How to find eigenvalues and eigenvectors of a three-by-three matrix

Finding eigenvalues in three dimensions follows the same process as in two dimensions. The characteristic determinant is larger, but once we factor it, we solve one homogeneous system for each eigenvalue. Let’s work through all three eigenspaces of a matrix.

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Find and factor the characteristic polynomial

$A=\left[\begin{array}{rrr}-1& 1& 0\\ 1& 2& 1\\ 0& 3& -1\end{array}\right]$

$\begin{array}{l}\det (\lambda I-A)=(\lambda +1)({\lambda }^2-\lambda -6)\\ =(\lambda +1)(\lambda -3)(\lambda +2)\end{array}$

Setting the determinant equal to zero gives the eigenvalues $-2$, $-1$, and $3$.

Solve the homogeneous system for each eigenvalue

For each value, solve $(A-\lambda I)v=0$. The solutions form its eigenspace. One nonzero vector from each space gives the following bases.

A basis vector for each eigenspace

Eigenvalue	Basis vector
$-2$	$(-1,1,-3)$
$-1$	$(1,0,-1)$
$3$	$(1,4,3)$

Check one eigenvector in the original matrix

For $v=(1,4,3)$, the product is $Av=(3,12,9)=3v$. So this vector is scaled by the eigenvalue $3$, as required.

Any nonzero multiple of a listed basis vector is another eigenvector for that same eigenvalue. The zero vector belongs to every eigenspace but is excluded from the definition of an eigenvector.

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