Computing Eigenvectors

Once you know an eigenvalue $\lambda$, finding the eigenvectors is a null space problem: row reduce $A - \lambda I$ and solve $(A - \lambda I)\mathbf{x} = \mathbf{0}$. The solution set — all vectors satisfying this equation, including $\mathbf{0}$ — forms a subspace called the eigenspace $E_\lambda$. Every nonzero vector in $E_\lambda$ is an eigenvector for $\lambda$.

To see this concretely, take $A = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix}$ with eigenvalues $\lambda_1 = 5$, $\lambda_2 = 2$. For $\lambda_1 = 5$: form $A - 5I = \begin{pmatrix}-1 & 1 \\ 2 & -2\end{pmatrix}$. Row reduce to $\begin{pmatrix}1 & -1 \\ 0 & 0\end{pmatrix}$, giving $x_1 = x_2$. So any vector of the form $t(1,1)^t$ is an eigenvector. For $\lambda_2 = 2$: $A - 2I = \begin{pmatrix}2 & 1 \\ 2 & 1\end{pmatrix}$ row reduces to $\begin{pmatrix}1 & 1/2 \\ 0 & 0\end{pmatrix}$, giving $x_1 = -\frac{1}{2}x_2$. Eigenvector: $t(1, -2)^t$.

An important property: eigenvectors for different eigenvalues are always linearly independent. You can never express one as a linear combination of the others. This makes sense geometrically — they point along completely different "natural axes" of the transformation.