Eigenvectors of Square Matrices

Eigenvectors of Square Matrices

Recall from the Eigenvalues of Square Matrices page that if $A$ is an $n \times n$ matrix and $\lambda \in \mathbb{R}$ then the characteristic polynomial for $A$ is the polynomial:

(1)
\begin{align} \quad \det (A - \lambda I) = 0 \end{align}

The roots/solutions of the characteristic polynomial are called the eigenvalues of $A$.

Now recall that we originally began with the matrix equation $Ax = \lambda x$ which is equivalent to the matrix equation $(A - \lambda I)x = 0$. We noted that this matrix equation has the trivial solution $x = 0$. If $\det (A - \lambda I) \neq 0$ the $(A - \lambda I)x = 0$ has only the trivial solution. However, we noted that if $\det (A - \lambda I) = 0$ (which happens when $\lambda$ is an eigenvalue of $A$) then there are infinitely many solutions $x$ corresponding to this $\lambda$. These nontrivial solutions are defined below.

 Definition: Let $A$ be an $n \times n$ matrix. If $\lambda$ is an eigenvalue of $A$ then a corresponding nonzero vector $v$ is called an Eigenvector of $A$ corresponding to $\lambda$ if $(A - \lambda I)v = 0$.

Note that eigenvectors of $A$ corresponding to an eigenvalue $\lambda$ are not unique as there are infinitely many.

Also note that the zero vector is not an eigenvector.

For example, consider the following matrix:

(2)
\begin{align} \quad A = \begin{bmatrix} 2 & 7 \\ -1 & -6 \end{bmatrix} \end{align}

We have previously seen that the eigenvalues of $A$ are $\lambda_1 = 1$ and $\lambda_2 = -5$. Let's find a corresponding eigenvalue to $\lambda_1 = 1$. We consider the following matrix equation:

(3)
\begin{align} \quad (A - \lambda_1I)x &= 0 \\ \quad \left ( \begin{bmatrix} 2 & 7 \\ -1 & -6 \end{bmatrix} - 1 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right ) \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ \quad \begin{bmatrix} 1 & 7 \\ -1 & -7 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \end{bmatrix} \end{align}

This matrix equation gives us the following system of equations:

(4)
\begin{align} \quad x_1 + 7x_2 &= 0 \\ -x_1 - 7x_2 &= 0 \end{align}

By letting $x_1 = 1$ we get that $x_2 = -\frac{1}{7}$. So a corresponding eigenvector of $A$ to the eigenvalue $\lambda_1 = 1$ is:

(5)
\begin{align} \quad v = \begin{bmatrix} 1 \\ -\frac{1}{7} \end{bmatrix} \end{align}

In fact, for each $t \in \mathbb{R} \setminus \{ 0 \}$, the following is a corresponding eigenvector of $A$ to the eigenvalue $\lambda_1 = 1$

(6)
\begin{align} \quad v_t = \begin{bmatrix} 7t \\ -t \end{bmatrix} \end{align}