Finding eigenvalues

When does a matrix flatten the plane? Slide a number λ and watch a parallelogram collapse onto a line, twice.

When the plane flattens

On Eigenvectors we found two lines that the matrix only stretches, by sweeping an arrow round and watching. That works when you can see the picture. For a bigger matrix you need a way to compute.

Start from the definition, $\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$. Move everything to one side. Writing $\lambda\mathbf{v}$ as $\lambda\mathbf{I}\mathbf{v}$, with the identity matrix $\mathbf{I}$, lets me take $\mathbf{v}$ out as a common factor:

So $\mathbf{v}$ is a nonzero vector in the null space of $\mathbf{A}-\lambda\mathbf{I}$. A matrix sends a nonzero vector to zero only when it crushes a whole line, and then it flattens the plane. By The determinant as area, flattening is exactly a determinant of zero. So $\lambda$ is an eigenvalue of $\mathbf{A}$ exactly when

That is what the picture at the top of the page hunts for. The parallelogram is $\mathbf{A}-\lambda\mathbf{I}$ applied to the unit square, and its signed area is $\det(\mathbf{A}-\lambda\mathbf{I})$. Drag $\lambda$ yourself and stop where the area reads 0.

The characteristic polynomial

For a $2\times2$ matrix the determinant is quick to write out. Subtracting $\lambda\mathbf{I}$ takes $\lambda$ off each diagonal entry:

The middle coefficient is the sum of the diagonal entries, the trace, written $\mathrm{tr}\,\mathbf{A}$. The last is $\det\mathbf{A}$. So for every $2\times2$ matrix

This is the characteristic polynomial of $\mathbf{A}$, and setting it to zero is the characteristic equation. The quadratic formula solves it:

For the matrix in the picture, $\mathrm{tr}\,\mathbf{A}=1.5+1=2.5$ and $\det\mathbf{A}=1.5\cdot1-1\cdot0.5=1$. The equation is $\lambda^2-2.5\lambda+1=0$, and

so $\lambda=2$ or $\lambda=0.5$: the two stretch factors of the eigenvectors page, found without a picture.

An $n\times n$ matrix gives a polynomial of degree $n$ in the same way, so it has $n$ eigenvalues, counted with repeats. Some of them may be complex numbers.

Two checks

Write the polynomial from its roots: $(\lambda-\lambda_1)(\lambda-\lambda_2)=\lambda^2-(\lambda_1+\lambda_2)\,\lambda+\lambda_1\lambda_2$. Match it with the form above, coefficient by coefficient:

The same holds for every size: the eigenvalues add up to the trace and multiply to the determinant. Here $2+0.5=2.5$ and $2\cdot0.5=1$. These two checks catch most slips in a hand calculation.

Eigenvectors from the null space

Each eigenvalue gives its eigenvectors as the null space of $\mathbf{A}-\lambda\mathbf{I}$, found as in The null space.

For $\lambda=2$:

The second row is minus the first, so one equation is left: $-0.5v_1+v_2=0$. Take $v_2=1$; then $v_1=2$, and $\mathbf{v}=(2,1)$.

For $\lambda=0.5$: $\mathbf{A}-0.5\mathbf{I}=\begin{bmatrix}1&1\\0.5&0.5\end{bmatrix}$ leaves $v_1+v_2=0$, so $\mathbf{v}=(1,-1)$. Both match the lines the sweep found.

Repeated and missing roots

The polynomial can surprise you in two ways.

A repeated root. The shear $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ has trace 2 and determinant 1, so its polynomial is $\lambda^2-2\lambda+1=(\lambda-1)^2$. The root 1 appears twice. But $\mathbf{A}-\mathbf{I}=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ has only the line of $(1,0)$ in its null space, so there is one line of eigenvectors, not two. Similar matrices and the Jordan form looks at matrices like this.

No real root. The quarter turn $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ has trace 0 and determinant 1, so $\lambda^2+1=0$. No real number solves it, which matches the picture: the turn keeps no line. The roots are the complex numbers $\pm j$, and Complex eigenvalues reads them as a turn.

A bigger matrix in blocks. When a matrix splits into blocks along its diagonal, each block keeps its own eigenvalues. Take

The first coordinate is left alone, stretched by 2, so 2 is an eigenvalue with eigenvector $(1,0,0)$. The block $\begin{bmatrix}3&4\\4&9\end{bmatrix}$ has trace 12 and determinant $27-16=11$, so $\lambda^2-12\lambda+11=(\lambda-1)(\lambda-11)$: eigenvalues 1 and 11. Check: $\mathbf{A}(0,1,2)=(0,11,22)=11\,(0,1,2)$. All three add to $2+1+11=14$, the trace, and multiply to $2\cdot1\cdot11=22$, the determinant.

Poles are eigenvalues

The filter $y[n]=1.5\,y[n-1]-0.56\,y[n-2]+x[n]$ remembers its last two outputs. Stack them as a state vector, $(y[n-1],\,y[n-2])$. With no input, one step of the filter turns it into $(y[n],\,y[n-1])$, and that step is the matrix

The first row is the filter’s equation; the second only passes $y[n-1]$ along. This is the companion matrix of the filter. Its trace is 1.5 and its determinant 0.56, so its characteristic polynomial is $\lambda^2-1.5\lambda+0.56=(\lambda-0.8)(\lambda-0.7)$.

Watch the two sets of marks: they sit at the same two numbers. The eigenvalues of the companion matrix are the filter’s poles.

Worked example

1. The matrix of the eigenvectors page. $\mathbf{A}=\begin{bmatrix}1.5&1\\0.5&1\end{bmatrix}$ has $\mathrm{tr}\,\mathbf{A}=2.5$ and $\det\mathbf{A}=1.5-0.5=1$. The characteristic equation $\lambda^2-2.5\lambda+1=0$ gives $\lambda=(2.5\pm1.5)/2$, so 2 and 0.5. The null space of $\mathbf{A}-2\mathbf{I}$ is the line of $(2,1)$, and that of $\mathbf{A}-0.5\mathbf{I}$ is the line of $(1,-1)$.

2. Elimination changes eigenvalues. Row 2 $-\ \tfrac13$ row 1 gives $\begin{bmatrix}1.5&1\\0&2/3\end{bmatrix}$, with eigenvalues 1.5 and $0.667$ but the same product, 1.

3. A repeated root. The shear has $(\lambda-1)^2$ and one line of eigenvectors, through $(1,0)$.

4. Blocks. $\begin{bmatrix}2&0&0\\0&3&4\\0&4&9\end{bmatrix}$ has eigenvalues 2, 1 and 11, with eigenvectors $(1,0,0)$, $(0,-2,1)$ and $(0,1,2)$.

5. No real root. The quarter turn gives $\lambda^2+1=0$.

6. Poles. $\lambda^2-1.5\lambda+0.56=(\lambda-0.8)(\lambda-0.7)$: check $0.8+0.7=1.5$ and $0.8\cdot0.7=0.56$.

Try these before you read the answers.

• Eigenvalues of $\begin{bmatrix}2&1\\1&2\end{bmatrix}$? Trace 4, determinant 3, so 3 and 1.
• A matrix has trace 5 and determinant 6. Its eigenvalues? 2 and 3.
• Eigenvalues of $\begin{bmatrix}3&5\\0&4\end{bmatrix}$? 3 and 4: it is triangular.
• If $\det\mathbf{A}\neq0$, can 0 be an eigenvalue? No: 0 is an eigenvalue exactly when $\det(\mathbf{A}-0\,\mathbf{I})=\det\mathbf{A}=0$.

Where you’ll meet this

Stable filters. A filter’s poles are the eigenvalues of its companion matrix. This filter is stable because both, 0.8 and 0.7, lie inside the unit circle, so the state dies away in the long run.

Vibrating structures. The natural frequencies of a structure built from masses and springs come from the eigenvalues of its stiffness matrix, as Symmetric matrices shows.

How software does it. Programs do not find eigenvalues by solving the characteristic polynomial. For large matrices its roots are extremely sensitive to rounding in the coefficients, so programs use repeated matrix steps that home in on the eigenvalues instead. The polynomial stays the right tool for a $2\times2$ by hand, and for understanding.

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