HW 4 - The Spectral Theorem

7.B: The Spectral Theorem

2

Theorem

Suppose $T$ is a self-adjoint operator on a finite-dimensional inner product space and $2, 3$ are the only eigenvalues of $T$ . Then:

T^{2} - 5 T + 6 I = 0

Proof
Notice:

T^{2} - 5 T + 6 I = (T - 2 I) (T - 3 I)

Let $v \in V$ be applied to these. Since $V$ is a finite-dimensional inner product space, we have to consider if $V$ is a complex or a real vector space.

For a complex vector space, using the Spectral Theorem yields that, since $T$ is self-adjoint and thus normal, that $V$ has an orthonormal basis consisting of eigenvectors of $T$ . Notice that since $2, 3$ are some eigenvalues, then their corresponding eigenvectors $v_{1}, v_{2}$ compose this basis. Notice since these are the only eigenvalues of $T$ , then $β = {v_{1}, v_{2}}$ is the only valid orthonormal eigenbasis for $V$ . As such, any $v \in V$ is spanned via:

v = α_{1} v_{1} + α_{2} v_{2}

so then:

\begin{aligned} (T^{2} - 5 T + 6 I) v & = (T - 2 I) (T - 3 I) (α_{1} v_{1} + α_{2} v_{2}) \\ = (T - 2 I) (T - 3 I) α_{1} v_{1} + (T - 2 I) \underset{= \vec{0}}{\underset{⏟}{(T - 3 I) α_{2} v_{2}}} \\ = (T - 3 I) \underset{= \vec{0}}{\underset{⏟}{(T - 2 I) α_{1} v_{1}}} + \vec{0} \\ = \vec{0} \end{aligned}

where the $\vec{0}$ 's come from the fact that $v_{1} \in null (T - 2 I)$ and $v_{2} \in null (T - 3 I)$ . Thus, since $v$ was arbitrary, then $T^{2} - 5 I + 6 I = 0$ .

For a real vector space, the same process applies since $T$ is adjoint, so using the Real Spectral Theorem gives that $V$ has the same orthonormal eigenbasis.
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3

Theorem

Give an example of an operator $T \in L (C^{3})$ such that $2$ and $3$ are the only eigenvalues of $T$ while $T^{2} - 5 T + 6 I \neq 0$ .

Proof
As the previous theorem implies, we should construct $T$ to not be self-adjoint. As such, define $T$ by it's actions on the standard basis, given by the matrix:

M (T) = [\begin{matrix} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 2 \end{matrix}]

Notice that there the only eigenvalues here are $3$ and $2$ . We can see this because there are only the eigenvectors:

v_{1} = (1, 1, 0), v_{2} = (1, 0, 0)

Now notice that:

\begin{aligned} M (T^{2} - 5 T + 6 I) & = {[\begin{array}{c} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 2 \end{array}]}^{2} - 5 [\begin{array}{c} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 2 \end{array}] + 6 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ = [\begin{array}{c} 4 & 0 & 0 \\ 0 & 9 & 5 \\ 0 & 0 & 4 \end{array}] - [\begin{array}{c} 10 & 5 & 0 \\ 0 & 15 & 5 \\ 0 & 0 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}] \\ = [\begin{array}{c} 0 & - 5 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] \\ \neq M (\vec{0}, β) \end{aligned}

Thus $T^{2} - 5 T + 6 I \neq \vec{0}$ .
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6

Theorem

A normal operator on a complex inner product space is self-adjoint iff all its eigenvalues are real.

Proof
Consider normal operator $T \in L (V)$ where $V$ is a complex inner product space. Then by the Complex Spectral Theorem, then $V$ has an orthonormal basis consisting of eigenvectors in $T$ .

If $T$ is self-adjoint then clearly by Chapter 7 - Operators on Inner Product Spaces#^6693ec then we have all eigenvalues of $T$ as real.

For the other direction, suppose that $T$ only has real eigenvalues $λ_{i} \in R$ with corresponding eigenvectors $v_{1}, . . ., v_{n}$ . We must have $n$ of them because we have an orthonormal basis of eigenvectors by $T$ (see previous paragraphs). And since $T$ only has real eigenvalues, then $v_{1}, . . ., v_{n}$ have only real corresponding eigenvalues.

Using Chapter 7 - Operators on Inner Product Spaces#^a30225, we can show $T$ is self-adjoint by showing $⟨ T v, v ⟩$ as real. Notice that if $v \in V$ is arbitrary, then since $v_{1}, . . ., v_{n}$ is a basis then:

v = \sum_{i = 1}^{n} α_{i} v_{i}

Now notice that:

\begin{aligned} ⟨ T v, v ⟩ & = ⟨ T (\sum_{i = 1}^{n} α_{i} v_{i}), \sum_{i = 1}^{n} α_{i} v_{i} ⟩ \\ = ⟨ \sum_{i = 1}^{n} α_{i} T (v_{i}), \sum_{i = 1}^{n} α_{i} v_{i} ⟩ \\ = ⟨ \sum_{i = 1}^{n} α_{i} λ_{i} v_{i}, \sum_{i = 1}^{n} α_{i} v_{i} ⟩ \\ = \sum_{i = 1}^{n} ⟨ α_{i} λ_{i} v_{i}, \sum_{j = 1}^{n} α_{j} v_{j} ⟩ \\ = \sum_{i = 1}^{n} λ_{i} ⟨ α_{i} v_{i}, \sum_{j = 1}^{n} α_{j} v_{j} ⟩ \\ = \sum_{i, j = 1}^{n} λ_{i} α_{i} \overset{―}{α_{j}} ⟨ v_{i}, v_{j} ⟩ \\ = \sum_{i = 1}^{n} λ_{i} α_{i} \overset{―}{α_{i}} \\ = \sum_{i = 1}^{n} λ_{i} | α_{i} |^{2} \\ \in R \end{aligned}

Where the last step is from the fact that $λ_{i} \in R$ and even though $α_{i} \in C$ we know that $| α_{i} |^{2} \in R$ regardless. Thus, we must have it that $T$ is self-adjoint as a result, since $v \in V$ was an arbitrary choice.
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7

Theorem

Suppose $V$ is a complex inner product space and $T \in L (V)$ is a normal operator such that $T^{9} = T^{8}$ . Then $T$ is self-adjoint and $T^{2} = T$ .

Proof
Since $T$ is normal, then $V$ has an orthonormal basis consisting of eigenvectors of $T$ , namely $β = {v_{1}, . . ., v_{n}}$ where each $v_{i}$ has corresponding eigenvalue $λ_{i} \in C$ (we don't know yet if $T$ is self-adjoint, so we'll assume they $λ_{i}$ 's are all complex for now). Suppose $T^{9} = T^{8}$ .

Via the complex spectral theorem then since $T$ is normal then $T$ has a diagonal matrix with respect to our orthonormal basis:

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

since $T^{8} = T^{9}$ :

\begin{aligned} {[\begin{array}{c} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n} \end{array}]}^{8} & = {[\begin{array}{c} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n} \end{array}]}^{9} \\ [\begin{array}{c} λ_{1}^{8} & 0 & \dots & 0 \\ 0 & λ_{2}^{8} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n}^{8} \end{array}] & = [\begin{array}{c} λ_{1}^{9} & 0 & \dots & 0 \\ 0 & λ_{2}^{9} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n}^{9} \end{array}] \end{aligned}

Thus each $λ_{i}^{8} = λ_{i}^{9}$ . As a result, then factoring out gives that $λ_{i} = 0, 1$ as possibilities. As such, then all $λ_{i} \in R$ so then $T$ must be self-adjoint. Furthermore, then notice that for all $λ_{i}$ 's that if $λ_{i} = 0$ then $λ_{i}^{2} = 0$ so $λ_{i} = λ_{i}^{2}$ . The same is said if $λ_{i} = 1$ . Hence, the following is true:

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

Thus $T^{2} = T$ .
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8

Theorem

Give an example of an operator $T$ on a complex vector space such that $T^{9} = T^{8}$ but $T^{2} \neq T$ .

Proof
We need to just give an operator $T$ that isn't normal, mainly one that isn't invertible. As such, have $T$ be defined by actions on whatever basis $β = {v_{1}, . . ., v_{n}}$ of $V$ (which is $n$ dimensional). For simplicity, let $V = C^{2}$ and $F = C$ . Choose $T$ such that:

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

Notice that:

M (T^{2}, β) = {[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]}^{2} = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] \neq M (T, β)

While:

M (T^{8}, β) = \vec{0} = M (T^{9}, β)

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9

Theorem

Suppose $V$ is a complex inner product space. Then every normal operator on $V$ has a square root. Namely, every normal operator $T \in L (V)$ has an operator $S \in L (V)$ where $S^{2} = T$ .

Proof
Since $V$ is a complex inner product space and $T \in L (V)$ is an arbitrary normal operator, then by the Complex Spectral Theorem then $T$ has a diagonal matrix with respect to some orthonormal basis of $V$ . Namely:

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

where $λ_{i} \in C$ . Choose $S \in L (V)$ to be where:

M (S, β) = [\begin{matrix} \sqrt{λ_{1}} & 0 & \dots & 0 \\ 0 & \sqrt{λ_{2}} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & 0 \\ 0 & 0 & \dots & \sqrt{λ_{n}} \end{matrix}]

Notice that:

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

Thus $S^{2} = T$ .
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10

Theorem

Give an example of a real inner product space $V$ and $T \in L (V)$ and real numbers $b, c$ with $b^{2} < 4 c$ such that $T^{2} + b T + c I$ is not invertible.

Proof
As the hint suggests, we should try to have $T$ not be self-adjoint. Namely, via the proof of Chapter 7 - Operators on Inner Product Spaces#^94dc7b, we should try to find some $T$ such that $⟨ T^{2} v, v ⟩ ≱ ‖ T v ‖^{2}$ . Have $V = R^{2}$ and $F = R$ . Define $T$ where $β$ is the standard basis by:

M (T, β) = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] $ $ N o t i c e t h a t w e c a n f i n d :

\begin{align}
\mathcal{M}(T^2 + bT + cI, \beta) &= \begin{bmatrix}
0 & -1 \1 & 0
\end{bmatrix}^2 + b\begin{bmatrix}
0 & -1 \1 & 0
\end{bmatrix} + c \begin{bmatrix}
1 & 0 \
0 & 1
\end{bmatrix}\
&= \begin{bmatrix}
-1 & 0 \
0 & -1
\end{bmatrix} + \begin{bmatrix}
b & b \
0 & b
\end{bmatrix} + \begin{bmatrix}
c & 0 \
0 & c
\end{bmatrix}\
&= \begin{bmatrix}
-1 + b + c & b \
0 & -1 + b + c
\end{bmatrix}
\end

C h o o s e a s a n e x a m p l e $ b = 0 $ a n d $ c = 1 $ . T h e n :

\mathcal{M}(T^2 + bT + cI, \beta) = \begin{bmatrix}
0 & 0 \
0 & 0
\end

Which is not invertible as it's the zero map. ☐ ### 11 > [!theorem] > Every self-adjoint operator on $V$ has a cube root, an operator $S \in \mathcal{L}(V)$ such that $S^3 = T$. *Proof* If $T \in \mathcal{L}(V)$ is a self-adjoint operator on $V$, then by either the Real or Complex Spectral Theorem, then $T$ has a diagonal matrix with respect to some orthonormal basis of $V$, called $\beta = \{v_1, ..., v_n\}$. As a result:

\mathcal{M}(T,\beta) = \begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \
0 & \lambda_2 & \dots & 0 \
\vdots & \vdots & \ddots & \vdots \
0 & 0 & \dots & \lambda_n
\end

T h e n c h o o s e $ S \in L (V) $ w h e r e :

\mathcal{M}(S,\beta) = \begin{bmatrix}
\sqrt[3]{\lambda_1} & 0 & \dots & 0 \
0 & \sqrt[3]{\lambda_2} & \dots & 0 \
\vdots & \vdots & \ddots & \vdots \
0 & 0 & \dots & \sqrt[3]{\lambda_n}
\end

W h i c h w o r k s s i n c e n o m a t t e r o u r r e a l o r c o m p l e x s p a c e, $ \sqrt[3]{λ_{i}} $ i s s t i l l i n t h a t s p a c e . T h e n n o t i c e t h a t :

\mathcal{M}(S^3,\beta) = \begin{bmatrix}
\sqrt[3]{\lambda_1} & 0 & \dots & 0 \
0 & \sqrt[3]{\lambda_2} & \dots & 0 \
\vdots & \vdots & \ddots & \vdots \
0 & 0 & \dots & \sqrt[3]{\lambda_n}
\end{bmatrix}^3 = \begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \
0 & \lambda_2 & \dots & 0 \
\vdots & \vdots & \ddots & \vdots \
0 & 0 & \dots & \lambda_n
\end{bmatrix} = \mathcal{M}(T,\beta)

Thus $S^3 = T$ as desired. ☐ ### 14 > [!theorem] > Suppose $U$ is a finite-dimensional real vector space and $T \in \mathcal{L}(U)$. Then $U$ has a basis consisting of eigenvectors of $T$ iff there is an inner product on $U$ that makes $T$ into a self-adjoint operator. *Proof* $(\rightarrow)$: Suppose $U$ has a basis $u_1, ..., u_m$ consisting of eigenvectors of $T$. Thus $Tu_i = \lambda_iu_i$ for all $i \in \{1, ..., m\}$ where $\lambda_i \in \mathbb{C}$. Create the inner product $\braket{\cdot , \cdot}_U$ such that:

\braket{u_i, u_j}_U = \chi(i = j)

s o t h a t a n y $ {⟨ u_{i}, u_{i} ⟩}_{U} = 1 $ a n d $ {⟨ u_{i}, u_{j} ⟩}_{U} = 0 $ f o r $ i \neq j $ . W e f i r s t s h o w t h a t t h i s i s a v a l i d i n n e r p r o d u c t . - P o s i t i v e - D e f i n i t e n e s s : B y t h e d e f i n i t i o n i t^{'} s p o s i t i v e - d e f i n i t e - C o n j u g a t e s y m m e t r y : S i n c e $ {⟨ \cdot, \cdot ⟩}_{U} \in R $ t h e n i t h a s c o n j u g a t e s y m m e t r y . - L i n e a r i t y : N o t i c e $ \vec{x}, \vec{y}, \vec{z} $ a r e s p a n n e d b y $ u_{1}, . . ., u_{m} $ s o t h e n f o r $ a, b \in F $ :

\begin{align}
\braket{a\vec{x} + b\vec{y}, \vec{z}} &= \left\langle a\sum_{i=1}^n \alpha_iu_i + b\sum_{i=1}^n \beta_iu_i, \sum_{i=1}^n \gamma_iu_i \right\rangle\
&= \left\langle \sum_{i=1}^n (a\alpha_i + b\beta_i)u_i, \sum_{i=1}^n \gamma_iu_i \right\rangle \
&= \sum_{i,j=1}^n (a\alpha_i + b\beta_i)\overline{\gamma_j} \left\langle u_i, u_j \right\rangle\
&= \sum_{i=1}^n (a\alpha_i + b\beta_i)\overline{\gamma_i} \
&= \sum_{i=1}^n a\alpha_i\braket{u_i, \gamma_iu_i} + \sum_{i=1}^n b\beta_i\braket{u_i, \gamma_iu_i}\
&= \sum_{i=1}^n a\braket{\alpha_iu_i, \gamma_iu_i} + \sum_{i=1}^n b\braket{\beta_iu_i, \gamma_iu_i}\
&= \braket{a\vec{x}, \vec{z}} + \braket{b\vec{y}, \vec{z}}
\end

Thus, notice that this forces all $u_1, ..., u_m$ to be specifically an *orthonormal* eigenbasis for $U$ (as the basis vectors are eigenvectors). Thus, using the Real Spectral Theorem, then $T$ must be self-adjoint. $(\leftarrow)$: Suppose there exists an inner product on $U$, denoted $\braket{\cdot, \cdot}_U$, that makes $T$ a self-adjoint operator. Since $U$ is a real vector space and $T$ is self-adjoint, then by the Real Spectral Theorem, then $U$ has an orthonormal basis $u_1, ..., u_m$ consisting of eigenvectors of $T$. ☐ ### 15 > [!theorem] > Find the matrix entry below that is covered up. > ![Pasted image 20240430205804.png](/img/user/1%20Attachments/12%20Images/Pasted%20image%2020240430205804.png) *Proof* Let:

A = \begin{bmatrix}
1 & 1 & 0 \
0 & 1 & 1 \
1 & 0 & a
\end

W e n e e d t o d e t e r m i n e $ a $ s u c h t h a t :

AA^* = A^*A

AA^* = \begin{bmatrix}
1 & 1 & 0 \
0 & 1 & 1 \
1 & 0 & a
\end{bmatrix}\begin{bmatrix}
1 & 0 & 1 \
1 & 1 & 0 \
0 & 1 & a
\end{bmatrix} = \begin{bmatrix}
2 & 1 & 1 \
1 & 2 & a \
1 & a & 1 + a^2
\end

A^*A = \begin{bmatrix}
1 & 0 & 1 \
1 & 1 & 0 \
0 & 1 & a
\end{bmatrix}\begin{bmatrix}
1 & 1 & 0 \
0 & 1 & 1 \
1 & 0 & a
\end{bmatrix} = \begin{bmatrix}
2 & 1 & a \
1 & 2 & 1 \
a & 1 & 1 + a^2
\end

E q u a t i n g e n t r i e s s h o w s t h a t :

\boxed

T h u s t h e n :

A = \begin{bmatrix}
1 & 1 & 0 \
0 & 1 & 1 \
1 & 0 & 1
\end

☐