Eigenvalues and Eigenspaces

A direction that only scales

For a square matrix $A$ , a nonzero vector $v$ is an eigenvector with eigenvalue $\lambda$ if $Av=\lambda v$ . Rearranging gives $(A-\lambda I)v=0$ . A nonzero solution exists exactly when $\det(A-\lambda I)=0$ , the characteristic equation.

For each eigenvalue, its eigenspace is $N(A-\lambda I)$ , including zero. Zero belongs to the eigenspace as a subspace but is never itself called an eigenvector. Any nonzero scalar multiple of an eigenvector represents the same eigen-direction.

Visual guide

VISUAL GUIDEAn eigenvector keeps its direction

A(1, 1)

For A = diag(2, 1), the horizontal eigenvector doubles and the vertical eigenvector stays unchanged. A diagonal input (1, 1) becomes (2, 1), changing direction, so it is not an eigenvector of this map.

Worked example: two preserved directions

For $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ , the characteristic polynomial is $(2-\lambda)^2-1=(\lambda-3)(\lambda-1)$ . For $\lambda=3$ , solving $(A-3I)v=0$ gives vectors proportional to $(1,1)^T$ . For $\lambda=1$ , they are proportional to $(1,-1)^T$ .

The matrix stretches one diagonal direction by three and leaves the other unchanged. A generic vector combines both behaviors. Solving only the characteristic polynomial does not provide the eigenvectors; each null-space calculation is still needed.

Worked example: a zero eigenvalue

For $A=\operatorname{diag}(4,0)$ , eigenvalues are $4$ and $0$ . The vertical eigenvectors map to zero. Therefore $A$ is singular. In general, zero is an eigenvalue exactly when a square matrix is not invertible.

Eigenvalues of a triangular matrix are its diagonal entries, counted with algebraic multiplicity. The sum of eigenvalues equals the trace and their product equals the determinant, counting multiplicity over the complex numbers. These are useful checks, not replacements for finding eigenspaces.

Practice

Find eigenpairs of $\operatorname{diag}(2,5)$ .
Is every nonzero vector an eigenvector of $3I$ ?
Can a real matrix have nonreal eigenvalues?

Show worked solutions

Eigenvalue $2$ has direction $(1,0)^T$ ; eigenvalue $5$ has direction $(0,1)^T$ .
Yes: $3Iv=3v$ for every nonzero $v$ .
Yes. A $90^\circ$ plane rotation has characteristic equation $\lambda^2+1=0$ , giving $\lambda=\pm i$ over the complex numbers.