Eigenvalues and Eigenvectors

Eigenvectors of a 2×2 Matrix

For each eigenvalue λ, row-reduce (A - λI) to find the null space. The null space basis is the eigenvector. Verify A·v = λ·v.

Example

Reduce each eigenspace matrix and read eigenvectors.

highlighted = computed this step

Step 1 — Set up

Set up the given matrix data.

A=[4123]A=\begin{bmatrix}4&1\\2&3\end{bmatrix}

Step 2 — Reduce lambda 1 matrix

Row-reduce A-lambda I for the eigenspace.

B=[2121]B=\begin{bmatrix}2&1\\\hl{2}&\hl{1}\end{bmatrix}

Step 3 — Eigenvector for lambda 1

Solve the eigenspace equation for the first eigenvector.

v=[1-2]v=\begin{bmatrix}\hl{1}\\\hl{-2}\end{bmatrix}

Step 4 — Reduce lambda 2 matrix

Row-reduce A-lambda I for the second eigenspace.

C=[−112-2]C=\begin{bmatrix}-1&1\\\hl{2}&\hl{-2}\end{bmatrix}

Step 5 — Eigenvector for lambda 2

Read eigenvectors for lambda 2 and lambda 5.

v=[1−2],w=[11]v=\begin{bmatrix}1\\-2\end{bmatrix},\quad w=\begin{bmatrix}1\\1\end{bmatrix}
eigenvector A nonzero vector v is an eigenvector of A for eigenvalue λ if Av = λv. Eigenvectors are found by solving (A - λI)v = 0 — equivalently, by row-reducing A - λI and reading off the free variable.