Computing Eigenvectors¶

Question

How do we find the eigenvectors of a matrix?

Example¶

Once we have the eigenvalues of a matrix, the next logical step is to find the eigenvectors for each of these eigenvalues. Let us continue with the example that we have been working with so far:

\[ A = \begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix} \]

We already know that the eigenvalues are $2$ and $3$. What are the eigenvectors corresponding to $\lambda = 2$? First, we note that if $x$ is an eigenvector corresponding to $\lambda = 2$, then:

\[ \begin{aligned} (A - 2I) x &= 0\\\\ \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix} \end{aligned} \]

From this, we have $x_1 + x_2 = 0$. $x_2$ is an independent variabe and $x_1$ is a dependent variable. The rank of this matrix is $1$, so every eigenvector is of the form:

\[ k\begin{bmatrix} -1\\ 1 \end{bmatrix} \]

Therefore, the eigenspace corresponding to the eigenvalue $\lambda = 2$ is given by:

\[ \text{span}\left( \left\{ \begin{bmatrix}-1\\1\end{bmatrix} \right\} \right) \]

If we repeat the same process for $\lambda = 3$, the eigenspace turns out to be:

\[ \text{span}\left( \left\{ \begin{bmatrix}1\\0\end{bmatrix} \right\} \right) \]

What have we done so far? We have just computed the nullspace of $A - \lambda I$. So, this is the recipe to follow: compute a basis for the nullspace of the matrix $A - \lambda I$. The span of this basis (exculding the zero vector) is the set of all eigenvectors of the matrix $A$ corresponding to the eigenvalue $\lambda$. We have already discussed an algorithm to compute the basis for the nullspace of a matrix.

Note

For any matrix $A$, the nullspace of $A$ (excluding the vector $0$) has all the eigenvectors of $A$ with eigenvalue $0$. This is because $Ax = 0 = 0x$ if $x \in N(A)$.

Special examples¶

Let us take up some special examples and see how the eigenvectors and eigenvalues look.

Diagonal matrix¶

If $D$ is a diagonal matrix, then the eigenvalues are the diagonal entries of the matrix. We can express it as: $$ D = \text{diag}(\lambda_1, \cdots, \lambda_n) $$ The characteristic polynomial is $(\lambda - \lambda_1) \cdots (\lambda - \lambda_n)$. The eigenvectors of $D$ are going to be elements of the standard basis $\{e_1, \cdots, e_n\}$. Diagonal matrices are the most interesting when it comes to the study of eigenvalues and eigenvectors because of the ease with which we can read off the eigenvalues.

Identity matrix¶

If $I$ is an identity matrix of size $n \times n$, then every vector in $\mathbb{R}^{n}$ is an eigenvector with eigenvalue $1$. The characteristic polynomial for the identity matrix is of the form $(\lambda - 1)^n$. There is exactly one eigenvalue that is repeated $n$ times. This is en extreme case of the diagonal matrix.

Projection matrix¶

If $P$ is a projection matrix corresponding to a matrix $A$, then every vector that is in the column space of $A$ is an eigenvector with eigenvalue $1$. To see why this is true, note that $Px = x$ for any vector in the column space of $A$. Also, every vector that is orthogonal to the column space of $A$ will be an eigenvector of $P$ with eigenvalue $0$. This is because $Px = 0$ for any vector in this space.