HW 1 - Products and Quotients, Duality

3.E

1

Theorem

Suppose $T$ is a function from $V$ to $W$ . The graph of $T$ is the subset of $V \times W$ given by:

graph of T = {(v, T v) \in V \times W : v \in V}

Then $T$ is a linear map iff the graph of $T$ is a subspace of $V \times W$ .

Proof
Consider $(\to)$ . Suppose $T$ is indeed a linear map. We'll show that the graph of $T$ is a subspace of $V \times W$ . First note that $\vec{0} = (0, 0) \in graph of T$ since $T 0 = 0$ , so then clearly it contains the zero vector. For additivity, let $v, w \in V$ , so then $(v, T v), (w, T w) \in G (T)$ , where $G (T)$ is a shorthand for the graph of $T$ . Notice that:

\begin{aligned} (v + w, T (v + w)) & = (v + w, T v + T w) & T is linear \\ \in G (T) \end{aligned}

since $T v \in W$ and $T w \in W$ and $W$ is closed under vector addition, and $v + w \in V$ since $V$ is also closed under vector addition. Hence $G (T)$ is closed under vector addition.

For scalar multiplication, let $λ \in F$ be arbitrary. Then, using $v$ from prior, notice that:

\begin{aligned} (λ v, T (λ v)) & = (λ v, λ T v) & T is linear \\ \in G (T) \end{aligned}

since $λ v \in V$ by closure under scalar multiplication, and $λ T v \in W$ for the same reason.

Therefore, $G (T)$ is a subspace of $V \times W$ .

Now consider $(\leftarrow)$ . As such, say that $G (T)$ is a subspace of $V \times W$ . We'll show that $T$ is a linear map. First to show additivity, let $v, w \in V$ be arbitrary. Then:

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

since $G (T)$ is a subspace of $V \times W$ and thus is closed under vector addition. Adding terms together:

(v + w, T (v + w)) = (v, T v) + (w, T w) = (v + w, T v + T w)

thus $T$ has additivity by equating the two tuples above.

Homogeneity is shown similarly for some $λ \in F$ :

(λ v, T (λ v)) = λ (v, T v) = (λ v, λ T v)

So $T (λ v) = λ T v$ showing homogeneity for $T$ .

As such, then $T$ is a linear map.
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5

Theorem

Suppose $W_{1}, . . ., W_{m}$ are vector spaces. Then $L (V, W_{1} \times \dots \times W_{m})$ and $L (V, W_{1}) \times \dots \times L (V, W_{m})$ are isomorphic vector spaces.

Proof

Finite Dimensional Case

Consider each vector space has it's own dimension:

dim (V) = n, \forall i (dim (W_{i}) = n_{i})

We know that two vector spaces are isomorphic if their dimensions equal. We also know for some $T \in L (V, W)$ that it's isomorphic to $F^{m, n}$ , so then the dimension of it is $m n$ . In a similar vein, there we know the dimension of the first space is:

dim (L (V, W_{1} \times \dots \times W_{m})) = dim (V) dim (W_{1} \times \dots \times W_{m}) = n (\sum_{i = 1}^{m} n_{i})

and for the other space:

dim (L (V, W_{1}) \times \dots \times L (V, W_{m})) = \sum_{i = 1}^{m} n \cdot n_{i} = n (\sum_{i = 1}^{m} n_{i})

hence, the dimensions are the same, so then the two spaces are isomorphic.

Infinite Dimensional Case

Now it's possible that $V$ isn't finite-dimensional, hence this argument wouldn't work. Notice that the former set is the set of transformations from $V$ to an $m$ tuple of $w$ 's, while the latter set is a tuple of transformations. We can make a bijection (isomorphism) between the two spaces.

Consider the isomorphism $T$ that maps from the latter to the former set, where for $T \in L (V, W_{1} \times \dots \times W_{m})$ we define it by:

T (T_{1} (v), . . ., T_{m} (v)) = (T_{1}, . . ., T_{m}) (v)

where all $T_{i} \in L (V, W_{i})$ . Let's notice that $T$ is linear by being the identity map, which itself is linear.

We'll show that this is an isomorphism by showing that $T$ is both injective and surjective.

For injectivity, suppose some $(T_{1}, . . ., T_{m}) \in null (T)$ is arbitrary. If we show that this must be the tuple of all zero transformations, then $T$ must be injective. Notice that because it's in the nullspace, then:

T (T_{1}, . . ., T_{m}) = ({\vec{0}}_{W}, . . ., {\vec{0}}_{W})

where each ${\vec{0}}_{W}$ is the zero vector of $W$ . As such, we can equate transformations when we plug in a vector $v \in V$ :

T (T_{1} v, . . ., T_{m} v) = ({\vec{0}}_{W}, . . ., {\vec{0}}_{W})

As such for all $i$ we have it that $T_{i} v = {\vec{0}}_{W}$ so then clearly each $T_{i}$ is the zero map. Hence, we've achieved the goals for injectivity.

Now for surjectivity. Let $τ \in L (V, W_{1} \times \dots \times W_{m})$ be arbitrary. We'll show that there always exists some item $T \in L (V, W_{1}) \times \dots \times L (V, W_{m})$ that can be constructed such that $T (T) = τ$ . Construct $T$ such that if:

τ (v) = (T_{1} v, . . ., T_{m} v)

where all $T_{i} \in L (V, W_{i})$ , then have $T$ be:

T = (T_{1}, . . ., T_{m})

because then:

T (T) (v) = T (T_{1} v, . . ., T_{m} v) = (T_{1} v, . . ., T_{m} v) = τ (v)

Hence $T (T) = τ$ as desired. Thus, $T$ is surjective.

Since $T$ is injective and surjective, then it's a bijection and thus a valid isomorphism from the two spaces, showing they're isomorphic.
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7

Theorem

Suppose $v, x \in V$ and $U, W$ are subspaces of $V$ such that $v + U = x + W$ . Then $U = W$ .

Proof
We'll do a subset argument here. First consider $(\subseteq)$ . Let $u \in U$ be arbitrary. We'll show that $u \in W$ . First, since $u \in U$ , then clearly:

v + u \in v + U \Rightarrow v + u \in x + W

Thus there's some vector $w \in W$ such that $v + u = x + w$ . But look!

u = \underset{\in W}{\underset{⏟}{(x - v)}} + w \in W

Where $x - v \in W$ because clearly $v = x + w^{'}$ since $v + {\vec{0}}_{U} \in v + U \Rightarrow x + W$ so then $v - x = w^{'} \in W$ for some $w^{'} \in W$ . As such, $u \in W$ , so then $U \subseteq W$ as $u$ was arbitrary.

The other direction for $(\supseteq)$ is similar. Let $w \in W$ be arbitrary. Clearly:

x + w \in x + W \Rightarrow x + w \in v + U

So there's a vector $u \in U$ such that $x + w = v + u$ so then:

w = \underset{\in U}{\underset{⏟}{(v - x)}} + u \in U

Where $v - x \in U$ via similar reasons prior. As such, then $U \supseteq W$ .

Hence $U = W$ .
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13

Theorem

Suppose $U$ is a subspace of $V$ and $v_{1} + U, . . ., v_{m} + U$ is a basis of $V / U$ and $u_{1}, . . ., u_{n}$ is a basis of $U$ . Then $v_{1}, . . ., v_{m}, u_{1}, . . ., u_{n}$ is a basis of $V$ .

Proof
Let's determine if it's a basis by equating dimension, and determining LI.

First, the dimension of $V / U$ is $m$ given by the first basis, and the dimension of $U$ is $n$ via the second basis. As such, then:

dim (V / U) = dim (V) - dim (U) \Rightarrow dim (V) = m + n

so the proposed basis has the right number of vectors from $V$ .

Now for LI. Consider when:

α_{1} v_{1} + \dots + α_{m} v_{m} + α_{m + 1} u_{1} + \dots + α_{m + n} u_{n} = \vec{0}

Notice that our first basis for $V / U$ implies that for any $a_{i} \in F$ that if:

a_{1} (v_{1} + U) + \dots + a_{m} (v_{m} + U) = ((a_{1} v_{1}) + U) + \dots + ((a_{m} v_{m}) + U) = \vec{0} = U

Then all $a_{i} = 0$ , which becomes:

(a_{1} v_{1} + \dots + a_{m} v_{m}) + U = U \Rightarrow a_{1} v_{1} + \dots + a_{m} v_{m} = \vec{0}

So namely we've found that $v_{1}, . . ., v_{m}$ must be LI. We already know that the same thing occurs for $u_{1}, . . ., u_{n}$ showing those vectors are LI.

Notice that if $α_{m + 1} u_{1} + \dots + α_{m + n} u_{n} = \vec{0}$ then we are done since then:

α_{1} v_{1} + \dots + α_{m} v_{m} = \vec{0}

So all $α_{i} = 0$ as required. Now instead, say that $α_{m + 1} u_{1} + \dots + α_{m + n} u_{n} \neq \vec{0}$ , call it $u \in U$ . We'll show that this is impossible via contradiction. So then:

α_{1} v_{1} + \dots + α_{m} v_{m} + u = \vec{0}

Clearly if $α_{1} v_{1} + \dots + α_{m} v_{m} = \vec{0}$ then we'd have $u = \vec{0}$ which is a contradiction, so suppose the sum equals $v$ . Then:

v + u = \vec{0} \Rightarrow v = - u

But this is a contradiction! This is because then $v \in U$ since $v = - u$ . That implies that $v + U = U$ , which implies that $\vec{0} = v$ which is a contradiction. As such, we must have one of $v, u = \vec{0}$ , so then we get that all $a_{i} = 0$ and thus the list of vectors are LI.

As such, the list of vectors is LI and of size $m + n$ , so then it's a valid basis for $V$ .
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18

Theorem

Suppose $T \in L (V, W)$ and $U$ is a subspace of $V$ . Let $π$ denote the quotient map from $V$ onto $V / U$ . Then there exists $S \in L (V / U, W)$ such that $T = S \circ π$ iff $U \subseteq null (T)$ .

Proof
Consider $(\to)$ first, so suppose there is some $S \in L (V / U, W)$ where $T = S \circ π$ . We'll show that $U \subseteq null (T)$ , so have $u \in U$ be arbitrary. We need to show that $T u = \vec{0}$ to then say $u \in null (T)$ . Just use our definition of $T$ :

T u = (S \circ π) u = S (π u) = S (u + U) = S (\vec{0} + U) = \vec{0}

Because here $u \in U$ so then $u + U = \vec{0} + U$ , which is the zero vector of the input vector space, hence why the whole thing becomes the zero vector. As such, then $T u = \vec{0}$ so then $u \in null (T)$ . Thus, $U \subseteq null (T)$ .

Now for ( $\leftarrow$ ). Suppose that $U \subseteq null (T)$ . We'll need to construct a map $S : V / U \to W$ such that $T = S \circ π$ . Notice that for such a map we'd need to take $v \in V$ as follows:

T v = (S \circ π) v = S (π v) = S (v + U)

Thus choose $S : V / U \to W$ to be $S (v + U) = T v$ ! Notice that clearly from our construction that $T = S \circ π$ , but is $S$ linear? We show additivity first. If $v, w \in V$ then:

\begin{array}{r} S ((v + w) + U) = T (v + w) = T v + T w = S (v + U) + S (w + U) \end{array}

and for homogeneity, we know that for some $λ \in F$ :

S (λ v + U) = T (λ v) = λ T v = λ S (v + U)

So $S$ is linear.

We need to show though that $S$ is well defined too. Namely if $v = w$ then $v + U = w + U$ so then $v - w \in U$ , which using our supposition implies that $v - w \in null (T)$ . As such, then:

T (v - w) = \vec{0} \Rightarrow T v = T w \Rightarrow S (v + U) = S (w + U)

which is what we want.
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20

Theorem

Suppose $U$ is a subspace of $V$ . Define $Γ : L (V / U, W) \to L (V, W)$ by:

Γ (S) = S \circ π

Then:

$Γ$ is a linear map.
$Γ$ is injective.
$range (Γ) = {T \in L (V, W) : T u = 0 \forall u \in U}$

Proof

We'll later use the lemma we proved in (18):

Theorem

We know that $Γ$ is well defined as it's just standard notation we've dealt with for a bit now.

Let's first show that $Γ$ is linear. Let $T_{1}, T_{2} \in L (V / U, W)$ be arbitrary, and $α \in F$ as well. For additivity:

Γ (T_{1} + T_{2}) = (T_{1} + T_{2}) \circ π

Let $v \in V$ be arbitrary. Then:

\begin{aligned} (T_{1} + T_{2}) \circ π (v) & = (T_{1} + T_{2}) (v + U) \\ = T_{1} (v + U) + T_{2} (v + U) & (T_{1}, T_{2} \in L (V / U, W)) \\ = (T_{1} \circ π) v + (T_{2} \circ π) v \\ = (T_{1} \circ π + T_{2} \circ π) v \\ \Rightarrow Γ (T_{1} + T_{2}) & = Γ (T_{1}) + Γ (T_{2}) \end{aligned}

Hence, $Γ$ is additive. A similar story holds for homogeneity:

\begin{aligned} Γ (α T_{1}) & = (α T_{1}) \circ π \\ \Rightarrow Γ (α T_{1}) v & = α T_{1} \circ π (v) \\ = α T_{1} (v + U) \\ = α (T_{1} \circ π) v \\ \Rightarrow Γ (α T_{1}) & = α (T_{1} \circ π) \\ = α Γ (T_{1}) \end{aligned}

Thus $Γ$ is homogeneous. As a result, then $Γ$ is linear.

Showing that the range of $Γ$ is as the theorem describes is really easy. Notice that the codomain is within the superset of $L (V, W)$ as the first condition requires. Let $u \in U$ now be arbitrary. Via HW 1 - Products and Quotients, Duality#^fe60ed, then since $T$ exists how the lemma describes, then it follows that $U \subseteq null (T)$ , so then $u \in null (T)$ . As a result, then $T u = 0$ as the theorem describes, hence creating the set for the range as required. It's not possible that there are other additional conditions, as the lemma shows that the only condition is the equivalence that we get.

Now to show that $Γ$ is injective. Let $T_{1}, T_{2} \in L (V / U, W)$ be arbitrary, and suppose that $Γ (T_{1}) = Γ (T_{2})$ . We'll show that $T_{1} = T_{2}$ as a result. Notice that:

\Rightarrow T_{1} \circ π = T_{2} \circ π

Now let $v \in V$ be arbitrary. Then:

\begin{aligned} (T_{1} \circ π) v & = T_{1} (v + U) \end{aligned}

And similar for $T_{2}$ . As such:

T_{1} (v + U) = T_{2} (v + U)

Thus $T_{1} = T_{2}$ .
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3.F

1

Theorem

Every linear functional $φ \in L (V, F)$ is either surjective or the zero map.

Proof
If $φ$ is the zero map then we are done, so suppose that $φ$ is not the zero map. As such, then there exists some vector $v \in V$ such that $φ (v) \neq 0$ . Call it $φ (v) = α$ . We'll show that then $φ$ is surjective.

Let $λ \in F$ be arbitrary. We need to construct some $v^{'} \in V$ where $φ (v^{'}) = λ$ . Choose $\frac{λ}{α} v$ as this vector. Notice that:

φ (\frac{λ}{α} v) = \frac{λ}{α} φ (v) = \frac{λ α}{α} = λ

where this works since we know $α \neq 0$ . Hence, then $φ$ is surjective.
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2

Theorem

Give three distinct examples of linear functionals on $R^{[0, 1]}$ .

Proof
One example is the functional $α : R^{[0, 1]} \to R$ where:

α (f) = \int_{0}^{1} f (x) d x

Another example is the functional $β : R^{[0, 1]} \to R$ defined as:

β (f) = f (0.5)

And again, one last example is the functional $γ : R^{[0, 1]} \to R$ defined as:

γ (f) = \frac{f (0) + f (1)}{2}

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4

Theorem

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ such that $U \neq V$ . Then there exists $φ \in V^{'}$ such that $φ (u) = 0$ for every $u \in U$ but $φ \neq 0$ .

Proof
Define $φ : V \to F$ by:

φ (v) = {\begin{cases} 0 & v \in U \\ 1 & v \notin U \end{cases}

Notice the following. Since $U \neq V$ and since $U$ is a subspace of $V$ then $U \subseteq V$ . As a result, then we must have some vector $v \in V$ while $v \notin U$ (notice if it weren't the case, then that would imply that $V \subseteq U$ which implies $V = U$ which is a contadiction). As such, then $φ (u) \neq 0$ so there's some vector to show that $φ \neq 0$ .

By definition, for all $u \in U$ , we have $φ (u) = 0$ as required.
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5

Theorem

Suppose $V_{1}, . . ., V_{m}$ are vector spaces. Then $(V_{1} \times \dots \times V_{m})^{'}$ and $V_{1}^{'} \times \dots \times V_{m}^{'}$ are isomorphic vector spaces.

Proof
We'll create an isomorphism $γ : V_{1}^{'} \times \dots \times V_{m}^{'} \to (V_{1} \times \dots \times V_{m})^{'}$ , by showing our constructed $γ$ is both injective and surjective. Construct $γ$ by defining it as:

γ (ϕ_{1}, . . ., ϕ_{m}) (v_{1}, . . ., v_{m}) = \sum_{i = 1}^{m} ϕ_{i} (v_{i})

where each $ϕ_{i} \in V_{i}^{'}$ for each $i$ . Really, without the $v_{i}$ 's:

γ (ϕ_{1}, . . ., ϕ_{m}) = \sum_{i = 1}^{m} ϕ_{i}

Showing linearity of $γ$ :

(additivity) Let $φ = (ϕ_{1}, . . ., ϕ_{m}), Ψ = (ψ_{1}, . . ., ψ_{m}) \in (V_{1}^{'} \times \dots \times V_{m}^{'})$ be arbitrary:

\begin{aligned} γ (φ + Ψ) & = γ (ϕ_{1} + ψ_{1}, . . ., ϕ_{m} + ψ_{m}) \\ = \sum_{i = 1}^{m} (ϕ_{i} + ψ_{i}) \\ = \sum_{i = 1}^{m} ϕ_{i} + \sum_{i = 1}^{m} ψ_{i} \\ = γ (φ) + γ (Ψ) \end{aligned}

(homogeneity) Let $λ \in F$ be arbitrary. Using $Ψ$ as above:

\begin{aligned} γ (λ Ψ) & = γ (λ ψ_{1}, . . ., λ ψ_{m}) \\ = \sum_{i = 1}^{m} λ ψ_{i} \\ = λ \sum_{i = 1}^{m} ψ_{i} \\ = λ γ (Ψ) \end{aligned}

Thus $γ$ is linear.

For injectivity, consider some vectors $(ϕ_{11}, . . ., ϕ_{1 m}), (ϕ_{21}, . . ., ϕ_{2 m}) \in V_{1}^{'} \times \dots \times V_{m}^{'}$ have it where their transformation after $γ$ are equal. We'll show that the two vectors are equivalent:

\begin{aligned} γ (ϕ_{11}, . . ., ϕ_{1 m}) & = γ (ϕ_{21}, . . ., ϕ_{2 m}) \\ (\sum_{i = 1}^{m} ϕ_{1 i}) & = (\sum_{i = 1}^{m} ϕ_{2 i}) \\ \sum_{i = 1}^{m} ϕ_{1 i} & = \sum_{i = 1}^{m} ϕ_{2 i} \\ \sum_{i = 1}^{m} (ϕ_{1 i} - ϕ_{2 i}) & = \vec{0} \end{aligned}

Hence each $ϕ_{1 i} - ϕ_{2 i} = \vec{0}$ (the zero vector here is of the tuple) so then $ϕ_{1 i} = ϕ_{2 i}$ . Thus, we've shown injectivity.

For surjectivity, let $ϕ \in (V_{1} \times \dots \times V_{m})^{'}$ be arbitrary. We'll show that we can construct some $ψ \in V_{1}^{'} \times \dots \times V_{m}^{'}$ such that $γ (ψ) = ϕ$ . We know that:

ϕ (v_{1}, . . ., v_{m}) \to F

For clarity, define:

ϕ_{i} (v_{i}) = ϕ \overset{i -th position}{\overset{⏞}{(\vec{0}, \vec{0}, . . ., v_{i}, . . ., \vec{0}, \vec{0})}}

then define $ψ$ as:

ψ = (ϕ_{1}, . . ., ϕ_{m})

Then notice:

\begin{aligned} γ (ψ) (v_{1}, . . ., v_{m}) & = γ (ϕ_{1}, . . ., ϕ_{m}) (v_{1}, . . ., v_{m}) \\ = \sum_{i = 1}^{m} ϕ_{i} v_{i} \\ = ϕ (v_{1}, . . ., v_{m}) \end{aligned}

where the last step comes from the fact that $ϕ$ is linear, and thus you can add the tuples together to get $ϕ (v_{1}, . . ., v_{m})$ :

\sum_{i = 1}^{m} ϕ_{i} = ϕ (v_{1}, . . ., \vec{0}) + \dots + ϕ (\vec{0}, . . ., v_{m}) = ϕ (v_{1}, . . ., v_{m})

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7

Theorem

Suppose $m \in Z^{+}$ . Show that the dual basis of the basis $1, x, . . ., x^{m}$ of $P_{m} (R)$ is $φ_{0}, . . ., φ_{m}$ where:

φ_{j} (p) = \frac{p^{(j)} (0)}{j!}

where here $p^{(j)}$ is the $j$ -th derivative of $p$ , with the understanding that the 0-th derivative of $p$ is $p$ .

Proof
Notice that we know the dimension of $P_{m} (R)$ is $m + 1$ . Clearly there are enough $ϕ$ 's so then all we have to do is show that $φ_{0}, . . . ., φ_{m}$ is LI.

Consider the functional:

a_{0} φ_{0} + \dots + a_{m} φ_{m} = \vec{0}

Now consider some input vector $p_{i}$ . Then:

(a_{0} φ_{0} + \dots + a_{m} φ_{m}) (p_{i}) = 0

Notice that for each $j$ that:

a_{j} φ_{j} (p_{i}) = a_{j} \frac{p_{i}^{(j)} (0)}{j!}

So then plugging in:

\sum_{k = 0}^{m} a_{k} φ_{k} (p_{i}) = \sum_{k = 0}^{m} a_{k} \frac{p_{i}^{(k)} (0)}{k!} = 0

where notice since $p_{i} (x) = β_{0} + β_{1} x + \dots + β_{m} x^{m}$ then:

p_{i}^{(k)} (0) = k β_{k} \forall k = 1, . . ., m; p_{i}^{(0)} (0) = β_{0}

Because, to clarify:

p_{i}^{(0)} (0) = β_{0}, p_{i}^{(1)} (0) = β_{1}, p_{i}^{(2)} (0) = 2 β_{2}, . . .

As such, then our sum simplifies to:

0 = a_{0} β_{0} + \sum_{k = 1}^{m} a_{k} \frac{k! β_{k}}{k!} = \sum_{k = 0}^{m} a_{k} β_{k}

Notice that if we only consider the basis vector $p_{i} \in {1, . . ., x^{m}}$ . When we get that then we have all $β_{i} = 0$ except for $β_{m} = 1$ . As such, then for all $i$ then for its corresponding $p_{i}$ :

0 = \sum_{k = 0}^{m} a_{k} β_{k} = a_{i} β_{i} = a_{i}

Thus all $a_{i} = 0$ , so then $φ_{0}, . . ., φ_{m}$ is LI, and thus is a basis.
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8

Theorem

Suppose $m \in Z^{+}$ . Then:

$1, x - 5, . . ., (x - 5)^{m}$ is a basis of $P_{m} (R)$ .
The dual basis of the basis above is $φ_{0}, . . ., φ_{m}$ where:

φ_{i} (p (x) (x - 5)^{j}) = χ (i = j)

where $j = 0, . . ., m$ depending on the number of factors of $(x - 5)$ is in the input vector $q (x) = p (x) (x - 5)^{j}$ .

Proof
For (a), notice that since the dimension of $P_{m} (R)$ is $m + 1$ we have the right number of basis vectors. Hence, let's show that this set of vectors is LI. Consider:

a_{0} (1) + a_{1} (x - 5) + \dots + a_{m} (x - 5)^{m} = \vec{0}

Notice that the degree of the expanded $x^{m}$ term on the LHS should be:

a_{m} (x^{m} + \dots) + \dots + a_{0} (1) = \vec{0}

Where notice that no other terms other than the $a_{m}$ term has $x^{m}$ . But equating sides shows that $a_{m} = 0$ since $x^{m} = 0 x^{m}$ .

Repeat this process for all $a_{m - 1}, . . ., a_{1} = 0$ . Once that occurs then at the end $a_{0} = 0$ . Thus, the set of vectors is LI, and thus is a valid basis.

For (b), we can check that for each $φ_{i} (p_{j}) = χ (i = j)$ where $p_{j}$ is each vector from the basis we defined prior in (a). As a result, notice that:

p_{j} = (x - 5)^{j} \Rightarrow φ_{i} (p_{j}) = φ_{i} ((x - 5)^{j}) = χ (i = j)

as expected.
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9

Theorem

Suppose $v_{1}, . . ., v_{n}$ is a basis of $V$ and $φ_{1}, . . ., φ_{n}$ is the corresponding dual basis of $V^{'}$ . Suppose $Ψ \in V^{'}$ . Then:

Ψ = Ψ (v_{1}) φ_{1} + \dots + Ψ (v_{n}) φ_{n}

Proof
Notice that since $Ψ \in V^{'} = L (V, F)$ then consider showing the equality by plugging in some arbitrary $v \in V$ . Notice that this $v = \sum_{i}^{n} a_{i} v_{i}$ by the basis, so:

\begin{aligned} Ψ (v) & = Ψ (a_{1} v_{1} + \dots + a_{n} v_{n}) \\ = \sum_{i = 1}^{n} a_{i} Ψ (v_{i}) \end{aligned}

and from the right side:

\begin{aligned} (Ψ (v_{1}) φ_{1} + \dots + Ψ (v_{n}) φ_{n}) (v) & = \sum_{i = 1}^{n} Ψ (v_{i}) φ_{i} (v) \\ = \sum_{i = 1}^{n} Ψ (v_{i}) φ_{i} (a_{1} v_{1} + \dots + a_{n} v_{n}) \\ = \sum_{i = 1}^{n} \sum_{j = 1}^{n} Ψ (v_{i}) φ_{i} (a_{j} v_{j}) \\ = \sum_{i = 1}^{n} a_{i} Ψ (v_{i}) \underset{χ (i = i) = 1}{\underset{⏟}{φ_{i} (v_{i})}} \\ = \sum_{i = 1}^{n} a_{i} Ψ (v_{i}) \end{aligned}

Thus clearly $Ψ (v) = (Ψ (v_{1}) φ_{1} + \dots + Ψ (v_{n}) φ_{n}) (v)$ . so then the two transformations are the same.
☐