Discover Linear Algebra: A first course in linear algebra

First, we form the matrix

Then we compute its determinant, to obtain the characteristic polynomial of $A\text{:}$

The eigenvalues are the roots of the characteristic polynomial, so we have two eigenvalues ${\lambda }_1=-1$ and ${\lambda }_2=3\text{.}$

The eigenspace $E_{{\lambda }_1}(A)$ is the same as the null space of the matrix ${\lambda }_{1}I-A\text{,}$ so we determine a basis for the eigenspace by row reducing:

This system requires one parameter to solve, as $x_2$ is free. Setting $x_2=t\text{,}$ the general solution in parametric form is

Associated to the single parameter we get a single basis vector, so that

In particular, we have

Now move on to the next eigenvalue. Again, we determine a basis for $E_{{\lambda }_2}(A)$ by row reducing ${\lambda }_{2}I-A\text{:}$

Again, $x_2$ is free. One parameter means one basis vector, so again

The first row of the reduced matrix says $x_1=-2x_2\text{,}$ so we have

Start with

and compute the characteristic polynomial,

The eigenvalues are ${\lambda }_1=0$ and ${\lambda }_2=2\text{.}$

The eigenspace $E_{{\lambda }_1}(A)$ is the null space of $0I-A=-A\text{,}$ so row reduce:

Notice that the null space of $0I-A=-A$ is the same as the null space of $A\text{,}$ since our first step in row reducing $-A$ could be to multiply each row by $-1\text{.}$ Since this homogeneous system has nontrivial solutions, $A$ must be singular.

The homogeneous system $({\lambda }_{1}I-A)\mathbf{x}=\mathbf{0}$ requires one parameter, so

The variable $x_3$ is free, and the nonzero rows of the reduced matrix tell us $x_1=(2/3)x_3$ and $x_2=(4/3)x_3\text{.}$ Setting $x_3=t\text{,}$ our general solution in parametric form is

However, to avoid fractions in our basis vector, we may wish to pull out an additional scalar:

giving us

Now row reduce ${\lambda }_{2}I-A\text{:}$

This time we have two free variables, so $\dim (E_{{\lambda }_2}(A))=2\text{.}$ Setting $x_1=s$ and $x_3=t\text{,}$ the general solution in parametric form is

giving us

For $\lambda =1\text{:}$

For $\lambda =2\text{:}$

For $\lambda =3\text{:}$

For ${\lambda }_1=2\text{:}$

For ${\lambda }_2=-1\text{:}$

Let’s do a $4\times 4$ example to demonstrate. Consider

To obtain the characteristic polynomial, we want to compute the determinant of

Let’s row reduce a bit first:

In our first step above, we performed two operations: swapping rows and multiplying a row by $-1\text{.}$ Both of these operations change the determinant by a factor of $-1\text{,}$ so the two effects cancel out. Our other operations in the second step above do not affect the determinant, so the determinant of this third matrix above will be equal to the characteristic polynomial of $A\text{.}$

Now, we cannot divide a row by zero. So we should not divide either the second or fourth rows by $\lambda -1$ in an attempt to obtain the next leading one, because we would inadvertently be dividing by zero in the case $\lambda =1\text{.}$ However, we can still simplify one step further, even without a leading one:

This last matrix is not quite upper triangular, but it’s close enough that we can proceed by cofactors from here.

We now see that the eigenvalues are ${\lambda }_1=1$ and ${\lambda }_2=-2\text{.}$

For ${\lambda }_1=1\text{:}$

For ${\lambda }_2=-2\text{,}$ again starting from (✶):