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Discover Linear Algebra

Jeremy Sylvestre

Section 34.5 Theory

Here we will record the facts that Jordan normal form answers Question 28.1.1 and can be used to distinguish between similarity classes of matrices.
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Proof outline.

The proof basically confirms that Procedure 34.3.1 does what it says it does, and follows the same calculation pattern as Example 34.4.2.
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If the characteristic polynomial \(c_A(\lambda)\) factors completely, then from Theorem 30.5.1 we know there exists an invertible matrix \(M\) such that \(U = \inv{M} A M\) has triangular block form. Each block \(U_j\) in \(U\) is a scalar-triangular block corresponding to eigenvalue \(\lambda = \lambda_j\) of \(A\) of algebraic multiplicity \(m = m_j\text{,}\) and can be decomposed as
\begin{equation*} U_j = \lambda_j I + N_j \text{,} \end{equation*}
where \(N_j\) is an \(m_j \times m_j\) nilpotent matrix. By Theorem 33.6.1, for each \(N_j\) there exists an invertible \(m_j \times m_j\) matrix \(Q_j\) such that \(N_j' = \inv{Q_j} N_j Q_j\) is in Jordan normal form, with all zeros down the diagonal.
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Aside: Note.

The matrix
\begin{equation*} A_j = \lambda_j I + \inv{Q_j} N_j Q_j = \lambda_j I + N_j' \end{equation*}
is then also in Jordan normal form, but with \(\lambda_j\) down the diagonal. Putting this all together, if we set \(P = Q M\text{,}\) where
\begin{equation*} Q = \begin{bmatrix} Q_1 \\ \amp Q_2 \\ \amp \amp \ddots \\ \amp \amp \amp Q_\ell \end{bmatrix} \text{,} \end{equation*}
then \(\inv{P} = \inv{M} \inv{Q}\text{,}\) and so
\begin{align*} \amp \inv{P} A P \\ \amp = \inv{Q} \inv{M} A M Q \\ \amp = \left[\begin{smallmatrix} \inv{Q_1} \\ \amp \inv{Q_2} \\ \amp \amp \ddots \\ \amp \amp \amp \inv{Q_\ell} \end{smallmatrix}\right] \left[\begin{smallmatrix} U_1 \\ \amp U_2 \\ \amp \amp \ddots \\ \amp \amp \amp U_\ell \end{smallmatrix}\right] \left[\begin{smallmatrix} Q_1 \\ \amp Q_2 \\ \amp \amp \ddots \\ \amp \amp \amp Q_\ell \end{smallmatrix}\right]\\ \amp = \left[\begin{smallmatrix} \inv{Q_1} U_1 Q_1 \\ \amp \inv{Q_2} U_2 Q_2 \\ \amp \amp \ddots \\ \amp \amp \amp \inv{Q_\ell} U_\ell Q_\ell \end{smallmatrix}\right]\\ \amp = \left[\begin{smallmatrix} \inv{Q_1}(\lambda_1 I + N_1)Q_1 \\ \amp \inv{Q_2}(\lambda_2 I + N_2)Q_2 \\ \amp \amp \ddots \\ \amp \amp \amp \inv{Q_\ell}(\lambda_\ell I + N_\ell)Q_\ell \end{smallmatrix}\right]\\ \amp = \left[\begin{smallmatrix} \lambda_1 \inv{Q_1}Q_1 + \inv{Q_1} N_1 Q_1 \\ \amp \lambda_2 \inv{Q_2}Q_2 + \inv{Q_2} N_2 Q_2 \\ \amp \amp \ddots \\ \amp \amp \amp \lambda_\ell \inv{Q_\ell}Q_\ell + \inv{Q_\ell} N_\ell Q_\ell \end{smallmatrix}\right]\\ \amp = \left[\begin{smallmatrix} \lambda_1 I + N_1' \\ \amp \lambda_2 I + N_2' \\ \amp \amp \ddots \\ \amp \amp \amp \lambda_\ell I + N_\ell' \end{smallmatrix}\right]\\ \amp = \left[\begin{smallmatrix} A_1 \\ \amp A_2 \\ \amp \amp \ddots \\ \amp \amp \amp A_\ell \end{smallmatrix}\right]\text{.} \end{align*}
Since each matrix \(A_j\) is in Jordan normal form, and each corresponds to a different eigenvalue \(\lambda_j\text{,}\) so also \(\inv{P} A P\) is in Jordan normal form.
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Proof.

We leave this proof as an exercise for you, the reader.
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