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