Lecture 24 - (finish) Decomposing Operators

Announcements:

Exam on Thursday
Wednesday will just be a review day (nothing new)
No stuff on 8.B on Square roots. Just the stuff before that (before pg. 258)

Recall from last week that we talked about this "main result":

Theorem

Given an operator $T \in L (V)$ where $V$ is a complex finite-dimensional vector space, then there is a basis of generalized eigenvectors of $T$ .

Namely:

V = G (λ_{1}, T) \oplus G (λ_{m}, T)

Further, we defined the multiplicity of an eigenvalue $λ_{i}$ :

multiplicity of an eigenvalue

For $T \in L (V)$ , the multiplicity of an eigenvalue is $\dim (G (λ, T)) = \dim (null (T - λ I)^{\dim (V)})$ .

And for the matrix representation of $M (T)$ using the basis of generalized eigenvectors? Suppose we have:

$v_{1}^{1}, . . ., v_{β_{1}}^{1}$ is the basis for $G (λ_{1}, T)$ .
$v_{1}^{2}, . . ., v_{β_{2}}^{2}$ is the basis for $G (λ_{2}, T)$
...
$v_{1}^{m}, . . ., v_{β_{m}}^{m}$ is the basis for $G (λ_{m}, T)$
Then the concatenation of all of these vectors is a basis, denoted $β$ , for $V$ as prior discussed. Here the notation suggest that each $β_{i}$ is the multiplicity of each $λ_{i}$ . Further note that $G (λ_{i}, T)$ is $T$ -invariant.

Then we have a bloch-diagonal structure:

M (T, β) = [\begin{matrix} M (T |_{G (λ_{1}, T)}) & 0 & \dots & 0 \\ 0 & M (T |_{G (λ_{2}, T)}) & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & M (T |_{G (λ_{n}, T)}) \end{matrix}]

Note that $T = (T - λ_{i} I) + λ_{i} I$ and here $T - λ_{i} I$ is nilpotent on $G (λ_{i}, T)$ specifically. What that means is that:

M (T) = [\begin{matrix} 0 & \dots & * \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & 0 \end{matrix}] + [\begin{matrix} λ_{i} & \dots & 0 \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & λ_{i} \end{matrix}] = [\begin{matrix} λ_{i} & \dots & * \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & λ_{i} \end{matrix}]

Thus:

M (T, β) = [\begin{matrix} [\begin{matrix} λ_{1} & \dots & * \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & λ_{1} \end{matrix}] & 0 & \dots & 0 \\ 0 & [\begin{matrix} λ_{2} & \dots & * \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & λ_{2} \end{matrix}] & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & [\begin{matrix} λ_{m} & \dots & * \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & λ_{m} \end{matrix}] \end{matrix}]

Note the $0$ 's in the main matrix are all matrices with corresponding entries all 0's.

Square Roots

We can do some nice things with nilpotent operators! Suppose $N$ is a nilpotent operator. Then there's some minimum positive integer $m \in Z^{+}$ such that $N^{m} = \vec{0}$ (the zero operator).

Let $p (x) \in P (F)$ , with:

p (x) = \sum_{i = 0}^{n} a_{i} x^{i}

then:

p (N) = \sum_{i = 0}^{n} a_{i} N^{i} = \sum_{i = 0}^{min (n, m - 1)} a_{i} N^{i}

where the second equality comes from the fact that $N^{m} = \vec{0}$ and same for higher powers. In fact, substituting $n \to \infty$ allows $p (N)$ to still be finite. So if you gave me some power series:

\sum_{i = 0}^{\infty} a_{i} x^{i}

then substituting $N$ gives:

\sum_{i = 0}^{\infty} a_{i} N^{i} = \sum_{i = 0}^{m - 1} a_{i} N^{i}

So using a nilpotent operator in a power series always gives a polynomial operator in $N$ . What can we do with fact? Look at the geometric series for example:

\sum_{i = 0}^{\infty} x^{i} = \frac{1}{1 - x}

(1 - x) (1 + x + x^{2} + \dots) = 1

Substituting $N$ in here:

I = (I - N) (I + N + N^{2} + \dots + \dots N^{m - 1})

because we have have a nilpotent operator. If we were to foil this out a lot of cancellation happens, and we'd only have the first $I$ term and the last $N^{m} = \vec{0}$ term which becomes irrelevant anyway. This all implies that $I - N$ is invertible and is given by:

(I - N)^{- 1} = \sum_{i = 0}^{m - 1} N^{i}

Looking at another series:

\frac{1}{1 + x} = \sum_{i = 0}^{\infty} (- 1)^{i} x^{i}

Implying that the inverse of $I + N$ is:

(I + N)^{- 1} = \sum_{i = 0}^{m - 1} (- 1)^{i} N^{i}

Now for a big one. We generalized:

(1 + x)^{k} = \sum_{i = 0}^{\infty} (\binom{k}{i}) x^{i}

We were allowed to choose things like $k = \frac{1}{2}$ though and it still worked:

\sqrt{1 + x} = \sum_{i = 0}^{\infty} (\binom{\frac{1}{2}}{i}) x^{i}

Now plugging $x := N$ to get:

an operator that squares to I + N = \sum_{i = 0}^{\infty} (\binom{\frac{1}{2}}{i}) N^{i} = \sum_{i = 0}^{m - 1} (\binom{\frac{1}{2}}{i}) N^{i}

For this specific case:

\begin{aligned} = I + (\binom{\frac{1}{2}}{1}) N + \dots + (\binom{\frac{1}{2}}{m - 1}) N^{m - 1} \\ = I + \frac{1}{2} N - \frac{1}{8} N^{2} + \frac{1}{16} N^{3} + \dots + (\binom{\frac{1}{2}}{m - 1}) N^{m - 1} \end{aligned}

This brings us to:

Theorem

Given a nilpotent $N \in L (V)$ , the operator $I + N$ has a square root

Proof
We can use the construction made above.
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Theorem

Every invertible operator $T \in L (V)$ on a complex vector space $V$ has a square root.

Proof
Let $λ_{1}, . . ., λ_{m}$ be the distinct eigenvalues of $T$ . As $V$ is complex, then:

V = ⨁_{i = 1}^{m} G (λ_{i}, T)

Let's construct an operator $R$ such that $R^{2} = T$ . Define $R$ by its action on a basis. In particular, using the basis $β = {v_{1}^{1}, . . ., v_{β_{1}}^{1}, . . ., v_{1}^{m}, . . ., v_{β_{m}}^{m}}$ .

Consider $G (λ_{i}, T) = span (v_{1}^{i}, . . ., v_{β_{i}}^{i})$ . On this space, notice $T = (T - λ_{i} I) + λ_{i} I$ whcih is a nilpotent plus the identity:

\begin{aligned} T & = (T - λ_{i} I) + λ_{i} I \\ = λ_{i} (I + \underset{nilpotent}{\underset{⏟}{\frac{1}{λ_{i}} (T - λ_{i} I)}}) & λ_{i} \neq 0 since T is invertible \\ = . . . see ya next time \end{aligned}

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