Discover Linear Algebra

Example 29.5.1. Scalar-triangularization procedure: a small matrix.

Is matrix

similar to some upper triangular matrix?

We begin by investigating the eigenvalues of $A\text{.}$ Compute the characteristic polynomial:

So $A$ has eigenvalue $\lambda = -1\text{,}$ with algebraic multiplicity $3\text{.}$ From our analysis in Subsection 29.4.3, since $A$ has a single eigenvalue, we are confident it can be made similar to a scalar triangular matrix.

Next, compute a basis for the eigenspace $E_{-1}(A)\text{.}$

As you can see, the geometric multiplicity of the eigenvalue $\lambda=-1$ is not equal to the algebraic multiplicity, so $A$ is not diagonalizable. So we continue with the scalar-triangularization procedure (Procedure 29.4.1).

Now compute a basis for the generalized eigensubspace of degree $2\text{:}$

Note that we have included our basis vector for $E_{-1}(A)$ as the first vector in our basis for $E_{-1}^2(A)\text{.}$

We are only up to two linearly independent vectors, so we need to keep going by computing a basis for $E_{-1}^3(A)\text{.}$ But if you compute $(-I-A)^3$ you will find that it is the zero matrix. What does this mean? Every vector in $\R^3$ is in the null space of the zero matrix, so we have $E_{-1}^3(A) = \R^3\text{.}$

However, we don't want to choose just any basis for $\R^3\text{,}$ as we need to maintain independence from the basis vectors for $E_{-1}^2(A)$ that we already have. That is, we need to choose a vector in $\R^3$ that is not already in the span of our existing two generalized eigenvectors (Proposition 18.5.6). But since $\dim E_{-1}^2(A) = 2\text{,}$ we know that it is not possible for all three of the standard basis vectors for $\R^3$ to be in $E_{-1}^2(A)\text{,}$ and so we can just choose any of them that is not. We will choose the third standard basis vector.

We now have our column vectors for a suitable transition matrix $P\text{:}$

Finally, let's determine the form matrix $\inv{P}AP\text{.}$ We already know that $A\uvec{p}_1 = -\uvec{p}_1\text{,}$ since $\uvec{p}_1$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda = -1\text{.}$ Next, compute $\bbrac{A-(-1)I}\uvec{p}_2$ and you will find that it is equal to $\uvec{p}_1\text{.}$ Thus, the entry above the diagonal in the second column must be $1\text{.}$ Lastly, compute

Thus, the entries above the diagonal in the third column are $2\text{,}$ $-1\text{.}$ Putting all this together yields

Example 29.5.2. Scalar-triangularization procedure: a larger matrix.

Is matrix

similar to some upper triangular matrix?

Again, we start by investigating eigenvalues. If you compute the characteristic polynomial for $A\text{,}$ you will find that it factors as $c_A(\lambda) = (\lambda - 3)^5\text{,}$ so that $A$ has a single eigenvalue $3$ of multiplicity $5\text{.}$ So from our analysis in Subsection 29.4.3, we are confident that $A$ can be made similar to a scalar triangular matrix.

Next, compute a basis for the eigenspace $E_3(A)\text{.}$

We need five vectors for the columns of the transition matrix $P\text{,}$ so this is not enough.

Next, compute a basis for $E_3^2(A)\text{.}$

The above spanning vectors are the ones arrived at from row reducing $(3 I - A)^2\text{.}$ But remember that we want to use our already obtained vectors from $E_3(A)\text{,}$ so we just need to choose two of these four that are independent from our eigenvectors. Those first two standard basis vectors will work, so take

Still not enough vectors, but $(3 I - A)^3 = \zerovec\text{,}$ so we can choose any vector in $\R^5$ that is not in the span of the four we already have. The third standard basis vector for $\R^5$ will work, and we now have

Now place these matrices in the transition matrix

To determine the form matrix, we can reduce $\left[\begin{array}{c|c} P \amp AP \end{array}\right]$ as

obtaining