Notes - Linear Algebra MT23, Triangular form theorem

Suppose:

• $V$ finite dimensional
• $T : V \to V$ linear transformation
• $\chi _ T(x)$ splits completely into (possibly repeated) linear factors

Then:

• There exists a basis in which $T$ is upper triangular.

Suppose:

• $V$ finite dimensional
• $T : V \to V$ linear transformation
• ($\chi _ T(x)$ splits completely into (possibly repeated) linear factors, but this is guaranteed by closure)

Then:

• There exists a basis in which $T$ is upper triangular.

The diagonal elements are eigenvalues, and this comes from the fact we use induction and need a $T$-invariant subspace at each step, which the lines spanned by eigenvectors are.

Two possible proofs:

• The matrix with respect to bases that contain invariant subspaces have a leading column $(\lambda, 0, \cdots, 0)^\intercal$
• Lines spanned by eigenvectors form invariant subspaces
• Use induction to whittle down the matrix by modding out the subspace spanned by each eigenvector, and using theorem that $\chi _ T(x) = \chi _ {T \vert _ U}(x) \chi _ {\overline T} (x)$

Or,

• First show that for matrices $T$ where $(T - \lambda I)^k = 0$ for some $k$, these are upper-triangular
• Under the assumption that $\chi _ T(x)$ splits into linear factors, use the primary decomposition theorem to split $V$ into subspaces of this form, and then consider a union of bases.

*Proof 1:

We proceed by induction on $\dim V$. The case where $\dim V = 1$ is trivial, so suppose $\dim V > 1$. By the assumption that $\chi _ T(x)$ splits completely, $\exists (\lambda, v)$ such that $v \ne 0$ is an eigenvector of $T$ with associated eigenvalue $\lambda$.

Consider $U = \langle v \rangle$, this is $T$-invariant. Then this induces a map of quotients

\[\overline T : V/U \to V/U\]

and by general properties of quotient maps,

\[\chi_{\overline T} (x) = \chi_T(x) / (\lambda - x)\]

So $\chi _ {\overline T}$ is also a product of linear factors. Also, note that $\dim V/U = \dim V - 1$, so by the inductive hypothesis there exists a basis $\{v _ 2 + U, \cdots, v _ n + U\}$ such that $\overline T$ is upper-triangular. But then by sticking $\{v\}$ and this basis together, we get that $T$ is similar to a matrix

\[T = \begin{pmatrix} \lambda & * \\ 0 & \overline T \end{pmatrix}\]

and so we are done.

Proof 2: (a more “concrete” argument using bases)

General idea: First show that for matrices $T$ where $(T - \lambda I)^k = 0$ for some $k$, these can be made upper-triangular, and then under the assumption that $\chi _ T(x)$ splits into linear factors, use the primary decomposition theorem to split $V$ into subspaces of this form.

First consider transformations $S$ such that $S^k = 0$ for some $k$. Note that

\[\\{\pmb 0\\} = \ker S^0 \subseteq \ker S^1 \subseteq \cdots \subseteq \ker S^k = V\]

Write

\[\ker S^k = \langle B_1\rangle \cup \langle B_2 \setminus B_1 \rangle \cup \langle B_3 \setminus (B_1 \cup B_2) \rangle \cup \cdots\]

where $B _ 1$ is a basis for the kernel of $S^1$, $B _ 2$ is a kernel for $S^2$, and so on, i.e. we have a basis for $B _ 1$ and we slowly extend to a basis for all of $V$, using a basis for each of the intermediate kernels.

Note that $\ker S^i \supseteq S^{-1}(\ker S^{i-1})$, so $S(\langle B _ i \rangle) \subseteq \langle B _ {i-1} \rangle$, so the matrix of $S$ with respect to the basis for $\ker S^k$ is strictly upper-triangular, since all entries corresponding to $B _ 1$ are $0$, and then all subsequent columns correspond to elements of $\ker S^{i+1} \setminus \ker S^i$. But these are sent to elements of $\ker S^i$ by $S$, hence are $0$ on and below the diagonal.

It then follows that any $T$ such that $(T - \lambda I)^k = 0$ for some $k$ is upper-triangular (since we just add a multiple of the identity matrix).

By our assumption,

\[m_T(x) = \prod^r_{i=1} (x - \lambda_i)^{k_i}\]

for some appropriate constants, and hence by the primary decomposition theorem

\[V = \bigoplus^r_{i = 1} \ker((T - \lambda_i I )^{k_i})\]

Applying the previous result to each of these, $T$ is an upper-triangular matrix with respect to the basis that makes each of these blocks upper-triangular.

(A proof combining both the above proofs: You could also show that all nilpotent matrices have a basis on which the diagonal entries are $0$ by using a similar argument to the first proof of the triangular form theorem, and then use the primary decomposition theorem).

If $v \in \langle e _ 1, \cdots, e _ n \rangle$, then

\[(A - \lambda_n I) v = \langle e_1, \ldots, e_{n-1}\rangle\]

\[\begin{aligned} 0 &= \text{Tr}(0) \\\\ &= \text{Tr}(m_T(T)) \\\\ &= \text{Tr}(T^n + a_{n-1} T^{n-1} + \cdots + a_0 I) \\\\ &= \text{Tr}(T^n) + a_{n-1} \text{Tr}(T^{n-1}) + \cdots + a_0 \text{Tr}(I) \\\\ &= 0 + 0 + \cdots + a_0N \end{aligned}\]

Then

\[a_0 N = 0\]

so $a _ 0 = 0$, since $N \ge 1$.