Example 29.5.1. A small matrix.
Is matrix
\begin{equation*}
A = \begin{abmatrix}{rrr} -6 \amp -14 \amp 5\\ 1 \amp 2 \amp -1\\ -2 \amp -5 \amp 1 \end{abmatrix}
\end{equation*}
similar to some upper triangular matrix?
We begin by investigating the eigenvalues of \(A\text{.}\) Compute the characteristic polynomial:
\begin{equation*}
\lambda I - A = \begin{bmatrix} \lambda+6 \amp 14 \amp -5 \\ -1 \amp \lambda-2 \amp 1 \\ 2 \amp 5 \amp \lambda-1 \end{bmatrix} \text{,}
\end{equation*}
\begin{align*}
c_A(\lambda) \amp= \det (\lambda I - A) \\
\amp= \lambda^3-3\lambda^2-3\lambda-1 \\
\amp = (\lambda+1)^3 \text{.}
\end{align*}
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 \(E_{-1}(A)\text{.}\)
\begin{gather*}
(-1) I - A =
\begin{abmatrix}{rrr}
5 \amp 14 \amp -5 \\
-1 \amp -3 \amp 1 \\
2 \amp 5 \amp -2
\end{abmatrix}
\qquad \rowredarrow \qquad
\begin{abmatrix}{ccr}
1 \amp 0 \amp -1 \\
0 \amp 1 \amp 0 \\
0 \amp 0 \amp 0
\end{abmatrix}\\
\\
\implies \qquad
E_{-1}(A) = \Span \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \right\}\text{.}
\end{gather*}
As you can see, the geometric multiplicity of the eigenvalue \(\lambda=-1\) is not equal to the algebraic multiplicity, so \(A\) is not diagonalizable and we continue with the scalar-triangularization procedure.
Now compute a basis for the generalized eigensubspace of degree \(2\text{:}\)
\begin{gather*}
(-I-A)^2 = \begin{abmatrix}{ccr} 1 \amp 3 \amp -1 \\ 0 \amp 0 \amp 0 \\ 1 \amp 3 \amp -1 \end{abmatrix}
\qquad \rowredarrow \qquad
\begin{abmatrix}{ccr} 1 \amp 3 \amp -1 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{abmatrix}\\
\\
\implies \qquad
E_{-1}^2(A) = \Span\left\{
\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},
\begin{abmatrix}{r} -3 \\ 1 \\ 0 \end{abmatrix}
\right\}\text{.}
\end{gather*}
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 \(P\text{:}\)
\begin{align*}
\uvec{p}_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \;
\uvec{p}_2 \amp = \begin{abmatrix}{r} -3 \\ 1 \\ 0 \end{abmatrix}, \;
\uvec{p}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix},
\amp \amp\implies \amp
P \amp= \begin{abmatrix}{crc} 1 \amp -3 \amp 0 \\ 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 1 \end{abmatrix}\text{.}
\end{align*}
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
\begin{equation*}
\begin{abmatrix}{cc|c} \uvec{p}_1 \amp \uvec{p}_2 \amp \bigl( A - (-1) I \bigr) \uvec{p}_3 \end{abmatrix}
= \begin{abmatrix}{cr|r} 1 \amp -3 \amp 5 \\ 0 \amp 1 \amp -1 \\ 1 \amp 0 \amp 2 \end{abmatrix}
\qquad \rowredarrow \qquad
\begin{abmatrix}{cc|r} 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \end{abmatrix}\text{.}
\end{equation*}
Thus, the entries above the diagonal in the third column are \(2\text{,}\) \(-1\text{.}\) Putting all this together yields
\begin{equation*}
\inv{P}AP = \begin{abmatrix}{rrr} -1 \amp 1 \amp 2 \\ 0 \amp -1 \amp -1 \\ 0 \amp 0 \amp -1 \end{abmatrix} \text{.}
\end{equation*}
Example 29.5.2. A larger matrix.
Is matrix
\begin{equation*}
A = \begin{abmatrix}{rrrrr}
3 \amp 1 \amp -8 \amp 8 \amp -3 \\
-2 \amp -5 \amp 0 \amp -2 \amp -5 \\
2 \amp 7 \amp 3 \amp 2 \amp 4 \\
3 \amp 11 \amp 1 \amp 5 \amp 7 \\
2 \amp 8 \amp 2 \amp 0 \amp 9
\end{abmatrix}
\end{equation*}
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 \(E_3(A)\text{.}\)
\begin{gather*}
3 I - A =
\begin{abmatrix}{rrrrr}
0 \amp -1 \amp 8 \amp -8 \amp 3 \\
2 \amp 8 \amp 0 \amp 2 \amp 5 \\
-2 \amp -7 \amp 0 \amp -2 \amp -4 \\
-3 \amp -11 \amp -1 \amp -2 \amp -7 \\
-2 \amp -8 \amp -2 \amp 0 \amp -6
\end{abmatrix}
\quad \rowredarrow \quad
\begin{abmatrix}{rrrrr}
1 \amp 0 \amp 0 \amp 1 \amp -\frac{3}{2} \\
0 \amp 1 \amp 0 \amp 0 \amp 1 \\
0 \amp 0 \amp 1 \amp -1 \amp \frac{1}{2} \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0
\end{abmatrix}\\
\\
\implies \qquad
E_3(A) = \Span\left\{
\begin{abmatrix}{r} -1 \\ 0 \\ 1 \\ 1 \\ 0 \end{abmatrix},
\begin{abmatrix}{r} 3 \\ -2 \\ -1 \\ 0 \\ 2 \end{abmatrix}
\right\}
\end{gather*}
We need five vectors for the columns of the transition matrix \(P\text{,}\) so this is not enough.
Next, compute a basis for \(E_3^2(A)\text{.}\)
\begin{gather*}
(3 I - A)^2 =
\begin{abmatrix}{rrrrr}
0 \amp 0 \amp 2 \amp -2 \amp 1 \\
0 \amp 0 \amp 4 \amp -4 \amp 2 \\
0 \amp 0 \amp -6 \amp 6 \amp -3 \\
0 \amp 0 \amp -8 \amp 8 \amp -4 \\
0 \amp 0 \amp -4 \amp 4 \amp -2
\end{abmatrix}
\quad \rowredarrow \quad
\begin{abmatrix}{rrrrr}
0 \amp 0 \amp 1 \amp -1 \amp \frac{1}{2} \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0
\end{abmatrix}\\
\\
\implies \qquad
E_3^2(A) = \Span\left\{
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix},
\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix},
\begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 0 \end{bmatrix},
\begin{abmatrix}{r} 0 \\ 0 \\ -1 \\ 0 \\ 2 \end{abmatrix}
\right\}
\end{gather*}
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
\begin{equation*}
E_3^2(A) = \Span\left\{
\begin{abmatrix}{r} -1 \\ 0 \\ 1 \\ 1 \\ 0 \end{abmatrix},
\begin{abmatrix}{r} 3 \\ -2 \\ -1 \\ 0 \\ 2 \end{abmatrix},
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix},
\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}
\right\}\text{.}
\end{equation*}
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
\begin{equation*}
E_3^3(A) = \Span\left\{
\begin{abmatrix}{r} -1 \\ 0 \\ 1 \\ 1 \\ 0 \end{abmatrix},
\begin{abmatrix}{r} 3 \\ -2 \\ -1 \\ 0 \\ 2 \end{abmatrix},
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix},
\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix},
\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}
\right\}\text{.}
\end{equation*}
Now place these matrices in the transition matrix
\begin{equation*}
P = \begin{abmatrix}{rrrrr}
-1 \amp 3 \amp 1 \amp 0 \amp 0 \\
0 \amp -2 \amp 0 \amp 1 \amp 0 \\
1 \amp -1 \amp 0 \amp 0 \amp 1 \\
1 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 2 \amp 0 \amp 0 \amp 0
\end{abmatrix}\text{.}
\end{equation*}
To determine the form matrix, we can reduce \(\begin{abmatrix}{c|c} P \amp AP \end{abmatrix}\) as
\begin{gather*}
\begin{abmatrix}{rrrrr|rrrrr}
-1 \amp 3 \amp 1 \amp 0 \amp 0 \amp -3 \amp 9 \amp 3 \amp 1 \amp -8 \\
0 \amp -2 \amp 0 \amp 1 \amp 0 \amp 0 \amp -6 \amp -2 \amp -5 \amp 0\\
1 \amp -1 \amp 0 \amp 0 \amp 1 \amp 3 \amp -3 \amp 2 \amp 7 \amp 3\\
1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 3 \amp 0 \amp 3 \amp 11 \amp 1\\
0 \amp 2 \amp 0 \amp 0 \amp 0 \amp 0 \amp 6 \amp 2 \amp 8 \amp 2
\end{abmatrix}\\
\\
\rowredarrow \qquad
\begin{abmatrix}{rrrrr|rrrrr}
1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 3 \amp 0 \amp 3 \amp 11 \amp 1 \\
0 \amp 1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 3 \amp 1 \amp 4 \amp 1 \\
0 \amp 0 \amp 1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 3 \amp 0 \amp -10 \\
0 \amp 0 \amp 0 \amp 1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 3 \amp 2 \\
0 \amp 0 \amp 0 \amp 0 \amp 1 \amp 0 \amp 0 \amp 0 \amp 0 \amp 3
\end{abmatrix}\text{,}
\end{gather*}
obtaining
\begin{equation*}
\inv{P} A P
= \begin{abmatrix}{rrrrr}
3 \amp 0 \amp 3 \amp 11 \amp 1 \\
0 \amp 3 \amp 1 \amp 4 \amp 1 \\
0 \amp 0 \amp 3 \amp 0 \amp -10 \\
0 \amp 0 \amp 0 \amp 3 \amp 2 \\
0 \amp 0 \amp 0 \amp 0 \amp 3
\end{abmatrix}\text{.}
\end{equation*}
