C10 - Eigenvalues and eigenvectors

Characteristic equation in 2D and 3D

Which directions does $\mathbf{A}=\begin{pmatrix}3&1\\1&3\end{pmatrix}$ stretch without turning them away from their original line? Try two vectors:

Each output is a scalar multiple of its input. The scale depends on the direction: 4 along one line and 2 along the other.

How can you find the scales without guessing vectors? Rearrange to $(\mathbf{M}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}$. If $\mathbf{M}-\lambda\mathbf{I}$ were invertible, multiplying by its inverse would force $\mathbf{v}=\mathbf{0}$. A nonzero solution therefore requires this matrix to be singular. Conversely, a singular square matrix has a nonzero solution to its homogeneous equations.

Use the determinant methods you already know. A row with zeros can make the three-dimensional calculation much shorter.

Keep that two-dimensional block and add a third coordinate scaled by $-1$. The extra direction adds one factor to the determinant.

Real matrices need not have only real eigenvalues. For example, a quarter-turn matrix $\mathbf{R}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has characteristic equation $\lambda^2+1=0$, with roots $\pm\mathrm{i}$. These are valid complex eigenvalues, but they do not describe real directions in the plane. For a real matrix, non-real roots occur in conjugate pairs because its characteristic polynomial has real coefficients. Geometrical directions below use real eigenvectors and real eigenvalues.

Find the characteristic equation for a fresh matrix.

Finding corresponding eigenspaces

An eigenvalue tells you the scale; solving $(\mathbf{M}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}$ tells you which vectors receive that scale. These are homogeneous equations, so choose a convenient free variable rather than seeking a unique vector.

The equations are dependent when $\lambda$ is an eigenvalue. An equation reducing to $0=0$ adds no restriction; it does not require a free variable to be zero.

Use the same eigenvalues and vectors in the first two coordinates, but now retain all three equations. The extra coordinate cannot simply be left arbitrary.

An eigenspace can be a plane. For $\operatorname{diag}(2,2,5)$, the repeated root 2 gives only $z=0$, so every $(s,t,0)^{\mathsf T}$ lies in its eigenspace. All except $(0,0,0)^{\mathsf T}$ are eigenvectors. The two free variables provide two independent directions. A repeated root does not automatically provide that many independent eigenvectors: you must solve the equations.

If complex vectors are allowed, the quarter-turn matrix from part 1 has eigenvectors too. Its equations give $-y=\mathrm{i}x$ for $\lambda=\mathrm{i}$, so $(1,-\mathrm{i})^{\mathsf T}$ works; conjugating gives $(1,\mathrm{i})^{\mathsf T}$ for $-\mathrm{i}$. Directly, $\mathbf{R}(1,-\mathrm{i})^{\mathsf T}=(\mathrm{i},1)^{\mathsf T}=\mathrm{i}(1,-\mathrm{i})^{\mathsf T}$. These are complex vectors, not arrows representing real points of the plane.

Find a two-dimensional eigenspace inside three-dimensional space.

Geometrical meaning of eigenvalues

For a real eigenvector, the equation $\mathbf{M}\mathbf{v}=\lambda\mathbf{v}$ describes motion along its original line. Length is multiplied by $|\lambda|$; the sign tells you what happens to the direction.

For instance, $\mathbf{H}=\begin{pmatrix}1&2\\2&1\end{pmatrix}$ sends $(1,1)^{\mathsf T}$ to $(3,3)^{\mathsf T}$, but sends $(1,-1)^{\mathsf T}$ to $(-1,1)^{\mathsf T}$. The same transformation stretches one direction and reverses the other without changing its length. Eigenvalues reveal that mixture far more clearly than the four entries alone.

[DIAGRAM: asset_name: Two directions under one matrix; asset_slug: aqa7367-eigen-directions; caption: One matrix stretches vectors on y equals x by three and reverses vectors on y equals minus x; description: Two equal-scale coordinate panels show the matrix with rows one two and two one. In the left panel a blue arrow from the origin to (1,1) and a clay arrow to (3,3) lie on the dashed line y equals x. In the right panel a blue arrow to (1,minus 1) and a clay arrow to (minus 1,1) lie on the dashed line y equals minus x. Labels identify the original and image and eigenvalues three and minus one.]

Linearity extends this description to every point $t\mathbf{v}$: its image is $\lambda t\mathbf{v}$. If $\lambda\ne0$, the whole line is mapped onto itself. When $\lambda=0$, the line collapses to the origin instead, so its image is contained in the original line but is not the whole line. Eigenvalue 1 gives a line of invariant points; another nonzero eigenvalue gives an invariant line on which nonzero points move.

Not every vector is an eigenvector. For the same $\mathbf{H}$, $(1,0)^{\mathsf T}\mapsto(1,2)^{\mathsf T}$, which is not a scalar multiple of $(1,0)^{\mathsf T}$. A quarter-turn has no real eigenvector at all: every nonzero real arrow is turned off its original line. Its complex eigenvalues are consistent with that geometry.

The opening matrix $\mathbf{A}$ has eigenvectors along $y=x$ and $y=-x$. These directions receive scales 4 and 2. Any nonzero multiple of either displayed vector therefore answers the opening question.

Distinguish a fixed direction from fixed points.

Quick Recap

• An eigenvector is nonzero and satisfies $\mathbf{M}\mathbf{v}=\lambda\mathbf{v}$; an eigenvalue may be zero.
• Solve $\det(\mathbf{M}-\lambda\mathbf{I})=0$, then solve the homogeneous equations separately for each root.
• An eigenspace includes zero; its nonzero members are the corresponding eigenvectors.
• For real vectors, $|\lambda|$ gives the length scale and a negative sign reverses direction. Complex eigenvectors do not represent real line directions.

The opening matrix stretches directions $(1,1)^{\mathsf T}$ and $(1,-1)^{\mathsf T}$ by 4 and 2, respectively, without turning them off their original lines.