4. How to find eigenvalues and eigenvectors

As a warmup, first recall that a square matrix $A$ is ``singular'' if many equivalent conditions occur: $\det A = 0$, $A$ has no inverse, $A$ is $n \times n$ but has rank less than $n$, the nullspace of $A$ has dimension $> 0$ (so $A$v$=$   0 has a nonzero solution v), and so on. We'll need the first and last of these conditions. In other words, remember that $A$v$=$   0 has a nonzero solution v if and only if $\det A = 0$.

Here is how to invent the method:

Given $A$, start with the equation $A$v$= \lambda$   v. At this point, we don't yet know what the possible values of $\lambda$ might be, or what nonzero vector v might go with $\lambda$.

As in ordinary algebra, put everything on one side: $A$   v$- \lambda$   v$=$   0.

Also as in ordinary algebra, it is tempting to try to factor out v, but this doesn't work because $(A - \lambda)$ doesn't make sense. You can't have a scalar minus a matrix!

A trick to get around this is to replace $\lambda$   v by $\lambda I$   v, where $I$ is an identity matrix. If $A$ is $2 \times 2$, for example, then $\lambda I = \matp{cc}{\lambda&0\\ 0&\lambda}$. So we can write $A$   v$- \lambda I$   v$=$   0 and then get $(A - \lambda I)$v$=$   0.

Example: For $A = \matp{rr}{4&1\\ 1&4}$, $A - \lambda I = \matp{rr}{4&1\\ 1&4}- \matp{cc}{\lambda&0\\ 0&\lambda}$
$= \matp{cc}{(4-\lambda)&1\\ 1&(4-\lambda)}$.

Next, from the warmup we notice that $(A - \lambda I)$v$=$   0 is possible for a nonzero vector v if and only if $\det(A - \lambda I) = 0$. In the example, we see that $\det(A - \lambda I) = (4 - \lambda) ^ 2 - 1 ^ 2 = \lambda ^ 2 - 8 \lambda + 16 - 1 = \lambda ^ 2 - 8 \lambda + 15$. We need to set this expression equal to 0 and solve.

In other words, the values of $\lambda$ we are looking for are the roots of the polynomial $\det(A - \lambda I)$. This polynomial is called the characteristic polynomial of $A$, denoted here by $p _ A(\lambda)$.

Find the roots using algebra: $\lambda ^ 2 - 8 \lambda + 15 = 0$ factors to $(\lambda - 5)(\lambda - 3) = 0$, so $\lambda = 5$ or $\lambda = 3$. Therefore the eigenvalues of $A$ are 5 and 3 (in this example).

To find eigenvectors, consider each eigenvalue separately. For the eigenvalue 5, for example, we need to find a nonzero vector v with $A$v$= 5$v, or equivalently, $(A - 5 I)$v$=$   0. This is just a set of linear equations!

In the example, for $\lambda = 5$ we have $A - 5I = \matp{cc}{4&1\\ 1&4}- \matp{cc}{5&0\\ 0&5}= \matp{rr}{-1&1\\ 1&-1}$, so if we write v$= \matp{c}{x\\ y}$, $(A - 5 I)$v$=$   0 becomes
$\matp{rr}{-1&1\\ 1&-1}\matp{c}{x\\ y}= \matp{c}{0\\ 0}$, or equivalently, $\left \{ \begin{array}{rrr} -x+y & = & 0\\ x-y & = & 0 \end{array} \right .$.

This set of equations is singular, with both equations giving the same information. Singular systems of equations may have seemed bad in the past, but here they are good, because we do want a nonzero solution. Also, it's no surprise, because we purposely chose $\lambda$ so $A - \lambda I$ is singular.

Choose any nonzero solution, say $x=1$, $y=1$. In other words, we have found that $\matp{c}{1\\ 1}$ is an eigenvector.

Similarly, for the eigenvalue $\lambda = 3$, we get $A - 3I = \matp{cc}{4&1\\ 1&4}- \matp{cc}{3&0\\ 0&3}= \matp{rr}{1&1\\ 1&1}$, so $(A-3I)$v$=$   0 becomes $\matp{rr}{1&1\\ 1&1}\matp{c}{x\\ y}= \matp{c}{0\\ 0}$, or equivalently, $\left \{ \begin{array}{lll} x+y & = & 0\\ x+y & = & 0 \end{array} \right .$. Let's choose the nonzero solution $x=-1$, $y=1$. We have found that $\matp{r}{-1\\ 1}$ is an eigenvector. ( $\matp{r}{1\\ -1}$ would also do.)

To summarize the method, for given $A$:

Step 1. Find the characteristic polynomial of $A$ by expanding $\det(A - \lambda I)$. (In the next section there is a shortcut for this when $A$ is $2 \times 2$.)

Step 2. Set the characteristic polynomial $= 0$ and find solutions (the roots of the characteristic polynomial). These are the eigenvalues of $A$.

Step 3. For each eigenvalue $\lambda$, treat $(A - \lambda I)$v$=$   0 as a set of linear equations and find a nonzero solution, which is a corresponding eigenvector.