Similar matrices and the Jordan form

Some matrices have two eigenvector lines that are almost one. Watch the two lines close in on each other and merge.

Two lines become one

The matrix in the picture is triangular, so its eigenvalues are its diagonal entries, 1 and $1+\varepsilon$, as on Finding eigenvalues. Its eigenvectors are $(1,0)$ and $(1,\varepsilon)$: check that $\mathbf{A}(\varepsilon)\,(1,\varepsilon)=(1+\varepsilon,\;\varepsilon+\varepsilon^2)=(1+\varepsilon)\,(1,\varepsilon)$.

As $\varepsilon$ shrinks, the second eigenvector tilts down onto the first. The eigenvector matrix

has determinant $\varepsilon$, so its inverse needs a division by $\varepsilon$ and grows without limit. At $\varepsilon=0$ the matrix is the shear, with the eigenvalue 1 twice and only one line of eigenvectors. Diagonalisation and matrix powers can’t handle it. This page shows what replaces the diagonal form.

Similar matrices

First, a way to compare matrices. If $\mathbf{V}$ is any invertible matrix, then

is similar to $\mathbf{A}$. Read it right to left: $\mathbf{V}$ turns coordinates on the grid of its columns into ordinary ones, $\mathbf{A}$ acts, and $\mathbf{V}^{-1}$ turns the result back. So $\mathbf{B}$ is the same map as $\mathbf{A}$, written on the grid of the columns of $\mathbf{V}$, as coordinates were on Independence, basis and dimension.

The same map has the same stretch factors, so similar matrices share their eigenvalues. In symbols, $\det(\mathbf{B}-\lambda\mathbf{I})=\det\big(\mathbf{V}^{-1}(\mathbf{A}-\lambda\mathbf{I})\mathbf{V}\big)=\det(\mathbf{A}-\lambda\mathbf{I})$. So they share the trace and determinant too. Their eigenvectors differ: an eigenvector $\mathbf{v}$ of $\mathbf{A}$ appears as $\mathbf{V}^{-1}\mathbf{v}$ on the new grid.

Diagonalising is the case where the new grid is made of eigenvectors, so that $\mathbf{B}$ comes out diagonal.

Compare the two panels. The grids move differently, and the eigenvector lines sit in different places, but the stretch factors are the same.

Too few eigenvectors

A matrix whose eigenvectors can’t fill a grid is called defective. Take

Its trace is 4 and its determinant $3+1=4$, so its polynomial is $\lambda^2-4\lambda+4=(\lambda-2)^2$: the eigenvalue 2, twice. But $\mathbf{A}-2\mathbf{I}=\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$ has only the line of $\mathbf{v}=(1,-1)$ in its null space. One line, where a diagonal form would need two.

The missing direction comes from one step further. Look for a vector $\mathbf{w}$ that $\mathbf{A}-2\mathbf{I}$ sends to the eigenvector, not to zero:

Here $\mathbf{w}=(1,0)$ works: $(\mathbf{A}-2\mathbf{I})(1,0)=(1,-1)$. Such a $\mathbf{w}$ is a generalised eigenvector. Use $\mathbf{v}$ and $\mathbf{w}$ as the new grid, $\mathbf{V}=\begin{bmatrix}1&1\\-1&0\end{bmatrix}$, with $\mathbf{V}^{-1}=\begin{bmatrix}0&-1\\1&1\end{bmatrix}$:

The first column says $\mathbf{A}\mathbf{v}=2\mathbf{v}$. The second says $\mathbf{A}\mathbf{w}=\mathbf{v}+2\mathbf{w}$: a stretch by 2, plus a push along $\mathbf{v}$.

The Jordan block

The matrix that came out is a Jordan block:

It is as close to diagonal as a defective matrix gets. The argument $(\lambda)$ tells it apart from the exchange matrix $\mathbf{J}$ of Transposes and permutations. Allowing complex numbers, every square matrix is similar to a matrix made of such blocks along its diagonal, a block of size 1 for each ordinary eigenvector, and longer blocks, with 1s just above the diagonal, where eigenvectors are missing. That arrangement is the Jordan form.

Its powers show what the extra 1 does. Squaring gives $\begin{bmatrix}\lambda^2&2\lambda\\0&\lambda^2\end{bmatrix}$, and in general

Each new step multiplies the corner by $\lambda$ and adds one more $\lambda^k$, so the corner counts the steps. The same happens in continuous time:

That corner $t$ is where solutions like $t\,e^{\lambda t}$ come from. For the defective $\mathbf{A}$ above, the equation $\mathrm{d}\mathbf{u}/\mathrm{d}t=\mathbf{A}\mathbf{u}$ of Linear differential equations, started at $(1,0)$, has the solution

Check the slope at $t=0$: $2\,(1,0)+(1,-1)=(3,-1)$, which is $\mathbf{A}\,(1,0)$.

Sensitivity of the Jordan form

A Jordan block lives exactly at a collision, and the smallest nudge breaks it apart. Change the bottom-left zero of the shear to a tiny number, $10^{-10}$, a change in the tenth decimal place. The eigenvalues of

The characteristic polynomial is $(\lambda-1)^2-10^{-10}$, so the eigenvalues move by the square root of the nudge, $10^{-5}$: a hundred thousand times more than the nudge itself. The picture shows the same thing from the other side. At $\varepsilon=0.01$ the matrix still has two eigenvector lines, but they are 0.57° apart, which is 0.0100 radians, and its tenth power has a top-right corner of $\big(1.01^{10}-1\big)/0.01=10.462$, against exactly 10 for the shear. The two are hard to tell apart.

For this reason programs do not compute the Jordan form of a measured matrix: rounding errors alone would change the answer. The Jordan form is a tool for understanding, not for computing.

Worked example

1. A similar copy. $\mathbf{A}=\begin{bmatrix}1.5&1\\0.5&1\end{bmatrix}$ and $\mathbf{V}=\begin{bmatrix}1&0\\1&1\end{bmatrix}$, so $\mathbf{V}^{-1}=\begin{bmatrix}1&0\\-1&1\end{bmatrix}$. Then $\mathbf{A}\mathbf{V}=\begin{bmatrix}2.5&1\\1.5&1\end{bmatrix}$ and $\mathbf{B}=\mathbf{V}^{-1}\mathbf{A}\mathbf{V}=\begin{bmatrix}2.5&1\\-1&0\end{bmatrix}$. Trace 2.5 and determinant 1, so the eigenvalues are 2 and 0.5 again. The eigenvector $(2,1)$ of $\mathbf{A}$ becomes $\mathbf{V}^{-1}(2,1)=(2,-1)$, and indeed $\mathbf{B}(2,-1)=(4,-2)=2\,(2,-1)$.

2. A defective matrix. $\begin{bmatrix}3&1\\-1&1\end{bmatrix}$: eigenvalue 2 twice, eigenvector $(1,-1)$, generalised eigenvector $(1,0)$, Jordan form $\begin{bmatrix}2&1\\0&2\end{bmatrix}$.

3. Powers. $\mathbf{J}(1)^{10}=\begin{bmatrix}1&10\\0&1\end{bmatrix}$. For $\varepsilon=0.01$, the top-right entry of $\mathbf{A}(\varepsilon)^{10}$ is 10.462.

Try these before you read the answers.

• Do similar matrices share their eigenvalues? Yes.
• Their eigenvectors? No: $\mathbf{v}$ becomes $\mathbf{V}^{-1}\mathbf{v}$.
• Eigenvectors of $\begin{bmatrix}2&1\\0&2\end{bmatrix}$? Only the line of $(1,0)$.
• As $\varepsilon$ falls to 0 in the picture, what happens to the angle between the eigenvector lines? It falls to 0°.

Where you’ll meet this

Critical damping. A door closer or a car suspension tuned to stop as fast as possible without swinging has a repeated eigenvalue. Its response is $(c_1+c_2t)\,e^{-at}$: the $t\,e^{-at}$ term is the Jordan block at work.

Double poles. In digital filters, a repeated pole gives responses like $n\,a^n$, as in the callout above.

What software does instead. Because the Jordan form is so fragile, numerical libraries use forms that rounding cannot upset, such as the Schur form, a triangular matrix reached through orthonormal coordinates, or the singular value decomposition.

Reference card