Linear Algebra/Jordan Canonical Form

This subsection moves from the canonical form for nilpotent matrices to the one for all matrices.

We have shown that if a map is nilpotent then all of its eigenvalues are zero. We can now prove the converse.

We have a canonical form for nilpotent matrices, that is, for each matrix whose single eigenvalue is zero: each such matrix is similar to one that is all zeroes except for blocks of subdiagonal ones. (To make this representation unique we can fix some arrangement of the blocks, say, from longest to shortest.) We next extend this to all single-eigenvalue matrices.

Observe that if $$ t $$'s only eigenvalue is $$ \lambda $$ then $$ t-\lambda $$'s only eigenvalue is $$ 0 $$ because $$ t(\vec{v})=\lambda\vec{v} $$ if and only if $$ (t-\lambda)\,(\vec{v})=0\cdot\vec{v} $$. The natural way to extend the results for nilpotent matrices is to represent $$t-\lambda$$ in the canonical form $$N$$, and try to use that to get a simple representation $$T$$ for $$t$$. The next result says that this try works.

An array that is all zeroes, except for some number $$\lambda$$ down the diagonal and blocks of subdiagonal ones, is a Jordan block. We have shown that Jordan block matrices are canonical representatives of the similarity classes of single-eigenvalue matrices.

We will now finish the program of this chapter by extending this work to cover maps and matrices with multiple eigenvalues. The best possibility for general maps and matrices would be if we could break them into a part involving their first eigenvalue $$ \lambda_1 $$ (which we represent using its Jordan block), a part with $$ \lambda_2 $$, etc.

This ideal is in fact what happens. For any transformation $$ t:V\to V $$, we shall break the space $$ V $$ into the direct sum of a part on which $$ t-\lambda_1 $$ is nilpotent, plus a part on which $$ t-\lambda_2 $$ is nilpotent, etc. More precisely, we shall take three steps to get to this section's major theorem and the third step shows that $$ V=\mathcal{N}_\infty(t-\lambda_1)\oplus\cdots\oplus \mathcal{N}_\infty(t-\lambda_\ell) $$ where $$ \lambda_1,\ldots,\lambda_\ell $$ are $$ t $$'s eigenvalues.

Suppose that $$ t:V\to V $$ is a linear transformation. Note that the restriction of $$ t $$ to a subspace $$ M $$ need not be a linear transformation on $$ M $$ because there may be an $$ \vec{m}\in M $$ with $$ t(\vec{m})\not\in M $$. To ensure that the restriction of a transformation to a "part" of a space is a transformation on the partwe need the next condition.

Two examples are that the generalized null space $$\mathcal{N}_\infty(t)$$ and the generalized range space $$\mathcal{R}_\infty(t)$$ of any transformation $$t$$ are invariant. For the generalized null space, if $$\vec{v}\in\mathcal{N}_\infty(t)$$ then $$t^n(\vec{v})=\vec{0}$$ where $$n$$ is the dimension of the underlying space and so $$t(\vec{v})\in\mathcal{N}_\infty(t)$$ because $$t^n(\,t(\vec{v})\,)$$ is zero also. For the generalized range space, if $$\vec{v}\in\mathcal{R}_\infty(t)$$ then $$\vec{v}=t^n(\vec{w})$$ for some $$\vec{w}$$ and then $$t(\vec{v})=t^{n+1}(\vec{w})=t^n(\,t(\vec{w})\,)$$ shows that $$t(\vec{v})$$ is also a member of $$\mathcal{R}_\infty(t)$$.

Thus the spaces $$\mathcal{N}_\infty(t-\lambda_i)$$ and $$\mathcal{R}_\infty(t-\lambda_i)$$ are $$t-\lambda_i$$ invariant. Observe also that $$t-\lambda_i$$ is nilpotent on $$\mathcal{N}_\infty(t-\lambda_i)$$ because, simply, if $$\vec{v}$$ has the property that some power of $$t-\lambda_i$$ maps it to zero&mdash; that is, if it is in the generalized null space&mdash; then some power of $$t-\lambda_i$$ maps it to zero. The generalized null space $$\mathcal{N}_\infty(t-\lambda_i)$$ is a "part" of the space on which the action of $$t-\lambda_i$$ is easy to understand.

The next result is the first of our three steps. It establishes that $$ t-\lambda_j $$ leaves $$ t-\lambda_i $$'s part unchanged.

The second step of the three that we will take to prove this section's major result makes use of an additional property of $$ \mathcal{N}_\infty(t-\lambda_i) $$ and $$ \mathcal{R}_\infty(t-\lambda_i) $$, that they are complementary. Recall that if a space is the direct sum of two others $$ V=\mathcal{N}\oplus \mathcal{R} $$ then any vector $$ \vec{v} $$ in the space breaks into two parts $$ \vec{v}=\vec{n}+\vec{r} $$ where $$ \vec{n}\in \mathcal{N} $$ and $$ \vec{r}\in \mathcal{R} $$, and recall also that if $$ B_{\mathcal{N}} $$ and $$ B_{\mathcal{R}} $$ are bases for $$ \mathcal{N} $$ and $$ \mathcal{R} $$ then the concatenation $$ B_{\mathcal{N}}\!\mathbin{{}^\frown}\!B_{\mathcal{R}} $$ is linearly independent (and so the two parts of $$ \vec{v} $$ do not "overlap"). The next result says that for any subspaces $$ \mathcal{N} $$ and $$ \mathcal{R} $$ that are complementary as well as $$ t $$ invariant, the action of $$ t $$ on $$ \vec{v} $$ breaks into the "non-overlapping" actions of $$ t $$ on $$ \vec{n} $$ and on $$ \vec{r} $$.

{{TextBox|1= Let $$ t:V\to V $$ be a transformation and let $$ \mathcal{N} $$ and $$ \mathcal{R} $$ be $$ t $$ invariant complementary subspaces of $$ V $$. Then $$ t $$ can be represented by a matrix with blocks of square submatrices $$T_1$$ and $$T_2$$
 * Lemma 2.8{{anchor|le:InvCompSubspSplitTrans}}:



\left(\begin{array}{c|c} T_1  &Z_2  \\  \hline Z_1 &T_2 \end{array}\right) \begin{array}{ll} \} \dim(\mathcal{N})\text{-many rows} \\ \} \dim(\mathcal{R})\text{-many rows} \end{array} $$

where $$ Z_1 $$ and $$ Z_2 $$ are blocks of zeroes. }}

To see that $$ t $$ has been decomposed into its action on the parts, observe that the restrictions of $$ t $$ to the subspaces $$ \mathcal{N} $$ and $$ \mathcal{R} $$ are represented, with respect to the obvious bases, by the matrices $$ T_1 $$ and $$ T_2 $$. So, with subspaces that are invariant and complementary, we can split the problem of examining a linear transformation into two lower-dimensional subproblems. The next result illustrates this decomposition into blocks.

From Lemma 2.9 we conclude that if two subspaces are complementary and $$ t $$ invariant then $$ t $$ is nonsingular if and only if its restrictions to both subspaces are nonsingular.

Now for the promised third, final, step to the main result.

{{TextBox|1= Because $$ \dim (V) $$ is the degree $$ p_1+\cdots+p_\ell $$ of the characteristic polynomial, to establish statement (1) we need only show that statement (2) holds and that $$ \mathcal{N}_\infty(t-\lambda_i)\cap\mathcal{N}_\infty(t-\lambda_j) $$ is trivial whenever $$ i\neq j $$.
 * Proof:

For the latter, by Lemma 2.7, both $$ \mathcal{N}_\infty(t-\lambda_i) $$ and $$ \mathcal{N}_\infty(t-\lambda_j) $$ are $$ t $$ invariant. Notice that an intersection of $$ t $$ invariant subspaces is $$ t $$ invariant and so the restriction of $$ t $$ to $$ \mathcal{N}_\infty(t-\lambda_i)\cap\mathcal{N}_\infty(t-\lambda_j) $$ is a linear transformation. But both $$ t-\lambda_i $$ and $$ t-\lambda_j $$ are nilpotent on this subspace and so if $$ t $$ has any eigenvalues on the intersection then its "only" eigenvalue is both $$ \lambda_i $$ and $$ \lambda_j $$. That cannot be, so this restriction has no eigenvalues: $$ \mathcal{N}_\infty(t-\lambda_i)\cap\mathcal{N}_\infty(t-\lambda_j) $$ is trivial (Lemma V.II.3.10 shows that the only transformation without any eigenvalues is on the trivial space).

To prove statement (2), fix the index $$ i $$. Decompose $$ V $$ as $$ \mathcal{N}_\infty(t-\lambda_i)\oplus\mathcal{R}_\infty(t-\lambda_i) $$

and apply Lemma 2.8.



T=

\left(\begin{array}{c|c} T_1  &Z_2  \\  \hline Z_1   &T_2 \end{array}\right) \begin{array}{ll} \} \dim(\,\mathcal{N}_\infty(t-\lambda_i)\,)\text{-many rows} \\ \} \dim(\,\mathcal{R}_\infty(t-\lambda_i)\,)\text{-many rows} \end{array} $$

By Lemma 2.9, $$ \left|T-xI\right|=\left|T_1-xI\right|\cdot\left|T_2-xI\right| $$. By the uniqueness clause of the Fundamental Theorem of Arithmetic, the determinants of the blocks have the same factors as the characteristic polynomial $$ \left|T_1-xI\right|=(x-\lambda_1)^{q_1}\dots(x-\lambda_\ell)^{q_\ell} $$ and $$ \left|T_2-xI\right|=(x-\lambda_1)^{r_1}\dots(x-\lambda_\ell)^{r_\ell} $$, and the sum of the powers of these factors is the power of the factor in the characteristic polynomial: $$ q_1+r_1=p_1 $$, ..., $$ q_\ell+r_\ell=p_\ell $$. Statement (2) will be proved if we will show that $$q_i=p_i$$ and that $$q_j=0$$ for all $$j\neq i$$, because then the degree of the polynomial $$\left|T_1-xI\right|$$&mdash; which equals the dimension of the generalized null space&mdash; is as required.

For that, first, as the restriction of $$ t-\lambda_i $$ to $$ \mathcal{N}_\infty(t-\lambda_i) $$ is nilpotent on that space, the only eigenvalue of $$ t $$ on it is $$ \lambda_i $$. Thus the characteristic equation of $$ t $$ on $$ \mathcal{N}_\infty(t-\lambda_i) $$ is $$ \left|T_1-xI\right|=(x-\lambda_i)^{q_i} $$. And thus $$q_j=0$$ for all $$j\neq i$$.

Now consider the restriction of $$ t $$ to $$ \mathcal{R}_\infty(t-\lambda_i) $$. By Note V.III.2.2, the map $$ t-\lambda_i $$ is nonsingular on $$ \mathcal{R}_\infty(t-\lambda_i) $$ and so $$ \lambda_i $$ is not an eigenvalue of $$ t $$ on that subspace. Therefore, $$ x-\lambda_i $$ is not a factor of $$ \left|T_2-xI\right| $$, and so $$ q_i=p_i $$. }}

Our major result just translates those steps into matrix terms.

Jordan form is a canonical form for similarity classes of square matrices, provided that we make it unique by arranging the Jordan blocks from least eigenvalue to greatest and then arranging the subdiagonal $$1$$ blocks inside each Jordan block from longest to shortest.

We close with the statement that the subjects considered earlier in this Chpater are indeed, in this sense, exhaustive.

Exercises
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