HW 3 - Self-Adjoint and Normal Operators

Doenjang jigae

7.A

1

Theorem

Suppose $n$ is a positive integer. Define $T \in L (F^{n})$ by:

T (z_{1}, . . ., z_{n}) = (0, z_{1}, . . ., z_{n - 1})

Find $T^{*} (z_{1}, . . ., z_{n})$ .

\begin{proof}
Fix a point $y = (y_{1}, . . ., y_{n}) \in F^{n}$ , and let $x = (x_{1}, . . ., x_{n}) \in F^{n}$ be arbitrary:

\begin{aligned} ⟨ x, T^{*} y ⟩ & = ⟨ T x, y ⟩ \\ = ⟨ (0, x_{1}, . . ., x_{n - 1}), (y_{1}, . . ., y_{n}) ⟩ \\ = x_{1} y_{2}^{*} + \dots + x_{n - 1} y_{n}^{*} \\ = x_{1} y_{2}^{*} + \dots + x_{n - 1} y_{n}^{*} + 0 x_{n} \\ = ⟨ x, (y_{2}, . . ., y_{n}, 0) ⟩ \end{aligned}

Thus:

T^{*} y = (y_{2}, . . ., y_{n}, 0)

\end{proof}

2

Theorem

Suppose $T \in L (V)$ and $λ \in F$ . Then $λ$ is an eigenvalue of $T$ iff $\overset{―}{λ}$ is an eigenvalue of $T^{*}$ .

\begin{proof}
Consider $v \in V$ is an eigenvector of $T$ (hence $v \neq 0$ ) where $λ$ is the corresponding eigenvalue. Then if we fix some $w \in V$ :

\begin{aligned} ⟨ v, T^{*} w ⟩ & = ⟨ T v, w ⟩ \\ = ⟨ λ v, w ⟩ \\ = λ ⟨ v, w ⟩ \\ = ⟨ v, \overset{―}{λ} w ⟩ \\ \Leftrightarrow ⟨ v, (T - \overset{―}{λ} I) w ⟩ = 0 \end{aligned}

Notice that $(T - \overset{―}{λ} I) w \neq v$ by contraction as then $v = 0$ , so then $T^{*} - \overset{―}{λ}$ has to be non-surjective, so there's some $w$ such that it makes the $(T - \overset{―}{λ} I) w = \vec{0}$ .

Thus $(T^{*} - \overset{―}{λ}) w = 0$ so $w$ was an eigenvector with eigenvalue $\overset{―}{λ}$ . The other way is proved in the same method.
\end{proof}

3

Theorem

Suppose $T \in L (V)$ and $U$ is a subspace of $V$ . Then $U$ is invariant under $T$ iff $U^{⊥}$ is invariant under $T^{*}$ .

\begin{proof}
$U$ is invariant under $T$ iff for all $u \in U$ we have $T u \in U$ , which iff for any $u^{'} \in U^{⊥}$ :

\begin{aligned} ⟨ \underset{\in U}{\underset{⏟}{u}}, \underset{\in U^{⊥}}{\underset{⏟}{T^{*} u^{'}}} ⟩ & = ⟨ \underset{\in U}{\underset{⏟}{T u}}, \underset{\in U^{⊥}}{\underset{⏟}{u^{'}}} ⟩ = 0 \end{aligned}

iff $T^{*} u^{'} \in U^{⊥}$ iff $T^{*}$ is invariant under $U^{⊥}$ .
\end{proof}

4

Theorem

Suppose $T \in L (V, W)$ :

$T$ is injective iff $T^{*}$ is surjective.
$T$ is surjective iff $T^{*}$ is injective.

\begin{proof}
Let $n = dim (V)$ and $m = dim (W)$ .

For (a), $T$ is injective iff $null (T) = {0}$ iff $dim (null (T)) = 0$ iff (by the FTOLM) $dim (range (T)) = m$ . This is iff $dim ((range (T))^{⊥}) = m - n$ (by the supposition $m - n \geq 0$ iff $dim (null (T^{*})) = m - n$ . Thus, again by the FTOLM iff $dim (range (T^{*})) = m - (m - n) = n$ iff $range (T^{*}) = W$ iff $T^{*}$ is surjective.

For (b), $T$ is surjective iff $range (T) = W$ iff $dim (range (T)) = m$ iff by FTOLM $dim (null (T)) = n - m$ iff $dim ((range (T^{*}))^{⊥}) = n - m$ (again $n - m \geq 0$ ) iff $dim (range (T^{*})) = n - (n - m) = m$ iff $dim (null (T^{*}) = 0$ iff $null (T^{*}) = {\vec{0}}$ iff $T^{*}$ is injective.
\end{proof}

5

Theorem

$dim (null (T^{*})) = dim (null (T)) + dim (W) - dim (V)$
$dim (range (T^{*})) = dim (range (T))$
for all $T \in L (V, W)$ .

\begin{proof}

\begin{aligned} dim (null (T^{*})) & = dim (W) - dim (range (T^{*})) & FTOLM \\ = dim (W) - dim ((null (T))^{⊥}) \\ = dim (W) - (dim (V) - dim (null (T))) \\ = dim (null (T)) + dim (W) - dim (V) \end{aligned}

\begin{aligned} dim (range (T^{*})) & = dim (W) - dim (null (T^{*})) & See (a) \\ = dim (W) - dim (range (T)^{⊥}) \\ = dim (W) - (dim (W) - range (T)) \\ = range (T) \end{aligned}

\end{proof}

6

Theorem

Consider the inner product over $P_{2} (R)$ via:

⟨ p, q ⟩ = \int_{0}^{1} p (x) q (x) d x

Define $T \in P_{2} (R)$ by $T (a_{0} + a_{1} x + a_{2} x^{2}) = a_{1} x$ .

$T$ is not self-adjoint
The matrix of $T$ w.r.t. to the standard basis is:

[\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}]

This matrix equals its conjugate transpose, even though $T$ is not self-adjoint. Explain why this not a contradiction.

\begin{proof}
For (a), notice that we can find $T^{*}$ . Fix $q = x + 1$ and let $p = x + 2$ be arbitrary:

⟨ p, T q ⟩ = ⟨ x + 1, x ⟩ = \int_{0}^{1} (x + 1) x d x = \frac{x^{3}}{3} + \frac{x^{2}}{2} |_{0}^{1} = \frac{5}{6}

while:

⟨ T p, q ⟩ = ⟨ x, x + 2 ⟩ = \int_{0}^{1} x (x + 2) d x = \frac{x^{3}}{3} + x^{2} |_{0}^{1} = \frac{4}{3}

so $⟨ p, T q ⟩ \neq ⟨ T p, q ⟩$ so $T \neq T^{*}$ by counterexample.

For (b), notice that:

T (1) = 0 + 0 x + 0 x^{2}, T (x) = 0 + 1 x + 0 x^{2}, T (x^{2}) = 0 + 0 x + 0 x^{2}

building the matrix:

M (T) = [\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}]

Notice that this matrix is $M (T)^{*}$ then:

M (T)^{*} = {[\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}]}^{*} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 1^{*} & 0 \\ 0 & 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}] = M (T)

This isn't a contradiction because (7.10) assumes that our basis $β$ is an orthonormal-basis, while here they are not orthogonal, nor normal.
\end{proof}

7

Theorem

Suppose $S, T \in L (V)$ are self-adjoint. Then $S T$ is self-adjoint iff $S T = T S$ .

\begin{proof}
We know $T = T^{*}$ and $S = S^{*}$ . $S T$ is self-adjoint iff $(S T) = (S T)^{*}$ which is:

\begin{aligned} S T & = (S T)^{*} \\ = T^{*} S^{*} \\ = T S \end{aligned}

\end{proof}

8

Theorem

Suppose $V$ is a real inner product space. That the set of self-adjoint operators on $V$ is a subspace of $L (V)$ .

\begin{proof}
Consider $S = {T \in L (V) : T = T^{*}}$ . We'll show $S$ is a subspace of $L (V)$ , since clearly it's a subset. Clearly the zero map $\vec{0} \in S$ since:

⟨ \vec{0} v, w ⟩ = ⟨ v, \vec{0} w ⟩ = 0

For additivity, let $T, S \in S$ . Then:

\begin{aligned} T + S & = T^{*} + S^{*} \\ = (T + S)^{*} \end{aligned}

thus $T + S \in S$ .

For homogeneity, let $λ \in F$ :

\begin{aligned} λ T & = λ T^{*} \\ = (\overset{―}{λ} T)^{*} \\ = (λ T)^{*} \end{aligned}

since $V$ is a real inner product space, so $λ = \overset{―}{λ}$ . Thus $λ T \in S$ . Thus $S$ is a subspace.
\end{proof}

9

Theorem

Suppose $V$ is a complex inner product space. Show that the set of self-adjoint operators on $V$ is NOT a subspace.

\begin{proof}
Notice that this is the same case as HW 3 - Self-Adjoint and Normal Operators#8 EXCEPT for homogeneity, where now $λ$ may not be $\overset{―}{λ}$ . As such, as a counterexample, let $λ = i$ , and have the self-adjoint operator be the identity operator on $V$ . Then:

i I = i I^{*} = (- i I)^{*} \neq (i I)^{*}

Showing $S$ is not a subspace.
\end{proof}

13

Theorem

There is an example of an operator $T \in L (C^{4})$ such that $T$ is normal but not self-adjoint.

\begin{proof}
Choose $T$ by:

T (e_{1}) = e_{2}, T (e_{2}) = - e_{1}, T (e_{3}) = 0, T (e_{4}) = 0

notice:

M (T) = [\begin{matrix} 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]

We can derive:

M (T^{*}) = {[\begin{matrix} 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]}^{*} = [\begin{matrix} 0 & 1 & 0 & 0 \\ - 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}] = - M (T)

Thus $T^{*} \neq T$ while $T \circ T^{*} = T^{*} \circ T$ since:

M (T) M (T^{*}) = - M (T)^{2} = M (T^{*}) M (T)

since
\end{proof}

14

Theorem

Suppose $T$ is a normal operator on $V$ . Suppose also that $v, w \in V$ satisfy the equations:

‖ v ‖ = ‖ w ‖ = 2, T v = 3 v, T w = 4 w

Then:

‖ T (v + w) ‖ = 10

\begin{proof}
Notice:

\begin{aligned} T (v + w) & = T v + T w \\ = 3 v + 4 w \end{aligned}

so since $T$ and $T^{*}$ have the same eigenvectors, so then $T^{*} v = 3 v$ and $T^{*} w = 4 w$ :

T^{*} (v + w) = T^{*} v + T^{*} w = 3 v + 4 w

and since $T = T^{*}$ :

\begin{aligned} ‖ T (v + w) ‖ & = ‖ T^{*} (v + w) ‖ \\ \sqrt{⟨ 3 v + 4 w ⟩} & = \sqrt{‖ 3 v ‖^{2} + ‖ 4 w ‖^{2} + \underset{= 0}{\underset{⏟}{2 ⟨ 3 v, 4 w ⟩}}} \\ = \sqrt{3^{2} \cdot 2^{2} + 4^{2} \cdot 2^{2}} \\ = 2 \sqrt{5^{2}} \\ = 10 \end{aligned}

Since $T$ is normal then $v, w$ as eigenvectors are orthogonal so $⟨ v, w ⟩ = 0$ above.
\end{proof}

16

Theorem

Suppose $T \in L (V)$ is normal. Then:

range (T) = range (T^{*})

\begin{proof}
Let's actually prove that $null (T) = null (T^{*})$ . Let $u \in null (T)$ . Then:

\begin{aligned} u & \in null (T) \\ \Leftrightarrow T u = 0 \\ \Leftrightarrow ‖ T u ‖ = 0 \\ \Leftrightarrow ⟨ T u, T u ⟩ = 0 \\ \Leftrightarrow ⟨ T^{*} T u, u ⟩ = 0 \\ \Leftrightarrow ⟨ T T^{*} u, u ⟩ = 0 & T T^{*} = T^{*} T \\ \Leftrightarrow ⟨ T^{*} u, T^{*} u ⟩ = 0 \\ \Leftrightarrow ‖ T^{*} u ‖ = 0 \\ \Leftrightarrow T^{*} u = 0 \\ \Leftrightarrow u \in null (T^{*}) \end{aligned}

We prove the range part using the properties of these ranges and nullspaces:

\begin{aligned} range (T) & = (null (T^{*}))^{⊥} \\ = (null (T))^{⊥} \\ = range (T^{*}) \end{aligned}

\end{proof}

17

Theorem

Suppose $T \in L (V)$ is normal. Then:

null (T^{k}) = null (T), range (T^{k}) = range (T)

for all $k \in N^{+}$ .

\begin{proof}
Let's prove this via induction. For $k = 1$ , the theorem is clearly true, so consider some $k - 1 \in N^{+}$ holds for the theorem. We'll prove the $k$ case. Notice:

\begin{aligned} v & \in null (T^{k - 1}) \\ \Rightarrow T^{k - 1} v = 0 \\ \Rightarrow T (T^{k - 1} v) = T 0 \\ \Rightarrow T^{k} v = 0 \\ \Rightarrow v \in null (T^{k}) \end{aligned}

Thus $v \in null (T^{k - 1}) \Rightarrow v \in null (T^{k})$ so $null (T^{k - 1}) \subseteq null (T^{k})$ . Since $null (T) = null (T^{k - 1})$ via the inductive hypothesis, then $null (T) \subseteq null (T^{k})$ . For the other way let $v \in null (T^{k})$ . Then:

\begin{aligned} T^{k} v & = 0 \\ T (T^{k - 1} v) & = 0 \\ T^{*} (T^{k - 1} v) & = 0 & null (T) = null (T^{*}) \\ ⟨ T^{*} T^{k - 1} v, T^{*} T^{k - 1} v ⟩ & = 0 \\ ⟨ T^{k - 1} v, T^{k - 1} v ⟩ & = 0 \\ \Leftrightarrow v \in null (T^{k - 1}) \\ \Leftrightarrow v \in null (T^{k}) \end{aligned}

Showing the range is easy since:

range (T^{k}) = null ((T^{k})^{*})^{⊥} = null ((T^{*})^{k})^{⊥} = null (T^{*})^{⊥} = range (T)

\end{proof}

19

Theorem

Suppose $T \in L (C^{3})$ is normal and $T (1, 1, 1) = (2, 2, 2)$ . Suppose $(z_{1}, z_{2}, z_{3}) \in null (T)$ . Then $z_{1} + z_{2} + z_{3} = 0$ .

\begin{proof}
Because $T$ is normal, then all eigenvectors of $T$ are orthogonal to one another. Notice that since $T (1, 1, 1) = 2 (1, 1, 1)$ then $v_{1} = (1, 1, 1)$ is an eigenvector of $T$ with $λ_{1} = 2$ . Further notice that since $(z_{1}, z_{2}, z_{3}) \in null (T)$ then $T (z_{1}, z_{2}, z_{3}) = 0 (z_{1}, z_{2}, z_{3})$ so $v_{2} = (z_{1}, z_{2}, z_{3})$ is an eigenvector with $λ_{2} = 0$ .

Therefore, $v_{1}, v_{2}$ are orthogonal, so $⟨ v_{1}, v_{2} ⟩ = 0 = z_{1} + z_{2} + z_{3}$ by doing the dot product.
\end{proof}