 &= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \end{align*}$


Leveraging this pattern, we can still write a formula for the Nth power of this matrix.

$\begin{align*} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \end{align*}$


When we conduct this same experiment with a $3 \times 3$ matrix of similar form, a more general pattern pops up.

$\begin{align*} \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^2 &= \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \\ \text{ } \\ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^3 &= \begin{pmatrix} 1 & 3 & 3 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} \\ \text{ } \\ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^4 &= \begin{pmatrix} 1 & 4 & 6 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix} \end{align*}$


The pattern is that the numbers are all just binomial coefficients taken from Pascal’s triangle! Writing this pattern more generally for a $k \times k$ square matrix, we have

$\begin{align*} \begin{pmatrix} 1 & 1 & 0 & \cdots & \cdots & 0 \\ & 1 & 1 & \cdots & \cdots & 0 \\ & & \ddots & \ddots & & \vdots \\ & & & \ddots & \ddots & \vdots \\ & & & & 1 & 1 \\ & & & & & 1 \end{pmatrix}^n &= \begin{pmatrix} 1 & \binom{n}{1} & \binom{n}{2} & \cdots & \cdots & \binom{n}{k−1} \\ & 1 & \binom{n}{1} & \cdots & \cdots & \binom{n}{k−2} \\ & & \ddots & \ddots & & \vdots \\ & & & \ddots & \ddots & \vdots \\ & & & & 1 & \binom{n}{1} \\ & & & & & 1 \end{pmatrix}. \end{align*}$


If we replace the diagonal with another number, say $\lambda$, then similar experimentation reveals the following formula:

$\begin{align*} \begin{pmatrix} \lambda & 1 & 0 & \cdots & \cdots & 0 \\ & \lambda & 1 & \cdots & \cdots & 0 \\ & & \ddots & \ddots & & \vdots \\ & & & \ddots & \ddots & \vdots \\ & & & & \lambda & 1 \\ & & & & & \lambda \end{pmatrix}^n &= \begin{pmatrix} \lambda^n & \binom{n}{1} \lambda^{n−1} & \binom{n}{2} \lambda^{n−2} & \cdots & \cdots & \binom{n}{k−1} \lambda^{n−k+1} \\ & \lambda^{n} & \binom{n}{1} \lambda^{n−1} & \cdots & \cdots & \binom{n}{k−2} \lambda^{n−k+2} \\ & & \ddots & \ddots & & \vdots \\ & & & \ddots & \ddots & \vdots \\ & & & & \lambda^{n} & \binom{n}{1} \lambda^{n−1} \\ & & & & & \lambda^{n} \end{pmatrix} \end{align*}$


These matrices consisting of a diagonal $\lambda$ directly below an off-diagonal of $1$s are called Jordan blocks, and a matrix consisting of Jordan blocks is called a Jordan matrix. For example, the matrix below consists of two Jordan blocks. (Note that blank entries correspond to $0$.)

$\begin{align*} \begin{pmatrix} 2 & 1 & & & \\ 0 & 2 & & & \\ & & 3 & 1 & 0 \\ & & 0 & 3 & 1 \\ & & 0 & 0 & 3 \end{pmatrix} \end{align*}$


The big question, then, is: which matrices $A$ can be expressed as

$\begin{align*} A=PJP^{-1} \end{align*}$


where $J$ is a Jordan matrix? The answer is quite satisfying: all of them! Thus, Jordan form provides a guaranteed backup plan for exponentiating matrices that are non-diagonalizable.

So, how do we construct the matrices $P$ and $J$? Let’s start out like we did with diagonalization, right-multiplying both sides of the equation by $P$.

$\begin{align*} AP = PJ \end{align*}$


To keep things simple but interesting enough to generalize our results, let’s assume the following two-block Jordan matrix.

$\begin{align*} J = \begin{pmatrix} \lambda_1 & 1 & & & \\ 0 & \lambda_1 & & & \\ & & \lambda_2 & 1 & 0 \\ & & 0 & \lambda_2 & 1 \\ & & 0 & 0 & \lambda_2 \end{pmatrix} \end{align*}$


Then we have

$\begin{align*} A \begin{pmatrix} | & | & | & | & | \\ v_{11} & v_{12} & v_{21} & v_{22} & v_{23} \\ | & | & | & | & | \end{pmatrix} &= \begin{pmatrix} | & | & | & | & | \\ v_{11} & v_{12} & v_{21} & v_{22} & v_{23} \\ | & | & | & | & | \end{pmatrix} \begin{pmatrix} \lambda_1 & 1 & & & \\ 0 & \lambda_1 & & & \\ & & \lambda_2 & 1 & 0 \\ & & 0 & \lambda_2 & 1 \\ & & 0 & 0 & \lambda_2 \end{pmatrix} \\ \begin{pmatrix} | & | & | & | & | \\ Av_{11} & Av_{12} & Av_{21} & Av_{22} & Av_{23} \\ | & | & | & | & | \end{pmatrix} &= \begin{pmatrix} | & | & | & | & | \\ \lambda_1 v_{11} & v_{11} + \lambda_1 v_{12} & \lambda_2 v_{21} & v_{21} + \lambda_2 v_{22} & v_{22} + \lambda_2 v_{23} \\ | & | & | & | & | \end{pmatrix}. \end{align*}$


First of all, since

$\begin{align*} Av_{11} &= \lambda_1 v_{11} \\ Av_{21} &= \lambda_2 v_{21} \end{align*}$


we see that $v_{11}$ and $v_{21}$ are eigenvectors corresponding to the eigenvalues $\lambda_1$ and $\lambda_2$, respectively. This makes intuitive sense because these columns mark the start of the Jordan blocks and thus don’t have a 1 above them – these columns are perfect diagonals.

Before we go on, notice that we can rearrange the above equations as follows:

$\begin{align*} \left( A − \lambda_1 I \right) v_{11} &= 0 \\ \left( A − \lambda_2 I \right) v_{21} &= 0 \end{align*}$


This will be helpful shortly. Now, we move into the more novel cases, beginning by equating the second columns.

$\begin{align*} Av_{12} = v_{11} + \lambda_1 v_{12} \end{align*}$


By rearranging the equation, we come up with an equation similar to those we found for the eigenvectors $v_{11}$ and $v_{21}$.

$\begin{align*} Av_{12} &= v_{11} + \lambda_1 v_{12} \\ \left( A−\lambda_1 I \right) v_{12} &= v_{11} \\ \left( A−\lambda_1 I \right)^2 v_{12} &= \left( A−\lambda_1 I \right) v_{11} \\ &= Av_{11} − \lambda_1 I v_{11} \\ &= \lambda_1 v_{11} − \lambda_1 v_{11} \\ &= 0 \end{align*}$


We call $v_{12}$ a generalized eigenvector of order $2$ for the eigenvalue $\lambda_1$ because it solves the equation $(A-\lambda_1 I)^2v=0$, whereas normal eigenvectors (i.e. generalized eigenvectors of order $1$) for the eigenvalue $\lambda_1$ solve the equation $(A-\lambda_1 I)v=0$. By the same reasoning, we conclude that $v_{22}$ is a generalized eigenvector of order $2$ for $\lambda_2$, and $v_{23}$ is a generalized eigenvector of order $3$ for $\lambda_2$.

To conclude this section, we walk through an example of exponentiating the non-diagonalizable matrix below by converting it to Jordan form.

$\begin{align*} \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix} \end{align*}$


First, we compute the eigenvalues. In the row manipulations, we use the symbol $\square$ to denote matrix entries that change but have no consequence when computing the determinant.

$\begin{align*} \mbox{det} \begin{pmatrix} 1−\lambda & 1 & −1 & 1 & −1 \\ 0 & −1−\lambda & 0 & 0 & 1 \\ 0 & 1 & −1−\lambda & 0 & 0 \\ 0 & 3 & −2 & 1−\lambda & −3 \\ 0 & 0 & 0 & 0 & −1−\lambda \end{pmatrix} &= 0 \end{align*}$


$\begin{align*} \mbox{det} \begin{pmatrix} 1−\lambda & 1 & −1 & 1 & −1 \\ 0 & −1−\lambda & 0 & 0 & 1 \\ 0 & 0 & −1−\lambda & 0 & \square \\ 0 & 0 & −2 & 1−\lambda &\square \\ 0 & 0 & 0 & 0 & −1−\lambda \end{pmatrix} &= 0 \end{align*}$


$\begin{align*} \mbox{det} \begin{pmatrix} 1−\lambda & 1 & −1 & 1 & −1 \\ 0 & −1−\lambda & 0 & 0 & 1 \\ 0 & 0 & −1−\lambda & 0 & \square \\ 0 & 0 & 0 & 1−\lambda &\square \\ 0 & 0 & 0 & 0 & −1−\lambda \end{pmatrix} &= 0 \end{align*}$


$\begin{align*} (1-\lambda)^2 (-1-\lambda)^3 = 0 \end{align*}$


$\begin{align*} \lambda = 1,1,-1,-1,-1,-1 \end{align*}$



Now that we have the eigenvalues $\lambda_1=1$ repeated twice and $\lambda_2=-1$ repeated three times, we solve for the first and second-order generalized eigenvectors for $\lambda_1$, and the first, second, and third-order generalized eigenvectors for $\lambda_2$.

First, we solve for the first-order generalized eigenvector $v_{11}$ of $\lambda_1=1$.


$\begin{align*} (A−\lambda_1 I)v_{11}=0 \end{align*}$

$\begin{align*} (A − (1)I)v_{11} = 0 \end{align*}$


$\begin{align*} (A−I) v_{11} = 0 \end{align*}$


$\begin{align*} \begin{pmatrix} 0 & 1 & −1 & 1 & −1 \\ 0 & −2 & 0 & 0 & 1 \\ 0 & 1 & −2 & 0 & 0 \\ 0 & 3 & −2 & 0 & −3 \\ 0 & 0 & 0 & 0 & −2 \end{pmatrix} v_{11} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 0 & 1 & −1 & 1 & 0 \\ 0 & −2 & 0 & 0 & 0 \\ 0 & 1 & −2 & 0 & 0 \\ 0 & 3 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{11} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 0 & 0 & −1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & −2 & 0 & 0 \\ 0 & 0 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{11} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{11} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{11} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} v_{11} &= \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


Next, we solve for the second-order generalized eigenvector $v_{12}$ of $\lambda_1=1$, which is independent of the first-order generalized eigenvector $v_{11}$.


$\begin{align*} (A-\lambda_1 I)^2 v_{12} = 0 \end{align*}$

$\begin{align*} \begin{pmatrix} 0 & 1 & −1 & 1 & −1 \\ 0 & −2 & 0 & 0 & 1 \\ 0 & 1 & −2 & 0 & 0 \\ 0 & 3 & −2 & 0 & −3 \\ 0 & 0 & 0 & 0 & −2 \end{pmatrix}^2 v_{12} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix}0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & −4 \\ 0 & −4 & 4 & 0 & 1 \\ 0 & −8 & 4 & 0 & 9 \\ 0 & 0 & 0 & 0 & 4 \end{pmatrix} v_{12} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix}0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 \\ 0 & −4 & 4 & 0 & 0 \\ 0 & −8 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{12} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix}0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{12} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix}0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{12} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} v_{12} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} \end{align*}$


Before we go on, let’s take inventory of what we have, filling in part of our Jordan form expression.

$\begin{align*} \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix} &= \begin{pmatrix} | & | & | & | & | \\ | & | & | & | & | \\ v_{11} & v_{12} & v_{21} & v_{22} & v_{23} \\ | & | & | & | & | \\ | & | & | & | & | \end{pmatrix} \begin{pmatrix} \lambda_1 & 1 & & & \\ 0 & \lambda_1 & & & \\ & & \lambda_2 & 1 & 0 \\ & & 0 & \lambda_2 & 1 \\ & & 0 & 0 & \lambda_2 \end{pmatrix} \begin{pmatrix} | & | & | & | & | \\ | & | & | & | & | \\ v_{11} & v_{12} & v_{21} & v_{22} & v_{23} \\ | & | & | & | & | \\ | & | & | & | & | \end{pmatrix}^{−1} \\ &\text{ } \\ \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix} &= \begin{pmatrix} 1 & 0 & | & | & | \\ 0 & 0 & | & | & | \\0 & 0 & v_{21} & v_{22} & v_{23} \\ 0 & 1 & | & | & | \\ 0 & 0 & | & | & | \end{pmatrix} \begin{pmatrix} 1 & 1 & & & \\ 0 & 1 & & & \\ & & \lambda_2 & 1 & 0 \\ & & 0 & \lambda_2 & 1 \\ & & 0 & 0 & \lambda_2 \end{pmatrix} \begin{pmatrix} 1 & 0 & | & | & | \\ 0 & 0 & | & | & | \\0 & 0 & v_{21} & v_{22} & v_{23} \\ 0 & 1 & | & | & | \\ 0 & 0 & | & | & | \end{pmatrix}^{−1} \end{align*}$



Continuing, we solve for the first-order generalized eigenvector $v_{21}$ of $\lambda_2=-1$.


$\begin{align*} (A- \lambda_2 I)v_{21} = 0 \end{align*}$

$\begin{align*} (A- (-1) I)v_{21} = 0 \end{align*}$


$\begin{align*} (A+ I)v_{21} = 0 \end{align*}$


$\begin{align*} \begin{pmatrix} 2 & 1 & −1 & 1 & −1 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 3 & −2 & 2 & −3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} v_{21} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 2 & 0 & −1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & −2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} v_{21} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & −1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} v_{21} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & −1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{21} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} v_{21} &= \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} \end{align*}$



Then, we solve for the second-order generalized eigenvector $v_{22}$ of $\lambda_2=-1$, which is independent of the first-order generalized eigenvector $v_{21}$.


$\begin{align*} (A- \lambda_2 I)^2v_{21} = 0 \end{align*}$

$\begin{align*} \begin{pmatrix} 2 & 1 & −1 & 1 & −1 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 3 & −2 & 2 & −3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}^2 v_{22} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 4 & 4 & −4 & 4 & −4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 4 & −4 & 4 & −3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} v_{22} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & −1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} v_{22} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & −1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} v_{22} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} v_{22} &= \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} \end{align*}$



Lastly, we solve for the third-order generalized eigenvector $v_{23}$ of $\lambda_2=-1$, which is independent of the first and second-order generalized eigenvectors $v_{21}$ and $v_{22}$.


$\begin{align*} (A−\lambda_2 I)^3 v_{23}=0 \end{align*}$

$\begin{align*} \begin{pmatrix} 2 & 1 & −1 & 1 & −1 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 3 & −2 & 2 & −3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}^3 v_{23} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 8 & 12 & −12 & 12 & −12 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 8 & −8 & 8 & −8 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} v_{23} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 8 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & −1 & 1 & −1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} v_{23} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & −1 & 1 & −1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} v_{23} &= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{align*}$


$\begin{align*} v_{23} &= \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} \end{align*}$



Finally, we can fill in the rest of our Jordan form expression.

$\begin{align*} \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & & & \\ 0 & 1 & & & \\ & & −1 & 1 & 0 \\ & & 0 &−1 & 1 \\ & & 0 & 0 & −1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}^{−1} \end{align*}$


$\begin{align*} \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & & & \\ 0 & 1 & & & \\ & & −1 & 1 & 0 \\ & & 0 &−1 & 1 \\ & & 0 & 0 & −1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & −1 & 1 & −1 \\ 0 & −1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & −1 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{align*}$


Exponentiating our matrix, we have

$\begin{align*} \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix}^n = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & & & \\ 0 & 1 & & & \\ & & −1 & 1 & 0 \\ & & 0 &−1 & 1 \\ & & 0 & 0 & −1 \end{pmatrix}^n \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & −1 & 1 & −1 \\ 0 & −1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & −1 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*}$


To exponentiate the middle matrix, it suffices to exponentiate the two blocks separately. The first block is simple and familiar:

$\begin{align*} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \end{align*}$


For the second block, we make use of the general formula

$\begin{align*} \begin{pmatrix} \lambda & 1 & 0 & \cdots & \cdots & 0 \\ & \lambda & 1 & \cdots & \cdots & 0 \\ & & \ddots & \ddots & & \vdots \\ & & & \ddots & \ddots & \vdots \\ & & & & \lambda & 1 \\ & & & & & \lambda \end{pmatrix}^n &= \begin{pmatrix} \lambda^n & \binom{n}{1} \lambda^{n−1} & \binom{n}{2} \lambda^{n−2} & \cdots & \cdots & \binom{n}{k−1} \lambda^{n−k+1} \\ & \lambda^{n} & \binom{n}{1} \lambda^{n−1} & \cdots & \cdots & \binom{n}{k−2} \lambda^{n−k+2} \\ & & \ddots & \ddots & & \vdots \\ & & & \ddots & \ddots & \vdots \\ & & & & \lambda^{n} & \binom{n}{1} \lambda^{n−1} \\ & & & & & \lambda^{n} \end{pmatrix} \end{align*}$


and find that

$\begin{align*} \begin{pmatrix} −1 & 1 & 0 \\ 0 & −1 & 1 \\ 0 & 0 & −1 \end{pmatrix}^n &= \begin{pmatrix} (−1)^n & \binom{n}{1}(−1)^{n−1} & \binom{n}{2}(−1)^{n−2} \\ 0 & (−1)^n & \binom{n}{1}(−1)^{n−1} \\ 0 & 0 & (−1)^n \end{pmatrix} \\ &= \begin{pmatrix} (−1)^n & n(−1)^{n−1} & \frac{n(n−1)}{2} (−1)^{n−2} \\ 0 & (−1)^n & n(−1)^{n−1} \\ 0 & 0 & (−1)^n \end{pmatrix}. \end{align*}$


Thus, we have

$\begin{align*} \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix}^n = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & n & & & \\ 0 & 1 & & & \\ & & (−1)^n & n(−1)^{n−1} & \frac{n(n−1)}{2}(−1)^{n−2} \\ & & 0 &(−1)^n & n(−1)^{n−1} \\ & & 0 & 0 & (−1)^n \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & −1 & 1 & −1 \\ 0 & −1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & −1 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*}$


Multiplying out and simplifying, we reach

$\begin{align*} \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix}^n = \begin{pmatrix} 1 & n & −n & n & −n \\ 0 & (−1)^n & 0 & 0 & n(−1)^{n+1} \\ 0 & n(−1)^{n+1} & (−1)^n & 0 & \frac{n(n−1)}{2} (−1)^n \\ 0 & (n+1)(−1)^{n+1}+1 & −1+(−1)^n & 1 & \frac{n(n+1)}{2}(−1)^n + (−1)^n − 1 \\ 0 & 0 & 0 & 0 & (−1)^n \end{pmatrix}. \end{align*}$


Lastly, let’s verify this formula on the case $n=2$.

$\begin{align*} \begin{pmatrix} 1 & 1 & −1 & 1 & −1 \\ 0 & −1 & 0 & 0 & 1 \\ 0 & 1 & −1 & 0 & 0 \\ 0 & 3 & −2 & 1 & −3 \\ 0 & 0 & 0 & 0 & −1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 2 & −2 & 2 & −2 \\ 0 & 1 & 0 & 0 & −2 \\ 0 & −2 & 1 & 0 & 1 \\ 0 & −2 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{align*}$


$\begin{align*} \begin{pmatrix} 1 & 2 & −2 & 2 & −2 \\ 0 & (−1)^2 & 0 & 0 & 2(−1)^{2+1} \\ 0 & 2(−1)^{2+1} & (−1)^2 & 0 & \frac{2(2−1)}{2} (−1)^2 \\ 0 & (2+1)(−1)^{2+1}+1 & −1+(−1)^2 & 1 & \frac{2(2+1)}{2}(−1)^2 + (−1)^2 − 1 \\ 0 & 0 & 0 & 0 & (−1)^2 \end{pmatrix} &= \begin{pmatrix} 1 & 2 & −2 & 2 & −2 \\ 0 & 1 & 0 & 0 & −2 \\ 0 & −2 & 1 & 0 & 1 \\ 0 & −2 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{align*}$


It checks out!


Practice Problems


For each matrix $A$, express $A=PJP^{-1}$ where $J$ is a Jordan matrix, and use this Jordan expression to compute $A^n$. Check your formula on the case $n=2$. (You can view the solution by clicking on the problem.)

$\begin{align*} 1) \hspace{.5cm} \begin{pmatrix} 2 & 1 \\ −1 & 0 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 1 & 0 \\ −1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \\ &A^n = \begin{pmatrix} n+1 & n \\ −n & 1−n \end{pmatrix} \end{align*}$


$\begin{align*} 2) \hspace{.5cm} \begin{pmatrix} 3 & −2 \\ 2 & −1 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ \end{align*}$


$\begin{align*} 3) \hspace{.5cm} \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} −1 & 2 \\ 1 & −1 \end{pmatrix} \\ &A^n = \begin{pmatrix} 2n+1 & −2n \\ 2n & 1−2n \end{pmatrix} \end{align*}$


$\begin{align*} 4) \hspace{.5cm} \begin{pmatrix} 0 & −1 \\ 1 & −2 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} −3 & 1 \\ 1 & 0 \end{pmatrix} \\ &&A^n = 2^{n-1} \begin{pmatrix} 2 & 0 \\ n & 2 \end{pmatrix} \end{align*}$


$\begin{align*} 5) \hspace{.5cm} \begin{pmatrix} 1 & 1 & 0 \\ −1 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 1 & −1 \\ 1 & −2 \end{pmatrix} \begin{pmatrix} −1 & 1 \\ 0 & −1 \end{pmatrix} \begin{pmatrix} 2 & −1 \\ 1 & −1 \end{pmatrix} \\ &A^n = (−1)^n \begin{pmatrix} 1−n & n \\ −n & n+1 \end{pmatrix} \end{align*}$


$\begin{align*} 6) \hspace{.5cm} \begin{pmatrix} 1 & −1 & −1 \\ 4 & −3 & −6 \\ 0 & 0 & 3 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ −1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ &A^n = \begin{pmatrix} 2^{n}−2^{n−1}n & 2^{n−1}n & 0 \\ −2^{n−1}n & 2^{n−1}n+2^{n} & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align*}$


$\begin{align*} 7) \hspace{.5cm} \begin{pmatrix} 3 & −1 & 1 \\ 4 & −1 & 2 \\ −2 & 1 & 1 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 1 & 1 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & −1 \end{pmatrix} \begin{pmatrix} −1 & 1 & 0 \\ 0 & −1 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} −1 & 1 & 1 \\ 2 & −1 & −1 \\ 0 & 0 & −1 \end{pmatrix} \\ &A^n = \begin{pmatrix} (−1)^n(1−2n) & (−1)^nn & (−1)^nn \\ −4(−1)^nn & (−1)^n(2n+1) & (2n+1)(−1)^n−3^n \\ 0 & 0 & 3^n \end{pmatrix} \end{align*}$


$\begin{align*} 8) \hspace{.5cm} \begin{pmatrix} 1 & −1 & 1 \\ 1 & 2 & 0 \\ 0 & −1 & 3 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 1 & 0 & 0 \\ 2 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 2 & −1 & 1 \\ −2 & 1 & 0 \end{pmatrix} \\ &A^n = \frac{1}{2} \begin{pmatrix} 2+6n−2n^2 & n^2−3n & 2n \\ 12n−4n^2 & 2n^2−6n+2 & 4n \\ −4n & 2n & 2 \end{pmatrix} \end{align*}$


$\begin{align*} 9) \hspace{.5cm} \begin{pmatrix} 2 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 1 & 2 & −1 \\ 0 & 0 & 1 & 2 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & −1 \\ 1 & 0 & 0 \\ −1 & −1 & 1 \end{pmatrix} \\ &A^n = 2^{n−3} \begin{pmatrix} 8−4n & −4n & 4n \\ 5n−n^2 & 8+n−n^2 & n^2−n \\ n−n^2 & −n^2−3n & n^2+3n+8 \end{pmatrix} \end{align*}$


$\begin{align*} 10) \hspace{.5cm} \begin{pmatrix} −3 & −4 & 4 & 2 \\ 3 & 3 & −2 & −2 \\ −1 & −2 & 3 & 1 \\ 4 & 4 & −4 & −3 \end{pmatrix} \end{align*}$
Solution:
$\begin{align*} &\text{answers may vary; one correct answer is} \\ &A= \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} −1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & −1 \\ 1 & 0 & 0 & 0 \end{pmatrix} \\ &A^n = \frac{2^{n−4}}{3} \begin{pmatrix} 48 & 0 & 0 & 0 \\ n^3+3n^2+20n & 6n^2−6n+48 & 24n & 6n−6n^2 \\ 6n^2+18n & 24n & 48 & −24n \\ n^3 + 3n^2 − 4n & 6n^2−6n & 24n & 48+6n−6n^2 \end{pmatrix} \end{align*}$


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