3.E: Products and Quotients of Vector Spaces

Products of Vector Spaces

For instance, $(5-6x+4x^2,(3,8,7))\in {\mathcal{P}}_2(\mathbb{R})\times {\mathbb{R}}^3$.

We don't prove that here (the book doesn't even prove it), but it's easy to show. Note that $\overrightarrow{0}$ of this space is the $\overrightarrow{0}$ for each vector space put in each slot:

The additive inverse is just the negatives of each vector:

An Example

Notice that ${\mathbb{R}}^2\times {\mathbb{R}}^3$ and ${\mathbb{R}}^5$ are not equal. They can't even be compared! However, they are isomorphic via:

as our isomorphism.

Also notice that the list:

is a valid basis for ${\mathcal{P}}_2(\mathbb{R})\times {\mathbb{R}}^2$. This paints the idea for the next lemma:

Proof
Choose a basis of each $V_j$. For each basis vector of each $V_j$, consider the element of $V_1\times \cdots \times V_m$ that equals the basis vector in the $j$-th slot and $0$ in the other slots. The list of all such vectors is linearly independent and spans $V_1\times \cdots \times V_m$, so it's a valid basis, of the corresponding length.
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Products and Direct Sums

Notice that being injective here is the same as invertible. That's because this map is surjective by the definition of $U_1+\cdots +U_m$, as it's range is the sum space itself by definition.

Proof
The linear map $\mathrm{\Gamma }$ is injective iff the only way to write $\overrightarrow{0}$ as a sum $u_1+\cdots +u_m$ where each $u_j\in U_j$ is to have each $u_j=0$. Thus, Chapter 1 - Vector Spaces#^038942 applies iff $\mathrm{\Gamma }$ is injective, so then $\mathrm{\Gamma }$ is injective iff $U_1+\cdots +U_m$ is a direct sum.
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Proof
We know that $\mathrm{\Gamma }$ is surjective. By the FTOLM, we know $\mathrm{\Gamma }$ is injective iff:

Combining Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^e94bef and Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^f0ae57, we get that it's a direct sum iff:

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Quotients of Vector Spaces

For instance, in the above example, all the lines in ${\mathbb{R}}^2$ with slope 2 are parallel to $U$, and each of these parallel lines are each affine subsets of ${\mathbb{R}}^2$.

Another example, if $U=\{(x,y,0)\in {\mathbb{R}}^3:x,y\in \mathbb{R}\}$ then the affine subset of ${\mathbb{R}}^3$ parallel to $U$ are the planes in ${\mathbb{R}}^3$ that are parallel to the $xy$-plane $U$ in the usual sense. Notice here that no line in ${\mathbb{R}}^3$ would be an affine subset while still being parallel to the plane $U$. That's because, to be affine, we require being allowed to span the whole plane $U$, not just a parallel line.

For instance, if $U$ was the line $\{(x,2x):x\in {\mathbb{R}}^2\}$ then ${\mathbb{R}}^2/U$ is set of all lines in ${\mathbb{R}}^2$ that have slope 2.

If instead $U$ is a line in ${\mathbb{R}}^3$ containing the origin, then ${\mathbb{R}}^3/U$ is the set of all lines in ${\mathbb{R}}^3$ parallel to $U$.

If $U$ is a plane in ${\mathbb{R}}^3$ containing the origin, then ${\mathbb{R}}^3/U$ is the set of all planes in ${\mathbb{R}}^3$ parallel to $U$.

Proof
First suppose (1) holds, so $v-w\in U$. If any $u\in U$ then:

So then $v+U\subset w+U$. A similar argument shows that $w+U\subset v+U$, so then (2) is proved.

We know that if (2) holds then (3) must hold, since the sets themselves can't be empty (worst case is $\{\overrightarrow{v},\overrightarrow{w}\}\ne \mathrm{\varnothing }$).

Suppose (3). Thus there are $u_1,u_2\in U$ such that:

Then $v-w=u_2-u_1\in U$, showing (1).
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We need these to show that $V/U$ is a vector space.

Proof
The problem might arise where since $V/U$ is the set of all affine subsets, it might possibly not be unique. For example, suppose $v,w\in V$. Suppose also that $\hat{v},\hat{w}\in V$ are such that $v+U=\hat{v}+U$ and $w+U=\hat{w}+U$. To show our definition of addition on $V/U$ makes sense, we'll show that $(v+w)+U=(\hat{v}+\hat{w})+U$.

Via Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^045b5b, we have:

Because $U$ is a subspace of $V$ and thus closed under addition, this implies that $(v-\hat{v})+(w-\hat{w})\in U$ so then $(v+w)-(\hat{v}+\hat{w})\in U$. Using our theorem again, we get that:

so then the definition we made makes sense.

In a similar vien, suppose $\lambda \in \mathbb{F}$. Since $U$ is closed under scalar multiplication, then $\lambda (v-\hat{v})\in U$ so then $\lambda v-\lambda \hat{v}\in U$, where using the theorem again suggest that $(\lambda v)+U=(\lambda \hat{v})+U$, as desired.

Note that actually showing vector space properties is pretty. We note though that the additive identity of $V/U$ is $0+U$ and additive inverse of $v+U$ is $-v+U$.
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It's easy to show that $\pi$ is a linear map. Most often $V,U$ are given from context.

Proof
Let $\pi$ be the quotient map from $V$ to $V/U$. From Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^045b5b, we see that $\text{null}(\pi )=U$. Further, $\text{range}(\pi )=V/U$, so using the FTOLM:

and as such giving the desired result.
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Each linear map $T$ on $V$ induces a linear map $\tilde{T}$ on $V/(\text{null}(T))$, which we define:

To show that the definition of $\tilde{T}$ makes sense, suppose $u,v\in V$ are such that $u+\text{null}(T)=v+\text{null}(T)$. By Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^045b5b, we have $u-v\in \text{null}(T)$. Thus $T(u-v)=0$ so then $Tu=Tv$.

If you think about the idea of $U$ being a plane at the origin in ${\mathbb{R}}^3$, we can think of $V/U$ as the line of planes that can be made possible by adding our $v$'s. Hence, this is why $\text{dim}(V/U)=1$. Further, $\pi$ gives the plane for a chosen $v$ as it's output $v+U$. Lastly, $\tilde{T}$ is the map of all parallel sets (affine) to it's own null space, and outputs just the vector that's not in that nullspace $v$.

Proof
It's easy to show (1) as routine. For (2), suppose $v\in V$ and $\tilde{T}(v+\text{null}(T))=0$. Then $Tv=0$ so then $v\in \text{null}(T)$. Then Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^045b5b says that $v+\text{null}(T)=0+\text{null}(T)$ so then $\text{null}(\tilde{T})=\{0\}$ so then $\tilde{T}$ is injective.

The definition shows how (3) is true. For (4), use the FTOLM to show they have the same dimension, and thus there is some isomorphism. Further, $\tilde{T}$ is this isomorphism.
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3.F: Duality

The Dual Space and the Dual Map

For instance, the trace of a matrix is a linear function on $\mathbb{F}$. Another example is $\phi :\mathcal{P}(\mathbb{R})\to \mathbb{R}$ where it's given as:

This is a linear functional. We give a special name to $\mathcal{L}(V,\mathbb{F})$:

Proof
$\text{dim}(V^{\prime })=\text{dim}(V)\text{dim}(\mathbb{F})=\text{dim}(V)$.
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As such, then these two spaces are also isomorphic too.

The next result shows that the dual basis is indeed a basis. Thus, the terminology "dual basis" is justified.

Proof
Suppose $v_1,...,v_n$ is a basis for $V$. Let ${\phi }_1,...,{\phi }_n$ denote the dual basis. To show that ${\phi }_1,...,{\phi }_n$ is LI, suppose $a_1,...,a_n\in \mathbb{F}$ such that:

Now $(a_1{\phi }_1+\cdots +a_n{\phi }_n)(v_j)=a_j$ for $j=1,...,n$. Thus $a_1=\cdots =a_n=0$ so then ${\phi }_1,...,{\phi }_n$ is LI.

We also have $n$ vectors, implying that it's a basis for $V^{\prime }$.
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If $T\in \mathcal{L}(V,W)$ and $\phi \in W^{\prime }$ then $T^{\prime }(\phi )$ is defined above as the composition of the linear maps $\phi$ and $T$, so $T^{\prime }(\phi )$ is a linear map from $V$ to $\mathbb{F}$, so $T^{\prime }(\phi )\in V^{\prime }$.

To show the linear $W^{\prime }\to V^{\prime }$ part, if $\phi ,\psi \in W^{\prime }$ then:

and for $\lambda \in \mathbb{F}$:

For example, define $D$ as the standard derivative operator over the polynomials $\mathcal{P}(\mathbb{R})$. Suppose $\phi$ is the linear function on this space given by $\phi (p)=p(3)$. Then $D^{\prime }(\phi )$ is the linear functional on our polynomial space given by:

So then $D^{\prime }(\phi )$ is the linear functional that takes $p$ to $p^{\prime }(3)$.

Or if $\phi (p)={\int }_0^{1}p$ then:

Proof
The reader (me) will prove the first two in a HW problem.

For the third one, suppose $\phi \in W^{\prime }$ then:

Thus $(ST{)}^{\prime }=T^{\prime }{S}^{\prime }$.
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The Null Space and Range of the Dual of a Linear Map

Our goal here is to describe $\text{null}(T^{\prime })$ and the range in terms of the nullspace and range of $T$:

This is a "nullspace lite" for the $\phi$, only applying for a subspace $U$. For example, suppose $U$ is the subspace of $\mathcal{P}(\mathbb{R})$ consisting of all polynomial multiples of $x^2$. If $\phi$ is the linear functional on the space defined by $\phi (p)=p^{\prime }(0)$, then $\phi \in U^0$.

As an example, let $e_1,...,e_5$ denote the standard basis of ${\mathbb{R}}^5$. Let ${\phi }_1,...,{\phi }_5$ denote the dual basis of $({\mathbb{R}}^5{)}^{\prime }$. Suppose:

Then $U^0=\text{span}({\phi }_3,{\phi }_4,{\phi }_5)$. Why? We'll recall from Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^aab354 that:

just selects the $j$-th coordinate.

First, suppose $\phi \in \text{span}({\phi }_3,{\phi }_4,{\phi }_5)$. Then there are $c_3,c_4,c_5\in \mathbb{R}$ such that:

If $u\in U$ then $u=(u_1,u_2,0,0,0)$, so:

as a result, since it selects either the 3rd, 4th, or 5th entry. Thus $\phi \in U^0$. So then $\text{span}({\phi }_3,...,{\phi }_5)\subset U$. For the other direction, suppose $\phi \in U^0$. Since the dual basis is a basis of $({\mathbb{R}}^5{)}^{\prime }$ then there are $c_i\in \mathbb{R}$ where:

Since $e_1\in U$ and $\phi \in U^0$ then:

The same is for $e_2$, so $c_2=0$. Hence $\phi \in \text{span}({\phi }_3,{\phi }_4,{\phi }_5)$ and thus we get a subset the other way.

Proof
Clearly $0\in U^0$ where $0$ is the zero-linear functional on $V$. This is because for all $u\in U$ we have it that $0u=0$ where the left $0$ is the zero functional and the right $0$ is in $U$.

Suppose $\phi ,\psi \in U^0$. Then $\phi ,\psi \in V^{\prime }$. Further, $\phi (u)=\psi (u)=0$ for every $u\in U$. If $u\in U$ then:

Thus we get closure under vector addition. A similar argument shows closure under scalar multiplication. Thus, $U^0$ is a subspace of $V^{\prime }$.
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The following proof could work as follows:

• Choose a basis $u_1,...,u_m$ of $U$
• Extend the basis to $u_1,...,u)m,...,u_n$ for $V$
• Let ${\phi }_1,...,{\phi }_m,...,{\phi }_n$ be the dual basis of $V^{\prime }$.
• Show that ${\phi }_{m+1},...,{\phi }_n$ is a basis of $U^0$.

But this proof will be quicker than that:

Proof
Let $i\in \mathcal{L}(U,V)$ be the inclusion map defined by $i(u)=u$ for $u\in U$. Thus $i^{\prime }$ is a linear map from $V^{\prime }\to U^{\prime }$. Using the FTOLM:

But $\text{null}(i^{\prime })=U^0$ and since $\text{dim}(V^{\prime })=\text{dim}(V)$ then:

If $\phi \in U^{\prime }$ then $\phi$ can be extended to a linear function $\psi$ on $V$ via HW 3 - Linear Maps#11. As such, then $i^{\prime }(\psi )=\phi$ so then $\phi \in \text{range}(i^{\prime })$. Thus, $\text{range}(i^{\prime })=U^{\prime }$ so the left dimension becomes that for $\text{dim}(U)$ as we want.
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Proof
(a) First, suppose $\phi \in \text{null}(T^{\prime })$. Thus $0=T^{\prime }(\phi )=\phi \circ T$. Hence:

for any $v\in V$. Thus $\phi \in (\text{range}(T){)}^0$ so then $\subset$ is implied.

For the other way, suppose $\phi \in (\text{range}(T){)}^0$. Then $\phi (Tv)=0$ for all $v\in V$. Thus, $0=\phi \circ T=T^{\prime }(\phi )$ so then $\phi \in \text{null}(T^{\prime })$. Thus $\supset$ is implied.

(b) We have:

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The following is very useful as sometimes it's easier to prove that something is injective, rather than surjective.

Proof
The map $T\in \mathcal{L}(V,W)$ is surjective iff $\text{range}(T)=W$, which happens iff $(\text{range}(T){)}^0=\{0\}$, which happens iff $\text{null}(T^{\prime })=\{0\}$ by Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^87da71, which happens iff $T^{\prime }$ is injective.
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Proof
(a) We have

(b) First suppose $\phi \in \text{range}(T^{\prime })$. Thus, there exists $\psi \in W^{\prime }$ such that $\phi =T^{\prime }(\psi )$. If $v\in \text{null}(T)$, then:

Hence $\phi \in (\text{null}(T){)}^0$. This implies that $\text{range}(T^{\prime })\subset (\text{null}(T){)}^0$.

We will complete the proof by showing that $\text{range}(T^{\prime })$ and $(\text{null}(T){)}^0$ have the same dimension, showing equality. Note that:

where the first equality comes from (a), the second from the FTOLM. The third comes from Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^dc9ef2.
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Proof
The map $T$ is injective iff $\text{null}(T)=\{\overrightarrow{0}\}$, iff $(\text{null}(T){)}^0=V^{\prime }$ using dimensions. This only is true iff $\text{range}(T^{\prime })=V^{\prime }$ via Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^6e8978, which only happens if $T^{\prime }$ is surjective.
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The Matrix of the Dual of a Linear Map

We can now define the transpose of a matrix:

Notice that the transpose operator itself is linear $(A+C{)}^T=A^T+C^T$ and $(\lambda A{)}^T=\lambda A^T$ for all $m\times n$ matrices $A,C$ and all $\lambda \in \mathbb{F}$, and thus is a linear operator.

Proof
Suppose $1\le k\le p$ and $1\le j\le m$. Then:

Therefore $(AC{)}^T=(C^T{A}^T)$.
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For the next lemma, we assume a $n$ dimensional $V$ and dual basis ${\phi }_i\in V^{\prime }$, an $m$ dimensional $W$ along with it's dual basis ${\psi }_i\in W^{\prime }$. So $\mathcal{M}(T)$ s computer with respect to the bases just mentioned of $V$ and $W$ and $\mathcal{M}(T^{\prime })$ is computed with respect to the dual bases of $W^{\prime }$ and $V^{\prime }$.

Note that usually $T$ in the exponent is not talking about the exponentiation of $T$, which is an allowed operation.

Proof
Let $A=\mathcal{M}(T)$ and $C=\mathcal{M}(T^{\prime })$. Suppose $1\le j\le m$ and $1\le k\le n$. From the definition of $\mathcal{M}(T^{\prime })$ we have:

The left side of the equation equals ${\psi }_j\circ T$. So applying both sides of the equation on $v_k$ gives:

or:

Comparing the last line of both equations, then $C_{k,j}=A_{j,k}$ so $C=A^T$. Thus, $\mathcal{M}(T^{\prime })=(\mathcal{M}(T){)}^T$ as desired.
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The Rank of a Matrix

For example, consider:

The row rank of $A$ is the dimension of:

notice neither two vectors listed above are LD, so this is dimension $2$. The row rank of $A$ is 2.

The column rank is similar just with the columns of $A$. the span of this list is length 4, and must have at least dimension 2 since the first two vectors are LI with each other. But the span of the list is in ${\mathbb{F}}^{2,1}$ so it's dimension must be at most 2. Thus, the dimension is exactly 2, so the column rank of $A$ is 2.

Proof
Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_n$ is a basis of $W$. The function that takes $w\in \text{span}(T{v}_1,...,T{v}_n)$ to $\mathcal{M}(w)$ is easily seen to be an isomorphism from $\text{span}(T{v}_1,...,T{v}_n)$ onto $\text{span}(\mathcal{M}(T{v}_1),...,\mathcal{M}(T{v}_n))$. Thus, their dimensions equal, where the last dimension equals the column rank of $\mathcal{M}(T)$.

It is easy to see that $\text{range}(T)=\text{span}(T{v}_1,...,T{v}_n)$. Thus we have $\text{dim}(\text{range}(T))=\text{dim}(\text{span}(T{v}_1,...,T{v}_n))$ which is the column rank of $\mathcal{M}(T)$ as desired.
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Proof
Define $T:{\mathbb{F}}^{n,1}\to {\mathbb{F}}^{m,1}$ by $Tx=Ax$. Thus $\mathcal{M}(T)=A$, where $\mathcal{M}(T)$ is computed with respect to the standard bases of ${\mathbb{F}}^{n,1}$ and ${\mathbb{F}}^{m,1}$. Now:

where:

• second equality from the previous lemma
• third equality comes from Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^6e8978
• fourth equality from the previous lemma
• fifth equality comes from Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^118717
• the last equality comes from the definitions
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  We dispense of separating the "row rank" from "column rank" and just call it "rank":