HW 7 - Inner Product Spaces

6.B - 1,3,4,5,6,9,12

6.A: Inner Products and Norms

1

Theorem

The function that takes $((x_{1}, x_{2}), (y_{1}, y_{2})) \in R^{2} \times R^{2}$ to $| x_{1} y_{1} | + | x_{2} y_{2} |$ is not an inner product on $R^{2}$ .

Proof
We'll show that one of the properties isn't satisfied. Namely, the first two are easily shown to be satisfied no matter what. But notice that the additivity in the first slot is not. Consider $u = (1, 2)$ and $v = (- 1, - 2)$ and $w = (1, 1)$ . Then:

⟨ u + v, w ⟩ = ⟨ (0, 0), (1, 1) ⟩ = | 0 \cdot 1 | + | 0 \cdot 1 | = 0

while:

⟨ u, w ⟩ + ⟨ v, w ⟩ = ⟨ (1, 2), (1, 1) ⟩ + ⟨ (- 1, - 2), (1, 1) ⟩ = 3 + 3 = 6

hence the property of additivity in the first slot is not satisfied, so the function is not a valid inner product of $R^{2}$ .
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2

Theorem

The function that takes $((x_{1}, . . ., x_{3}), (y_{1}, . . ., y_{3})) \in R^{3} \times R^{3}$ to $x_{1} y_{1} + x_{3} y_{3}$ is not an inner product of $R^{3}$ .

Proof
We easily can show that if $v = (0, 1, 0) \in R^{3}$ that clearly $v \neq \vec{0}$ while:

⟨ v, v ⟩ = 0 \cdot 0 + 0 \cdot 0 = 0

so the property of definiteness is not satisfied.
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3

Consider when $F = R$ and $V \neq {\vec{0}}$ . We replace positivity ( $\forall v \in V (⟨ v, v ⟩ \geq 0)$ ) with the condition that $⟨ v, v ⟩ > 0$ for some $v \in V$ . We'll show that the set of functions from $V \times V \to R$ that are inner products on $V$ do not change.

Proof
Since $F = R$ then the conjugate symmetry property says that $⟨ u, v ⟩ = ⟨ v, u ⟩$ for any $u, v \in V$ . We also still have the additivity and homogeneity in the first slot, as well as definiteness.

Let $⟨ \cdot, \cdot ⟩$ be this arbitrary inner product operation. If we can show, using the properties prior to the change, that we get the new property, and if we use the new property along with the others to prove the switched-out property, then the proof is complete.

First, consider our properties before the change. Notice that since $V \neq {\vec{0}}$ then clearly $V \neq {}$ so then there must be some other non-zero vector $v \in V$ . Hence, by positivity then $⟨ v, v ⟩ \geq 0$ . Notice that $v \neq \vec{0}$ so then by definiteness we must have $⟨ v, v ⟩ \neq 0$ so then we must have $⟨ v, v ⟩ > 0$ as required.

Now, consider the properties after the change (so we cannot use positivity like we did before). Hence, there is some $v \in V$ where $⟨ v, v ⟩ > 0$ . Since this is the case, clearly then $v \neq \vec{0}$ . As such, then $V$ must contain the zero-vector and at least one other non-zero vector $v$ . Notice that if we let $u \in V$ be arbitrary, if $u = \vec{0}$ then clearly $⟨ u, u ⟩ = 0 \geq 0$ and if $u \neq \vec{0}$ then assume for contradiction that $⟨ u, u ⟩ < 0$ .

Consider the following cases. If $u$ is a linear combination of $v$ , then $u = λ v$ so then:

\begin{aligned} ⟨ u, u ⟩ & = ⟨ λ v, λ v ⟩ \\ = λ ⟨ v, λ v ⟩ \\ = λ \overset{―}{⟨ λ v, v ⟩} \\ = λ ⟨ λ v, v ⟩ \\ = λ^{2} ⟨ v, v ⟩ \\ > 0 & (λ^{2} > 0 \land ⟨ v, v ⟩ > 0) \end{aligned}

which contradicts our assumption. Suppose instead that $u$ is not a linear combination of $v$ , so then the vector $u + λ v \neq 0$ since if it was we'd get a linear combination. As such, notice that:

\begin{aligned} ⟨ u + λ v, u + λ v ⟩ & = ⟨ u, u ⟩ + ⟨ λ v, λ v ⟩ + ⟨ u, λ v ⟩ + ⟨ λ v, u ⟩ \\ = ⟨ u, u ⟩ + λ^{2} ⟨ v, v ⟩ + 2 λ ⟨ u, v ⟩ \end{aligned}

But this equals 0, so use the quadratic equation to try to solve for $λ$ in terms of everything else:

λ = \frac{- 2 ⟨ u, v ⟩ + \sqrt{(⟨ v, v ⟩)^{2} - 4 ⟨ v, v ⟩ ⟨ u, u ⟩}}{2 ⟨ u, u ⟩}

Since $⟨ v, v ⟩ > 0$ and $⟨ u, u ⟩ < 0$ then:

(⟨ v, v ⟩)^{2} - 4 ⟨ v, v ⟩ ⟨ u, u ⟩ > 0

so then $λ \in R$ specifically. So choosing this $λ$ gives:

⟨ u + λ v, u + λ v ⟩ = 0

which contradicts $u + λ v \neq 0$ , so then our assumption was wrong. Hence $⟨ u, u ⟩ \geq 0$ .
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4

Suppose $V$ is a real inner product space.

a

Theorem

$⟨ u + v, u - v ⟩ = ‖ u ‖^{2} - ‖ v ‖^{2}$ for all $u, v \in V$

Proof
Notice:

\begin{aligned} ⟨ u + v, u - v ⟩ & = ⟨ u, v ⟩ + ⟨ u, - v ⟩ + ⟨ u, u ⟩ + ⟨ v, - v ⟩ \\ = ⟨ u, v - v ⟩ + ‖ u ‖^{2} - ⟨ v, v ⟩ \\ = ⟨ u, \vec{0} ⟩ + ‖ u ‖^{2} - ‖ v ‖^{2} \\ = ‖ u ‖^{2} - ‖ v ‖^{2} \end{aligned}

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b

Theorem

If $u, v \in V$ have the same norm, then $u + v$ is orthogonal to $u - v$ .

Proof
Suppose $u, v$ have the same norm, so then $‖ u ‖ = ‖ v ‖$ . We'll show that $⟨ u + v, u - v ⟩ = 0$ :

\begin{aligned} ⟨ u + v, u - v ⟩ & = ‖ u ‖^{2} - ‖ v ‖^{2} & (Last Theorem) \\ = (‖ u ‖ - ‖ v ‖) (‖ u ‖ + ‖ v ‖) \\ = 0 (‖ u ‖ + ‖ v ‖) & (‖ u ‖ = ‖ v ‖) \\ = 0 \end{aligned}

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c

Theorem

The diagonals of a rhombus are perpendicular to each other.

Proof
Consider the following arbitrary rhombus:

We need to show that $⟨ v + u, v - u ⟩ = 0$ . But we showed that via HW 7 - Inner Product Spaces#^428e80.
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5

Theorem

Suppose $T \in L (V)$ is such that $‖ T v ‖ \leq ‖ v ‖$ for all $v \in V$ . Then $T - \sqrt{2} I$ is invertible.

Proof
If we can show that $T$ cannot have an eigenvalue of $\sqrt{2}$ , then we've shown that therefore $T - \sqrt{2} I$ is invertible. As such, assume for contradiction that $\sqrt{2}$ is an eigenvalue of $T$ . As such, there's some associated eigenvector $v_{o} \in V$ . As such, then $T v_{o} = \sqrt{2} v$ , but notice then that:

‖ T v_{o} ‖ = ‖ \sqrt{2} v_{o} ‖ = | \sqrt{2} | \cdot ‖ v_{o} ‖ > ‖ v_{o} ‖

which contradicts our supposition that $\forall v \in V (‖ T v ‖ \leq ‖ v ‖)$ , so then our assumption was false. Hence, $\sqrt{2}$ cannot be an eigenvalue of $T$ , so then $T - \sqrt{2} I$ is invertible.
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7

Theorem

Suppose $u, v \in V$ . Then $‖ a u + b v ‖ = ‖ b u + a v ‖$ for all $a, b \in R$ iff $‖ u ‖ = ‖ v ‖$ .

Proof

The forward direction is easiest. Notice if $a = 1$ and $b = 0$ then we get our desired result:

‖ a u + b v ‖ = ‖ b u + a v ‖ \Rightarrow ‖ u ‖ = ‖ v ‖

Let's prove the reverse direction. Suppose $‖ u ‖ = ‖ v ‖$ . Let $a, b \in R$ be arbitrary. Notice that:

\begin{aligned} ‖ a u + b v ‖^{2} & = ⟨ a u + b v, a u + b v ⟩ \\ = ⟨ a u, a u ⟩ + ⟨ b v, b v ⟩ + ⟨ a u, b v ⟩ + ⟨ b v, a u ⟩ \\ = ‖ a u ‖^{2} + ‖ b v ‖^{2} + a b (⟨ u, v ⟩ + ⟨ v, u ⟩) & (a, b \in R \to ⟨ a u, b v ⟩ = ⟨ b v, a u ⟩) \\ = | a |^{2} ‖ u ‖^{2} + | b |^{2} ‖ v ‖^{2} + a b (⟨ u, v ⟩ + ⟨ v, u ⟩) \\ = | a |^{2} ‖ v ‖^{2} + | b |^{2} ‖ u ‖^{2} + a b (⟨ u, v ⟩ + ⟨ v, u ⟩) \\ = ‖ a v ‖^{2} + ‖ b u ‖^{2} + a b (⟨ u, v ⟩ + ⟨ v, u ⟩) \\ = ⟨ a v, a v ⟩ + ⟨ b u, b u ⟩ + ⟨ a v, b u ⟩ + ⟨ b u, a v ⟩ \\ = ⟨ a v + b u, a v + b u ⟩ \\ = ‖ b u + a v ‖^{2} \end{aligned}

Square rooting both sides gives the desired result.
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8

Theorem

Suppose $u, v \in V$ and $‖ u ‖ = ‖ v ‖ = 1$ and $⟨ u, v ⟩ = 1$ . Then $u = v$ .

Proof
Suppose the necessary assumptions. Notice that if we prove that $u - v = \vec{0} \leftrightarrow ‖ u - v ‖ = 0$ then we've proved the above theorem:

\begin{aligned} ‖ u - v ‖^{2} & = ⟨ u - v, u - v ⟩ \\ = ⟨ u, u ⟩ + ⟨ - v, - v ⟩ + ⟨ u, - v ⟩ + ⟨ u, - v ⟩ \\ = 1 + (- 1) (- 1) (1) + 2 ⟨ u, - v ⟩ \\ = 1 + 1 + 2 (- 1) ⟨ u, v ⟩ \\ = 2 - 2 (1) \\ = 0 \end{aligned}

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10

Theorem

There exist vectors $u, v \in R^{2}$ such that $u$ is a scalar multiple of $(1, 3)$ and $v$ is orthogonal to $(1, 3)$ and $(1, 2) = u + v$ .

Proof
Using the idea of orthogonalization, here have $u = c (1, 3) \Rightarrow u_{1} = c, u_{2} = 3 c$ . Notice then that since $v$ is orthogonal to $(1, 3)$ then:

⟨ v, (1, 3) ⟩ = 0 \Rightarrow v_{1} + 3 v_{2} = 0

Further we know that:

(1, 2) = u + v = (u_{1} + v_{1}, u_{2} + v_{2}) \Rightarrow {\begin{cases} c + v_{1} = 1 \\ 3 c + v_{2} = 2 \end{cases}

Add $R_{1} + 3 R_{2}$ to get:

\begin{aligned} c + v_{1} + 9 c + 3 v_{2} & = 1 + 3 \cdot 2 = 7 \\ 10 c + v_{1} + 3 v_{2} & = 7 \\ 10 c + 0 & = 7 \\ c & = \frac{7}{10} \end{aligned}

Thus:

u = (\frac{7}{10}, \frac{21}{10}), v = (\frac{3}{10}, \frac{- 1}{10})

Where our properties above are validated by these vectors.
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12

Theorem

For all $n \in Z^{+}$ and all $x_{1}, . . ., x_{n} \in R$ :

(x_{1} + \dots + x_{n})^{2} \leq n (x_{1}^{2} + \dots + x_{n}^{2})

Proof
Consider $x = (x_{1}, . . ., x_{n}) \in R^{n}$ . Notice that:

‖ x ‖ = \sqrt{x_{1}^{2} + x_{2}^{2} + \dots + x_{n}^{2}}

Furthermore, consider $e = (1, 1, . . ., 1) \in R^{n}$ . Here, by Cauchy-Schartz:

\begin{aligned} | ⟨ x, e ⟩ | & \leq ‖ x ‖ ‖ e ‖ \\ | x_{1} + \dots + x_{n} | & \leq \sqrt{x_{1}^{2} + \dots + x_{n}^{2}} \sqrt{n} \\ (x_{1} + \dots + x_{n})^{2} & \leq n (x_{1}^{2} + \dots + x_{n}^{2}) \end{aligned}

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13

Theorem

Suppose $u, v \neq \vec{0}$ in $R^{2}$ . Then:

⟨ u, v ⟩ = ‖ u ‖ ‖ v ‖ \cos (θ)

where $θ$ is the angle between $u, v$ .

Proof
We draw out our triangle:

Using the law of cosines:

\begin{aligned} ‖ u - v ‖^{2} & = ‖ u ‖^{2} + ‖ v ‖^{2} - 2 ‖ u ‖ ‖ v ‖ \cos (θ) \\ ⟨ u - v, u - v ⟩ & = ⟨ u, u ⟩ + ⟨ v, v ⟩ - 2 ‖ u ‖ ‖ v ‖ \cos (θ) \\ ⟨ u, u ⟩ + ⟨ v, v ⟩ - 2 ⟨ u, v ⟩ & = ⟨ u, u ⟩ + ⟨ v, v ⟩ - 2 ‖ u ‖ ‖ v ‖ \cos (θ) \\ ⟨ u, v ⟩ & = ‖ u ‖ ‖ v ‖ \cos (θ) \end{aligned}

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17

Theorem

There is no inner product on $R^{2}$ such that the associated norm is given by:

‖ (x, y) ‖ = max {x, y}

Proof
Notice that if $x = (x_{1}, x_{2}), y = (y_{1}, y_{2}) \in R^{2}$ then:

\begin{aligned} ‖ x + y ‖^{2} & = ⟨ x, x ⟩ + ⟨ y, y ⟩ + 2 ⟨ x, y ⟩ \\ = ‖ x ‖^{2} + ‖ y ‖^{2} + 2 ⟨ x, y ⟩ \\ max {x_{1} + y_{1}, x_{2} + y_{2}}^{2} & = max {x_{1}, x_{2}}^{2} + max {y_{1}, y_{2}}^{2} + 2 ⟨ x, y ⟩ \\ \Rightarrow ⟨ x, y ⟩ & = \frac{max {x_{1} + y_{1}, x_{2} + y_{2}}^{2} - max {x_{1}, x_{2}}^{2} - max {y_{1}, y_{2}}^{2}}{2} \end{aligned}

Must be such an inner product. We can try to verify each property as follows.

Positivity:

⟨ x, x ⟩ = \frac{max {x_{1} + x_{1}, x_{2} + x_{2}}^{2} - max {x_{1}, x_{2}}^{2} - max {x_{1}, x_{2}}^{2}}{2} = \frac{(2 x^{'})^{2}}{2} - x^{' 2} = \frac{x^{' 2}}{2} \geq 0

Where here $x^{'}$ is the maximum of $x_{1}, x_{2}$ . Thus, this inner product would have positivity.

Definiteness: Suppose $⟨ x, x ⟩ = 0$ . Then that means that $\frac{x^{' 2}}{2} = 0$ so then $x^{'} = 0$ . As such, then $max {x_{1}, x_{2}} = 0$ . It's possible though that $x_{1} = 0$ and $x_{2} = - 1$ to allow for this. Notice that plugging that in gives that $⟨ x, x ⟩ = 0$ while $x \neq 0$ so then there's no definiteness.

Hence, if there was some $⟨ \cdot, \cdot ⟩$ where the norm is defined above, then it would fail to have at least one of the properties of the inner product, proving the theorem.
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18

Theorem

Suppose $p > 0$ . Then there is an inner product on $R^{2}$ such that the associated norm is given by:

‖ (x, y) ‖ = (x^{p} + y^{p})^{1 / p}

for all $(x, y) \in R^{2}$ iff $p = 2$ .

Proof
We have to prove both direction. The reverse direction is easier. Consider $p = 2$ first. Then choose the dot product giving that, for all $(x, y) \in R^{2}$ that:

\begin{aligned} ‖ (x, y) ‖^{2} & = ⟨ (x, y), (x, y) ⟩ \\ = x \cdot x + y \cdot y \\ = x^{2} + y^{2} \end{aligned}

Thus then we get that $‖ (x, y) ‖ = \sqrt{x^{2} + y^{2}}$ as we wanted.

Now consider the other direction, so there is an inner product on $R^{2}$ where the associated norm is given by:

‖ (x, y) ‖ = (x^{p} + y^{p})^{1 / p}

We'll have to show that $p = 2$ . Notice that we can plug in values for $(x, y)$ . Have $u = (1, 0)$ and $v = (0, 1)$ . Notice by the Parallelogram Equality:

‖ u + v ‖^{2} + ‖ u - v ‖^{2} = ‖ (1, 1) ‖^{2} + ‖ (0, 0) ‖^{2} = (1^{p} + 1^{p})^{1 / p} = 2^{1 / p}

which would be the same as:

2 (‖ u ‖^{2} + ‖ v ‖^{2}) = 2 (‖ (1, 0) ‖^{2} + ‖ (0, 1) ‖^{2}) = 2 (1 + 1) = 4

Hence:

4 = 2^{1 / p} \Rightarrow 4^{p} = 2 \Rightarrow p \ln (4) = \ln (2) \Rightarrow p = 2

as required.
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19

Theorem

Suppose $V$ is a real inner product space. Then:

⟨ u, v ⟩ = \frac{‖ u + v ‖^{2} - ‖ u - v ‖^{2}}{4}

Proof

\begin{aligned} ‖ u + v ‖^{2} - ‖ u - v ‖^{2} & = 2 (‖ u ‖^{2} + ‖ v ‖^{2}) - 2 ‖ u - v ‖^{2} \\ = 2 (⟨ u, u ⟩ + ⟨ v, v ⟩ - ⟨ u - v, u - v ⟩) \\ = 2 (⟨ u, u ⟩ + ⟨ v, v ⟩ + ⟨ v - u, u - v ⟩) \\ = 2 (⟨ u, u ⟩ + ⟨ v, v ⟩ + ⟨ v, u ⟩ + ⟨ u, v ⟩ - ⟨ u, u ⟩ - ⟨ v, v ⟩) \\ = 2 (⟨ 2 ⟨ u, v ⟩) \\ = 4 ⟨ u, v ⟩ \end{aligned}

Dividing by 4 gives the desired result.
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20

Theorem

Suppose $V$ is a complex inner product space. Then:

⟨ u, v ⟩ = \frac{‖ u + v ‖^{2} - ‖ u - v ‖^{2} + ‖ u + i v ‖^{2} i - ‖ u - i v ‖^{2} i}{4}

Proof

\begin{aligned} ‖ u + v ‖^{2} - ‖ u - v ‖^{2} + ‖ u + i v ‖^{2} i - ‖ u - i v ‖^{2} i & = ⟨ u + v, u + v ⟩ - ⟨ u - v, u - v ⟩ + i ⟨ u + i v, u + i v ⟩ - i ⟨ u - i v, u - i v ⟩ \\ = 2 ⟨ u, v ⟩ + 2 ⟨ v, u ⟩ + (2 ⟨ u, v i ⟩ + 2 ⟨ i v, u ⟩) i \\ = 2 ⟨ u, v ⟩ + 2 ⟨ v, u ⟩ + 2 i ⟨ u, v ⟩ - 2 ⟨ v, u ⟩ \\ = 4 ⟨ u, v ⟩ \end{aligned}

Dividing by 4 on both sides gives the result.
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24

Theorem

Suppose $S \in L (V)$ is an injective operator on $V$ . Define $⟨ \cdot, \cdot ⟩_{1}$ by:

⟨ u, v ⟩_{1} = ⟨ S u, S v ⟩

for $u, v \in V$ . Then $⟨ \cdot, \cdot ⟩_{1}$ is an inner product on $V$ .

Proof
We'll show all the properties of it being an inner product.

(Positivity): Let $v \in V$ . Then:

⟨ v, v ⟩_{1} = ⟨ S v, S v ⟩ \geq 0

by the definition of $⟨ \cdot, \cdot ⟩$ positivity.

(Definiteness): Let $v \in V$ . Then clearly if $v = \vec{0}$ as we showed then $S v = \vec{0}$ so we get $⟨ S v, S v ⟩ = ⟨ \vec{0}, \vec{0} ⟩ = 0$ . If $⟨ S v, S v ⟩ = 0$ then by definiteness from $⟨ ⟩$ then $S v = 0$ . And since $S v = S \vec{0} = \vec{0}$ then $v = \vec{0}$ by $S$ being injective. Hence, the new inner product is definite.

(Additivity in the first slot): Let $u, v, w \in V$ be arbitrary. Then:

⟨ u + v, w ⟩_{1} = ⟨ S (u + v), S (w) ⟩ = ⟨ S u + S v, S w ⟩ = ⟨ S u, S w ⟩ + ⟨ S v, S w ⟩ = ⟨ u, w ⟩_{1} + ⟨ v, w ⟩_{1}

(Homogeneity in the first slot): Let $v, u \in V$ and $λ \in F$ . Then:

⟨ λ u, v ⟩_{1} = ⟨ S (λ u), S v ⟩ = ⟨ λ S u, S v ⟩ = λ ⟨ S u, S v ⟩ = λ ⟨ u, v ⟩_{1}

(Conjugate Symmetry): Let $u, v \in V$ :

⟨ u, v ⟩_{1} = ⟨ S u, S v ⟩ = \overset{―}{⟨ S v, S u ⟩} = \overset{―}{⟨ v, u ⟩_{1}}

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25

Theorem

Suppose $S \in L (V)$ is not injective. Define $⟨ \cdot, \cdot ⟩_{1}$ as in the prior exercise. $⟨ \cdot, \cdot ⟩_{1}$ is not an inner product.

Proof
Since $S$ is not injective, then there are two $u, v \in V$ such that $u \neq v$ but $S u = S v$ . As such, then notice that if $⟨ v, v ⟩_{1} = ⟨ S v, S v ⟩ = 0$ (for the forward part of definiteness), then by definiteness from $⟨ ⟩$ then $S v = 0$ . Thus, then $S u = 0$ , so $⟨ S u, S u ⟩ = 0$ . But $u \neq v$ so then:

⟨ v, v ⟩_{1} = 0 = ⟨ u, u ⟩_{1}

while $u = v = \vec{0}$ but $u \neq v$ , which is a contradiction. As such, then our inner product isn't an inner product.
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6.B: Orthonormal Bases

1

a

Theorem

Suppose $θ \in R$ . Then $(\cos θ, \sin θ), (- \sin θ, \cos θ)$ and $(\cos θ, \sin θ), (\sin θ, - \cos θ)$ are orthonormal bases of $R^{2}$ .

Proof
We'll show that these are orthonormal, by showing their norms are 1 and that they are orthogonal vectors.

First, for norm:

‖ (\cos θ, \sin θ) ‖ = \sqrt{\cos^{2} θ + \sin^{2} θ} = \sqrt{1} = 1

| (- \sin θ, \cos θ) ‖ = \sqrt{(- \sin θ)^{2} + \cos^{2} θ} = 1

‖ (\sin θ, - \cos θ) ‖ = \sqrt{\sin^{2} θ + (- \cos θ)^{2}} = 1

Thus all vectors in question have norm 1.

For orthogonality, notice that:

⟨ (\cos θ, \sin θ), (- \sin θ, \cos θ) ⟩ = - \cos θ \sin θ + \sin θ \cos θ = 0

and:

⟨ (\cos θ, \sin θ), (\sin θ, - \cos θ) ⟩ = \cos θ \sin θ - \sin θ \cos θ = 0

thus each pair of vectors in question are orthogonal to each other.
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b

Theorem

Each orthonormal basis of $R^{2}$ is of the form given by the possibilities from (a)

Proof
Let $u, v \in R^{2}$ be arbitrary, where we want $u, v$ to form an arbitrary orthonormal basis of $R^{2}$ . Hence, we need $‖ u ‖ = ‖ v ‖ = 1$ and $⟨ u, v ⟩ = 0$ . Since their norms are all 1, that means $u, v$ must lie somewhere on the unit circle:

There's some $θ$ between the x-axis and $u$ , so for simplicity say that $u_{x} = \cos (θ)$ and $u_{y} = \sin (θ)$ . As a result:

u = (\cos (θ), \sin (θ))

Which is the first vector of both candidate bases! Notice that $v$ would have to follow similar logic, via some other angle $α$ :

v = (\cos α, \sin α)

Now we'll show $v$ has to be one of the other two forms. Notice that:

⟨ u, v ⟩ = 0 \Rightarrow \cos α \cos θ + \sin α \sin θ = 0

Using the $\cos (α - β)$ trigonometry identity:

\cos (α - θ) = 0 \Rightarrow α - θ = \frac{π}{2} + n π

There's only two really different possibilities, when $n = 0$ , and $n = 1$ . If $n = 0$ :

α - θ = \frac{π}{2} \Rightarrow α = \frac{π}{2} + θ \Rightarrow \sin α = \cos θ, \cos α = - \sin θ

Showing that $v = (- \sin θ, \cos θ)$ as the first candidate shows.

Using $n = 1$ instead:

α - θ = \frac{π}{2} + π \Rightarrow α = θ - \frac{π}{2} \Rightarrow \sin α = - \cos α, \cos θ = \sin θ

Thus $v = (\sin θ, - \cos θ)$ as the second case shows.

Notice that $u, v$ and everything were arbitrary, so then these must be the only two candidates!
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3

Theorem

Suppose $T \in L (R^{3})$ has an UT matrix with respect to the matrix $(1, 0, 0), (1, 1, 1), (1, 1, 2)$ . There's an northonormal basis of $R^{3}$ with respect to which $M (T, β^{'})$ is upper triangular.

Proof
We apply Gram-Schmidt to the given basis to product our orthonormal basis. We are guaranteed that $M (T, β^{'})$ is upper-triangular via Chapter 6 - Inner Product Spaces#^b22ab5.

As such, start with $v_{1} = (1, 0, 0)$ . Here $e_{1} = (1, 0, 0)$ as it's the scalar multiple of $v_{1}$ where it's norm is 1.

Now consider $e_{2}$ . Here it's numerator is:

\begin{aligned} (1, 1, 1) & - ⟨ (1, 1, 1), e_{1} ⟩ e_{1} \\ = (1, 1, 1) - 1 (1, 0, 0) \\ = (0, 1, 1) \end{aligned}

we'll normalize to:

e_{2} = (0, \sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}})

Notice that the numerator for $e_{3}$ is:

\begin{aligned} (1, 1, 2) & - ⟨ (1, 1, 2), e_{1} ⟩ e_{1} - ⟨ (1, 1, 2), e_{2} ⟩ e_{2} \\ = (1, 1, 2) - 1 (1, 0, 0) - 3 \sqrt{\frac{1}{2}} \sqrt{\frac{1}{2}} (0, 1, 1) \\ = (0, \frac{- 1}{2}, \frac{1}{2}) \end{aligned}

Then just normalize our vectors:

\Rightarrow (1, 0, 0), (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (0, \frac{- \sqrt{2}}{2}, \frac{\sqrt{2}}{2})

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4

Theorem

Suppose $n$ is a positive integer. Then:

\frac{1}{\sqrt{2 π}}, \frac{\cos (x)}{\sqrt{π}}, \dots, \frac{\cos (n x)}{\sqrt{π}}, \frac{\sin (x)}{\sqrt{π}}, . . ., \frac{\sin (n x)}{\sqrt{π}}

is an orthonormal list of vectors in $C [- π, π]$ , the space of continuous real-valued functions on $[- π, π]$ with inner product:

⟨ f, g ⟩ = \int_{- π}^{π} f (x) g (x) d x

Proof
Notice that we'll generalize all the $n \in N$ . Notice that:

⟨ \frac{1}{\sqrt{2 π}}, \frac{\cos (n x)}{\sqrt{π}} ⟩ = \int_{- π}^{π} \frac{\cos (n x)}{\sqrt{2} π} d x = \frac{1}{\sqrt{2} π n} (\sin (n x)) |_{- π}^{π} = 0

And a similar proof shows the same for the sine part:

⟨ \frac{1}{\sqrt{2 π}}, \frac{\sin (n x)}{\sqrt{π}} ⟩ = \int_{- π}^{π} \frac{\sin (n x)}{\sqrt{2} π} d x = \frac{1}{\sqrt{2} π n} (- \cos (n x)) |_{- π}^{π} = 0

And lastly we compare both the cosine and sine parts:

⟨ \frac{\cos (m x)}{\sqrt{π}}, \frac{\sin (n x)}{\sqrt{π}} ⟩ = \int_{- π}^{π} \frac{\cos (m x) \sin (n x)}{π} = 0

since the function in the integral is an even function (no matter if $m = n$ or not, as when there's equality we have $\cos (m x) \sin (n x) = 2 \sin (n x)$ which is also an odd function).
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5

Theorem

On $P_{2} (R)$ , consider the inner product given by:

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

Apply the Gram-Schmidt Procedure to the basis $1, x, x^{2}$ to produce an orthonormal basis of $P_{2} (R)$

Proof
Start with $v_{1} = 1$ . Notice that:

‖ 1 ‖ = \int_{0}^{1} 1^{2} d x = 1

So $e_{1} = 1$ . Next:

\begin{aligned} e_{2}^{'} = v_{2} - ⟨ v_{2}, e_{1} ⟩ e_{1} \\ = x - \int_{0}^{1} x d x \cdot 1 \\ = x - \frac{1}{2} \end{aligned}

Where to normalize $e_{2}^{'}$ :

‖ e_{2}^{'} ‖^{2} = \int_{0}^{1} (x - 1 / 2)^{2} d x = \frac{(x - 1 / 2)^{3}}{3} |_{0}^{1} = \frac{2}{2^{3} \cdot 3} = \frac{1}{12}

So $e_{2} = 2 \sqrt{3} (x - \frac{1}{2})$ . Now for $e_{3}$ :

\begin{aligned} e_{3}^{'} & = v_{3} - ⟨ v_{3}, e_{1} ⟩ e_{1} - ⟨ v_{3}, e_{2} ⟩ e_{2} \\ = x^{2} - \int_{0}^{1} x^{2} d x - (2 \sqrt{3}) (x - \frac{1}{2}) \int_{0}^{1} (2 \sqrt{3}) (x - \frac{1}{2}) x^{2} d x \\ = x^{2} - \frac{1}{3} - 12 (x - 1 / 2) \int_{0}^{1} x^{3} - 0.5 x^{2} d x \\ = x^{2} - \frac{1}{3} - 12 (x - \frac{1}{2}) (\frac{1}{4} - \frac{1}{6}) \\ = x^{2} - \frac{1}{3} - (x - \frac{1}{2}) \\ = x^{2} - x + \frac{1}{6} \end{aligned}

Where this norm is:

‖ e_{3}^{'} ‖ = \sqrt{\int_{0}^{1} (x^{2} - x + 1 / 6)^{2} d x} = \frac{1}{\sqrt{180}} = \frac{1}{6 \sqrt{5}}

Thus we have:

e_{3} = 6 \sqrt{5} (x^{2} - x - \frac{1}{6})

☐

6

Theorem

Find an orthonormal basis of our previous questions vector space, such that the differentiation operator on $V$ has an upper-triangular matrix with respect to this basis.

Proof
Notice that the standard basis of $1, x, x^{2}$ has the differentiation operator of:

D (1) = 0, D (x) = 1, D (x^{2}) = 2 x \Rightarrow M (D) = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{matrix}]

which is already upper-triangular. One can show that the current basis isn't orthonormal (just orthogonal), so we'll apply Gram-Schmidt to the problem, which we did in HW 7 - Inner Product Spaces#5, giving us the basis:

{e_{1}, e_{2}, e_{3}} = {1, 2 \sqrt{3} (x - \frac{1}{2}), 6 \sqrt{5} (x^{2} - x - \frac{1}{6})}

which by Chapter 6 - Inner Product Spaces#^b22ab5, then our basis has $M (D, β^{'})$ is UT.
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9

Theorem

What happens if the Gram-Schmidt Procedure is applied to a list of vectors that is not linearly independent?

Proof
Since we'll get $n$ vectors, given some starting number of $n$ vectors $v_{1}, . . ., v_{n}$ , we'll get $e_{1}, . . ., e_{n}$ . If we have that some of the $v_{i}$ 's are linear combinations of the others, then when we linearly combine previous vectors while applying the procedure, we may create vectors that are just linearly dependent on each other. Hence, $e_{1}, . . ., e_{n}$ will be linearly dependent. As a result, then some $e_{1}, . . ., e_{n}$ may not be orthogonal to each other (since there'll be linear combinations to get from one set of $e_{i}$ to a new $e_{j}$ ).
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12

Theorem

Suppose $V$ is finite-dimensional and $⟨ \cdot, \cdot ⟩_{1}, ⟨ \cdot, \cdot ⟩_{2}$ are inner products on $V$ with corresponding norms $‖ \cdot ‖_{1}$ and $‖ \cdot ‖_{2}$ . Then there is a positive number $c$ such that:

‖ v ‖_{1} \leq c ‖ v ‖_{2}

for all $v \in V$ .

Proof
Let $v \in V$ be arbitrary. Consider first:

⟨ v, v ⟩_{1} = ‖ v ‖_{1}

and similar for the second norm.

Consider the following cases. If $‖ v ‖_{1} = ‖ v ‖_{2}$ then $c = 1$ and we are done. Hence, consider if $‖ v ‖_{1} \neq ‖ v ‖_{2}$ . Since both are real, positive numbers, then by the Ordering of the Reals, then we have either $‖ v ‖_{1} < ‖ v ‖_{2}$ or $‖ v ‖_{1} > ‖ v ‖_{2}$ .

First, consider the former case. Hence:

‖ v ‖_{1} < ‖ v ‖_{2}

If $‖ v ‖_{2} = 0$ then because $⟨ v, v ⟩_{2} = 0$ then we have it that $v = 0$ as a property of the inner product. As such, then $⟨ v, v ⟩ = 0$ so then $‖ v ‖_{1} = 0$ , which contradicts them being non-equal

If instead $‖ v ‖_{2} \neq 0$ then notice that:

\frac{‖ v ‖_{1}}{‖ v ‖_{2}} < 1

But since both numerator and denominator are positive, then it's greater than 0. Hence, choose $c = \frac{‖ v ‖_{1}}{‖ v ‖_{2}} \in (0, 1)$ ; so:

‖ v ‖_{1} \leq c ‖ v ‖_{2} = \frac{‖ v ‖_{1}}{‖ v ‖_{2}} ‖ v ‖_{2} = ‖ v ‖_{1}

which is true.

Hence, no matter what for our former case, we can always choose $c$ . If instead we have:

‖ v ‖_{1} > ‖ v ‖_{2}

Again, we know that $‖ v ‖_{2} \neq 0$ , so then:

\frac{‖ v ‖_{1}}{‖ v ‖_{2}} > 1

So choose $c$ as the left-hand side of this inequality, since:

‖ v ‖_{1} \leq c ‖ v ‖_{2} = ‖ v ‖_{1}

Hence, by proof by exhaustion, the theorem always holds.
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6.A: Inner Products and Norms

1

2

3

4

a

b

c

5

7

8

10

12

13

17

18

19

20

24

25

6.B: Orthonormal Bases

1

a

b

3

4

5

6

9

12

6.C: