Lecture 24 - Finishing Eigenstuff

Recall the theorem from last time:

Theorem

Every operator $T \in L (V)$ , where $V$ is a finite-dimensional complex vector space has a matrix representation which is upper-triangular.

This just means that there is a basis $v_{1}, . . ., v_{n}$ such that $M (T, β)$ is U.T.

Proof
We'll prove this by induction on the size of the dimension $dim (V) = n$ .

For our base case, if $n = 1$ , then we have a $A_{1 \times 1}$ so then $A_{1 \times 1} = [a]$ which is already upper triangular.

Now let the theorem hold for all cases $n < k$ for some $k \in N$ . We'll try to prove the $k$ case. Via Lecture 23 - Polynomial Operator#^9bac66, $T$ has an eigenvalue $λ$ . Consider $U = range (T - λ I)$ .

Note

A thing about $U$ is that it is $T$ -invariant, so if $u \in U$ then $T u \in U$ . Let's prove this. Let $u \in U$ be arbitrary. Then $u \in range (T - λ I)$ so then there is some $v \in V$ such that $(T - λ I) v = u$ . Then $T u = T (T - λ I) v = (T - λ I) (T v) \in U$ completing the proof.

Also note that since $dim (null (T - λ I)) > 1$ since we have an eigenvector already (from it associated eigenvalue $λ$ ) so then by the FTOLM then $dim (U) < dim (V)$ .

So then $T |_{U}$ is an operator on $U$ since $dim (U) < n$ , so then by induction there is some basis $u_{1}, . . ., u_{k}$ such that $M (T |_{U}, (u_{1}, . . ., u_{k}))$ is upper triangular. Next, we'll extend this basis $u_{1}, . . ., u_{k}$ to a basis for $V$ , namely $u_{1}, . . ., u_{k}, v_{1}, . . ., v_{n - k}$ . Call it $β$ . We claim that this will give me an U.T. matrix representation.

M (T, β) = [\begin{matrix} upper triangular & don’t care \\ 0’s & λ ’s on the diagonal \end{matrix}]

where the columns come from $T (u_{1}), . . ., T (u_{k}), T (v_{1}), . . ., T (v_{n - k})$ and the rows come form $u_{1}, . . ., u_{k}, v_{1}, . . ., v_{n - k}$ . This is a weird matrix, so let's explain it:

The bottom right matrix follows that:

\begin{aligned} T (v_{i}) & = T (v_{i}) + λ v_{i} - λ v_{i} \\ = \underset{\in U}{\underset{⏟}{(T - λ I) v_{i}}} + λ v_{i} \end{aligned}

so then we have to have a $λ$ on the diagonal for this matrix, which is upper triangular.

The bottom left matrix is all 0's since $U$ is $T$ invariant, so no vectors of $v_{i}$ are needed to describe $T u$ since the basis of $U$ is only needed.
The top left matrix is upper triangular by our inductive hypothesis
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Okay, pause and take a break to get the proof above.

Theorem

Suppose $T \in L (V)$ and there is a basis $v_{1}, . . ., v_{n}$ denoted $β$ , for which $M (T, β)$ is U.T. Then $T$ is invertible iff every diagonal entry in $M (T, β)$ is non-zero.

So if you see a 0 on the diagonal, then you immediately know that $T$ isn't invertible.

Proof
We'll do a proof by contradiction for $\to$ . Assume $T$ is invertible and that $a_{k, k}$ entry in $M (T, β)$ is 0. That means that:

T (v_{k}) = a_{1 k} v_{1} + \dots + a_{k k} v_{k} + \dots

but since it upper triangular than all $a_{j, k}$ for $j > k$ are all 0's:

T (v_{k}) = a_{1 k} v_{1} + \dots + a_{k k} v_{k}

but also we assumed $a_{k k} = 0$ :

T (v_{k}) = a_{1 k} v_{k} + \dots + a_{(k - 1) k} v_{k - 1} \in span (v_{1}, . . ., v_{k - 1})

Since $T$ is invertible, we have it that $T (v_{1}), . . ., T (v_{k})$ is LI since $v_{1}, . . ., v_{k}$ are LI.

We have a LI list of length $k$ given by $T (v_{1}), . . ., T (v_{k})$ in a vector space of dimension $k - 1$ given by $span (v_{1}, . . ., v_{k - 1})$ which is a contradiction.

The other direction $\leftarrow$ requires some material that we haven't covered yet, and we'll look at that tomorrow.
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