aThe note here aims to provide shallow references to important aspects of linear maps that relate to the juicy part of linear algebra, matrix. Proofs are mostly omitted in the note as the author was in a hurry and lazy. Like the previous notes, we use V${\displaystyle V}$ to denote vector space over filed F${\displaystyle \mathcal{F}}$.1. Linear Map and Its Properties 1.1. Injectivity and surjectivity of linear maps 1.2. Invertibility and Isomorphism of linear maps 2. Matrix, A Representation of Linear Map 2.1. Rank of a matrix 2.2. Notations for matrix manipulations 3. Duality 3.1. Null space and range of the dual of a linear map 4. Things missed out in this note 1. Linear Map and Its PropertiesIn mathematics, mapping builds the correspondence from one set to another set. In linear algebra, the sets are vector spaces. A map is called linear if the following conditions are satisfied:

Definition1A map T${\displaystyle T}$ from V${\displaystyle V}$ to W${\displaystyle W}$ is called linear ifT(v+u) = Tv+Tu${\displaystyle T(v+u)=Tv+Tu}$ for v,u∈V${\displaystyle v,u\in V}$, andT(𝜆v) = 𝜆Tv${\displaystyle T(𝜆v)=𝜆Tv}$ for 𝜆∈F${\displaystyle 𝜆\in \mathcal{F}}$ and v∈V${\displaystyle v\in V}$

For the simplicity, we use T:V→W${\displaystyle T:V\to W}$ to denote a linear map from vector space V${\displaystyle V}$ to W${\displaystyle W}$. T${\displaystyle T}$ here stands for transformation as you will soon see why. Similar to the definition of vector space, the definition of linear map is accompanied by the definition of "zero" or "null":

Definition2The null space of linear map T:V→W${\displaystyle T:V\to W}$ is defined asnull T={Tv=0: v∈V}${\displaystyle nullT=\{Tv=0:v\in V\}}$

To give an example of null space, let V=R3${\displaystyle V={\mathbb{R}}^3}$, and for arbitrary (x1,x2,x3)∈V${\displaystyle (x_1,x_2,x_3)\in V}$, we have T(x1,x2,x3)=(x1,x2)${\displaystyle T(x_1,x_2,x_3)=(x_1,x_2)}$, i.e. T:R3→R2${\displaystyle T:{\mathbb{R}}^3\to {\mathbb{R}}^2}$. In this case, nullT={(0,0,x3):x3∈R}${\displaystyle nullT=\{(0,0,x_3):x_3\in \mathbb{R}\}}$. We also have a name for the set of vectors that lead to nonzero vectors upon application of T${\displaystyle T}$:

Definition3the range of a linear map T:V→W${\displaystyle T:V\to W}$, denoted as rangeT${\displaystyle rangeT}$, is a subset of W${\displaystyle W}$ containing vectors that satisfies the following conditionrange T={w∈W: ∃v∈V, Tv=w }${\displaystyle rangeT=\{w\in W:\exists v\in V,Tv=w\}}$

With these definitions introduced, we are ready to give the theorem that relates T${\displaystyle T}$, rangeT${\displaystyle rangeT}$, and nullT${\displaystyle nullT}$. Because we use it so frequently in the proofs of linear algebra, we give it a dramatic name as

Theorem1Fundemental theorem of linear maps: Suppose V${\displaystyle V}$ is finite-dimensional and T${\displaystyle T}$ is a linear map from V${\displaystyle V}$ to W${\displaystyle W}$, then we havedimV=dim nullT + dim rangeT${\displaystyle dimV=dimnullT+dimrangeT}$.

Definition 1 indicates that linear maps are closed under addition and multiplication, so it should not be a surprise that all the linear maps T:V→W${\displaystyle T:V\to W}$ form a linear space, denoted as L(V,W)${\displaystyle \mathcal{L}(V,W)}$. Before we discuss special properties of linear maps, we need to clarify two important concepts related to linear maps: injectivity and surjectivity.1.1. Injectivity and surjectivity of linear mapsA linear map is called injective if either one of the following conditions holds:

Theorem2Injectivity of a linear map T:V→W${\displaystyle T:V\to W}$ means:a. for v,u∈V${\displaystyle v,u\in V}$, Tv=Tu${\displaystyle Tv=Tu}$ iff v=u${\displaystyle v=u}$, andb. null T={0}${\displaystyle nullT=\{0\}}$ and dim null T=0${\displaystyle dimnullT=0}$

With Theorem 2, it should not be hard to imagine that a map is not injective when cordiality of V${\displaystyle V}$ is larger than W${\displaystyle W}$. For vector spaces, cordinal numbers are usually infinite. Thus, a formal way of saying this is that a linear map is not injective when dim V>dimW${\displaystyle dimV>dimW}$. To see this, we notice that range T⊂W${\displaystyle rangeT\subset W}$ in Theorem 2. According to Theorem 1, we have dim null T=dim V-dim range T≥dimeV-dim W>0${\displaystyle dimnullT=dimV-dimrangeT\ge dimeV-dimW>0}$, contradicting to Theorem 2b. Thus, there is at least one pair of v${\displaystyle v}$ and u${\displaystyle u}$ satisfying Tv=Tu${\displaystyle Tv=Tu}$ with u≠v${\displaystyle u\ne v}$. As for the surjectivity, we have the following theorem holds

Theorem3Surjectivity of a linear map T:V→W${\displaystyle T:V\to W}$ means range T=W${\displaystyle rangeT=W}$.

Theorem 3 indicates that a linear map is not surjective if dim V≤ dim W${\displaystyle dimV\le dimW}$, as can be proved by using Theorem 1. In other words, there is at least one w∈W${\displaystyle w\in W}$ which no v∈V${\displaystyle v\in V}$ can map to when dim V≤ dim W${\displaystyle dimV\le dimW}$. So what is so special about injectivity and surjectivity? How can we solve problems by exploiting the two properties of a linear map? Arguably the easiest application is to check if (in)homogeneous system of linear equations have a solution or not. A homogeneous system of m${\displaystyle m}$ linear equations can be written in the following form:a

a1,1x1+a1,2x2+⋯a1,nxn	=0
a2,1x1+a2,2x2+⋯a2,nxn	=0
⋮
am,1x1+am,2x2+⋯am,nxn	=0

   

(1)

$$\begin{array}{rl}{a}_{1,1}{x}_1+a_{1,2}{x}_2+\cdots a_{1,n}{x}_n& =0\\ a_{2,1}{x}_1+a_{2,2}{x}_2+\cdots a_{2,n}{x}_n& =0\\ ⋮& \\ a_{m,1}{x}_1+a_{m,2}{x}_2+\cdots a_{m,n}{x}_n& =0\end{array}\text{anchor-tag}$$If we let V=Fn${\displaystyle V={\mathcal{F}}^n}$, then we can define a linear map T:Fn→Fm${\displaystyle T:{\mathcal{F}}^n\to {\mathcal{F}}^m}$ as the followingT(x1,x2,...,xn)=( ∑j=1na1,jxj, ∑j=1na2,jxj,..., ∑j=1nam,jxj)  

(2)

$$T(x_1,x_2,...,x_n)=({\sum _{j=1}^{}}^n{a}_{1,j}{x}_j,{\sum _{j=1}^{}}^n{a}_{2,j}{x}_j,...,{\sum _{j=1}^{}}^n{a}_{m,j}{x}_j)\text{anchor-tag}$$So having a solution to the system above is equavalent to having nonzero element in null T${\displaystyle nullT}$, i.e., dim null T>0${\displaystyle dimnullT>0}$. Acoording to Theorem 1, such condition can only be satisfied when dim Fn>dim range T${\displaystyle dim{\mathcal{F}}^n>dimrangeT}$. Because dim range T≤dim Fm${\displaystyle dimrangeT\le dim{\mathcal{F}}^m}$, we must have solutions to the equation system when n>m${\displaystyle n>m}$. Thus, a homogeneous system of linear equations has solutions when its corresponding map T:Fn→Fm${\displaystyle T:{\mathcal{F}}^n\to {\mathcal{F}}^m}$ is not injective.On the other hand, for inhomogeneous system of linear equations, having a solution means there is at least one vector v=(x1,x2,...,xn)∈Fn${\displaystyle v=(x_1,x_2,...,x_n)\in {\mathcal{F}}^n}$ satisfying the condition of Tv=(c1,c2,...,cm)${\displaystyle Tv=(c_1,c_2,...,c_m)}$, where some ci≠0${\displaystyle c_i\ne 0}$. Given the linear map defined in Eqn.(2), we claim that solution for the inhomogeneous sytem does not exist for some choices of ci${\displaystyle c_i}$ when m>n${\displaystyle m>n}$ as the linear map T${\displaystyle T}$ is not surjective.Other than this simple application, surjectivity and injectivity also dictate two properties of linear maps, invertibility and isomorphism, as we shall now see.1.2. Invertibility and Isomorphism of linear mapsWe first define the invertibility of a linear map as the following

Definition4A linear map T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$ is invertible if there exists another linear map S∈L(W,V)${\displaystyle S\in \mathcal{L}(W,V)}$ that make S(Tv)=Iv=v${\displaystyle S(Tv)=Iv=v}$ hold for all v∈V${\displaystyle v\in V}$, where I${\displaystyle I}$ is identity map.

As a matter of fact, the inverted map S${\displaystyle S}$ in Definition 4 is usually written as T-1${\displaystyle T^{-1}}$ (i.e. inverse of T${\displaystyle T}$). If a linear map has an inverse, we have

Proposition1Inverse of a linear map is unique.

as can be shown by assuming two inverses, S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$, of a linear map T${\displaystyle T}$:S1=S1I=S1TS2=IS2=S2$$S_1=S_{1}I=S_{1}T{S}_2=I{S}_2=S_2$$According to Definition 4, STv=v${\displaystyle STv=v}$ for all v∈V${\displaystyle v\in V}$, which indicates

Proposition2Invertibility of a linear map indicates injectivity and surjectivity.

Being invertible for a linear map is very important to quantum-mechanical calculations, as an invertible map, if it is unitary, is equivalent to a reversible operator whose physical realization is a logic gate on quantum circuit. Unlike quantum computer, classical logic gates, such as AND, OR, XOR are irreversible as numbers of their inputs are larger than numbers of their outputs. From the perspective of information theory, being reversible means that we can recover input information by simply knowing the outputs, which can be achieved by gates that follow invertible linear map from inputs to outputs. To be streamlined with possible later discussion of quantum operators(gates). Here we give the definition of operator in linear algebra:

Definition5An operator is a linear map from a vector space onto itself, i,e, T:V→V${\displaystyle T:V\to V}$, and the set of operators on vector space V${\displaystyle V}$ forms a vector space as well, denoted as L(V)${\displaystyle \mathcal{L}(V)}$.

The concept of isomorphism is a natural extension of inversibility of vector space, as we shall introduce now

Definition6An isomorphism is a invertible linear map, and thus, two vector spaces are called isomorphic if an invertible linear map exists from one to another.

Isomorphism gives the equivalence of two vector spaces. This is very powerful as we can now make a problem, being hard in one vector space, much easier by simply reformulating it onto another vector space, if we know the two vector spaces are isomorphic. In linear algebra, arguably the most important isomorphism is between the vector space of linear map L(V,W)${\displaystyle \mathcal{L}(V,W)}$, and the vector space of matrices, Fm,n${\displaystyle {\mathcal{F}}^{m,n}}$. We will come back to the proof of this proposition once we introduce the concept of matrix.2. Matrix, A Representation of Linear MapThe title of this section indicates on intention of explaining representation theory used in quantum theories, even though the author would much love to. Instead, we will introduce the form of matrix, and the interpretation of it. However, we will not focus on standard contents such as matrix addition, multiplication, etc., for which readers can refer to introductory textbooks or Wikipedia pages. The primary purpose of this section is to reveal how one can understand matrix in terms of linear map, and mechanistic calculation of matrices is never the fun part.To begin with, suppose we have a vector space V${\displaystyle V}$ with its basis being v1,v2,...,vn${\displaystyle v_1,v_2,...,v_n}$, and another vector space W${\displaystyle W}$ with its basis of w1,w2,...,wm${\displaystyle w_1,w_2,...,w_m}$. Therefore, dimV=n${\displaystyle dimV=n}$ and dimW=m${\displaystyle dimW=m}$. Now we define a linear map T:V→W${\displaystyle T:V\to W}$ asTvi=a1,iw1+a2,iw2+⋯+am,iwm for i=1,2,...,n  

(3)

$$T{v}_i=a_{1,i}{w}_1+a_{2,i}{w}_2+\cdots +a_{m,i}{w}_m\text{for }i=1,2,...,n\text{anchor-tag}$$where ai,j∈F${\displaystyle a_{i,j}\in \mathcal{F}}$. Then, the linear map can be represented by a m-row, n-column matrix as shown belowv1 ⋯ vi ⋯ vna

w1

w2

⋮

wm

a

a1,1	⋯	a1,i	⋯	a1,n
a2,1		a2,i		a2,n
⋮		⋮		⋮
am,1		am,i		am,n

  

(4)

$$\begin{array}{c}{v}_1\cdots v_i\cdots v_n\\ \begin{array}{c}{w}_1\\ w_2\\ ⋮\\ w_m\end{array}\left(\begin{array}{ccccc}{a}_{1,1}& \cdots & a_{1,i}& \cdots & a_{1,n}\\ a_{2,1}& & a_{2,i}& & a_{2,n}\\ ⋮& & ⋮& & ⋮\\ a_{m,1}& & a_{m,i}& & a_{m,n}\end{array}\right)\text{anchor-tag}\end{array}$$By multiplying elements in each column with its associated wj${\displaystyle w_j}$ outside of the matrix, we recover the pesentation of Tvi${\displaystyle T{v}_i}$ in W${\displaystyle W}$. To see how an arbitrary vector v∈V${\displaystyle v\in V}$ is mapped to W${\displaystyle W}$ using the matrix in (4), let us first represent v${\displaystyle v}$ as a conlumn vector, i.e.,M(v)=a

c1

c2

⋮

cn

  

(5)

$$\mathcal{M}(v)=\left(\begin{array}{c}{c}_1\\ c_2\\ ⋮\\ c_n\end{array}\right)\text{anchor-tag}$$which indicates v=c1v1+c2v2+⋯+cnvn${\displaystyle v=c_1{v}_1+c_2{v}_2+\cdots +c_n{v}_n}$ and ci∈F${\displaystyle c_i\in \mathcal{F}}$ for i=1,2,...,n${\displaystyle i=1,2,...,n}$. M${\displaystyle \mathcal{M}}$ here maps v∈V${\displaystyle v\in V}$ to column vector, and it is also a linear map. We will discuss more about this later. Given Eqn.(3), we can calculate Tv${\displaystyle Tv}$ asa

Tv	=c1Tv1+c2Tv2+⋯+cnTvn
	=c1m∑j=1aj,1wj+c2m∑j=1aj,2wj+⋯+cnm∑j=1aj,nwj
	= (n∑i=1cia1,i)w1+(n∑i=1cia2,i)w2+⋯+(n∑i=1ciam,i)wm

  

(6)

$$\begin{array}{rl}Tv& =c_{1}T{v}_1+c_{2}T{v}_2+\cdots +c_{n}T{v}_n\\ & =c_1\sum _{j=1}^m{a}_{j,1}{w}_j+c_2\sum _{j=1}^m{a}_{j,2}{w}_j+\cdots +c_n\sum _{j=1}^m{a}_{j,n}{w}_j\\ & =\left(\sum _{i=1}^n{c}_i{a}_{1,i}\right)w_1+\left(\sum _{i=1}^n{c}_i{a}_{2,i}\right)w_2+\cdots +\left(\sum _{i=1}^n{c}_i{a}_{m,i}\right)w_m\end{array}\text{anchor-tag}$$If we use the rule of matrix multiplication to multiply matrix in (4) with the vector in (5), we obtain a m-row column vector beingw=a

(n∑i=1cia1,i)

(n∑i=1cia2,i)

⋮

(n∑i=1ciam,i)

  

(7)

$$w=\left(\begin{array}{c}\left(\sum _{i=1}^n{c}_i{a}_{1,i}\right)\\ \left(\sum _{i=1}^n{c}_i{a}_{2,i}\right)\\ ⋮\\ \left(\sum _{i=1}^n{c}_i{a}_{m,i}\right)\end{array}\right)\text{anchor-tag}$$Comparing (7) with (6), it should be obvious that the resulted vector w${\displaystyle w}$ is indeed the image of v${\displaystyle v}$ in W${\displaystyle W}$. In other words, v${\displaystyle v}$ is mapped to w${\displaystyle w}$ through the linear map T${\displaystyle T}$ represented as the matrix in (4). To be formal, we can define the matrix of a linear map as the following

Definition7Suppose T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$, and v1,v2,...,vn${\displaystyle v_1,v_2,...,v_n}$ is a basis of V${\displaystyle V}$, and w1,...,wm${\displaystyle w_1,...,w_m}$ is a basis of W${\displaystyle W}$. The matrix of T${\displaystyle T}$ with respect to these bases is a m-by-n matrix M(T)${\displaystyle \mathcal{M}(T)}$ whose entries ai,j${\displaystyle a_{i,j}}$ are defined by Eqn.(3).

Definition 7 emphasizes the use of bases, implying that matrix is always related to the change of basis. If we treat M${\displaystyle \mathcal{M}}$ as a map that maps T${\displaystyle T}$ to a matrix, then it's easy to show that M${\displaystyle \mathcal{M}}$ itself is a linear map as it is closed under addition and multiplication. Also, For the simplicity in later discussion, we use Fm,n${\displaystyle {\mathcal{F}}^{m,n}}$ to denote a m-by-n matrix.So far, we have taken for granted that linear map can always be represented by a matrix. But the validity of such a claim need to be examined by proving the following proposition:

Proposition3Suppose v1,v2,...,vn${\displaystyle v_1,v_2,...,v_n}$ is a basis of V${\displaystyle V}$ and w1,...,wm${\displaystyle w_1,...,w_m}$ a basis of W${\displaystyle W}$, then M${\displaystyle \mathcal{M}}$ is an isomorphism between L(V,W)${\displaystyle \mathcal{L}(V,W)}$ and Fm,n${\displaystyle {\mathcal{F}}^{m,n}}$.

Proof:M${\displaystyle \mathcal{M}}$ is a linear map as for any T,S∈L(V,W)${\displaystyle T,S\in \mathcal{L}(V,W)}$, M(T+S)=M(T)+M(S)${\displaystyle \mathcal{M}(T+S)=\mathcal{M}(T)+\mathcal{M}(S)}$, M(𝜆T)=𝜆M(T)${\displaystyle \mathcal{M}(𝜆T)=𝜆\mathcal{M}(T)}$ for 𝜆∈F${\displaystyle 𝜆\in \mathcal{F}}$. So we only need to prove it is injective and surjective because of Proposition 2. According to Theorem 2, if M(T)=0${\displaystyle \mathcal{M}(T)=0}$ then Tvi=0${\displaystyle T{v}_i=0}$ for any of the basis vector of V${\displaystyle V}$. This is only possible when T=0${\displaystyle T=0}$, i.e., null M={0}${\displaystyle null\mathcal{M}=\{0\}}$. Thus, M${\displaystyle \mathcal{M}}$ is injective. To prove M${\displaystyle \mathcal{M}}$ is surjective, suppose A∈Fm,n${\displaystyle A\in {\mathcal{F}}^{m,n}}$ is an arbitrary matrix. we can always define a linear map T${\displaystyle T}$ from V${\displaystyle V}$ to W${\displaystyle W}$ asTvi=m∑j=1Aj,iwj for i=1,2,...,n.$$T{v}_i=\sum _{j=1}^m{A}_{j,i}{w}_j\text{for}i=1,2,...,n.$$from which M(T)=A${\displaystyle \mathcal{M}(T)=A}$ holds. In other words, range M=Fm,n${\displaystyle range\mathcal{M}={\mathcal{F}}^{m,n}}$, and M${\displaystyle \mathcal{M}}$ is surjective.□2.1. Rank of a matrixSo far we only considered matrices that have each row or each column associated with a basis vector in vector spaces. What if we add to those matrices more rows or columns? Will the expanded matrices still be isomorphic to some linear maps? This is a weird question, but there is no limitation for expanding the size of matrices at all. Let's first add some rows to matrix A${\displaystyle A}$ isomorphic to a linear map T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$, and adding columns follows the same logic. Suppose v1,v2,v3${\displaystyle v_1,v_2,v_3}$ is a basis of V${\displaystyle V}$ and w1,w2,w3${\displaystyle w_1,w_2,w_3}$ a basis of W${\displaystyle W}$. If we define the map asTvi=A1,iw1+A2,iw2+A3,iw3$$T{v}_i=A_{1,i}{w}_1+A_{2,i}{w}_2+A_{3,i}{w}_3$$then we can write down A${\displaystyle A}$ as a 3-by-3 matrix. Now, if we rewrite the definition above asTvi=A′1,iw1+A′2,iw2+A′3,iw3+A′4,i(2w2+w3) with A′4,i≠0$$T{v}_i=A_{1,i}^{\prime }{w}_1+A_{2,i}^{\prime }{w}_2+A_{3,i}^{\prime }{w}_3+A_{4,i}^{\prime }(2w_2+w_3)\text{with }{A}_{4,i}^{\prime }\ne 0$$then it is obvious that A1,i=A′1,i${\displaystyle A_{1,i}=A_{1,i}^{\prime }}$, A2,i=A′2,i+2A′4,i${\displaystyle A_{2,i}=A_{2,i}^{\prime }+2A_{4,i}^{\prime }}$, A3,i=A′3,i+A′4,i${\displaystyle A_{3,i}=A_{3,i}^{\prime }+A_{4,i}^{\prime }}$. The two definitions will map arbitrary vector v=c1v1+c2v2+c3v3∈V${\displaystyle v=c_1{v}_1+c_2{v}_2+c_3{v}_3\in V}$ to the same vector w∈W${\displaystyle w\in W}$, but A′${\displaystyle A^{\prime }}$ is NOT isomorphic to T${\displaystyle T}$. To see this, we simply note that we can give different values to A′2,i${\displaystyle A_{2,i}^{\prime }}$, A′3,i${\displaystyle A_{3,i}^{\prime }}$, and A′4,i${\displaystyle A_{4,i}^{\prime }}$ and still make the relations A2,i=A′2,i+2A′4,i${\displaystyle A_{2,i}=A_{2,i}^{\prime }+2A_{4,i}^{\prime }}$ and A3,i=A′3,i+A′4,i${\displaystyle A_{3,i}=A_{3,i}^{\prime }+A_{4,i}^{\prime }}$ hold. In other words, T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$ corresponds to multiple A′∈F4,3${\displaystyle A^{\prime }\in {\mathcal{F}}^{4,3}}$, resulting in the map not injective or surjective. There are matrices that have rows or columns linearly dependent. In those cases, we define the row rank and the column rank of the matrix as the following

Definition8Suppose A∈Fm,n${\displaystyle A\in {\mathcal{F}}^{m,n}}$, then the row(column) rank of A${\displaystyle A}$ is the dimension of the span of the rows(columns) of A${\displaystyle A}$.

The rank numbers of a matrix can be related to a linear map through the following proposition

Proposition4Dimension of range T${\displaystyle rangeT}$ equals column rank of M(T)${\displaystyle \mathcal{M}(T)}$.

and

Proposition5Row rank of a matrix equals to column rank.

Thus, the rank of a matrix A∈Fm,n${\displaystyle A\in {\mathcal{F}}^{m,n}}$ is the column rank of A${\displaystyle A}$.2.2. Notations for matrix manipulationsWe end this section by introducing algebraic notations for matrix manipulations that will be used in the following sections. By following the notation in Axler's book, the transpose of a matrix is defined as the following

Definition9Transpose of a matrix A${\displaystyle A}$, denoted as At${\displaystyle A^t}$, has its element satisfying the relation below:Ati,j=Aj,i$$A_{i,j}^t=A_{j,i}$$

For people who have encountered linear algebra before, the following rule should not be alien:

Proposition6For two matrices A${\displaystyle A}$ and B${\displaystyle B}$, we have (AB)t=BtAt${\displaystyle (AB{)}^t=B^t{A}^t}$.

Proof: To see this, suppose A${\displaystyle A}$ is a m-by-n matrix and B${\displaystyle B}$ is n-by-q. So the element in AB${\displaystyle AB}$ can be represented as(AB)i,j=n∑kAi,kBk,j$$(AB{)}_{i,j}=\sum _k^n{A}_{i,k}{B}_{k,j}$$where Ai,k${\displaystyle A_{i,k}}$ and Bk,j${\displaystyle B_{k,j}}$ are elements in A${\displaystyle A}$ and B${\displaystyle B}$, respectively. Definition 9 gives a

(AB)ti,j	= (AB)j,i
	=n∑kAj,kBk,i=n∑kAtk,jBti,k=(BtAt)i,j

□$$\begin{array}{rl}(AB{)}_{i,j}^t& =(AB{)}_{j,i}\\ & =\sum _k^n{A}_{j,k}{B}_{k,i}=\sum _k^n{A}_{k,j}^t{B}_{i,k}^t=(B^t{A}^t{)}_{i,j}\end{array}$$Because of Proposition 3, we have invertibility of matrices well defined. Let A-1${\displaystyle A^{-1}}$ to be the inverse of a m-by-n A${\displaystyle A}$, then A-1A=I${\displaystyle A^{-1}A=I}$, with I${\displaystyle I}$ being n-by-n identity matrix. We can also use row and column indices to represent the i${\displaystyle i}$th row of a matrix A${\displaystyle A}$ as Ai,⋅${\displaystyle A_{i,\cdot }}$, and the j${\displaystyle j}$th column as A⋅,j${\displaystyle A_{\cdot ,j}}$. With this, the following identities hold:a

(AB)k,⋅	=Ak,⋅B
(AB)⋅,r	=AB⋅,r

$$\begin{array}{rl}(AB{)}_{k,\cdot }& =A_{k,\cdot }B\\ (AB{)}_{\cdot ,r}& =A{B}_{\cdot ,r}\end{array}$$3. DualityFor quantum computing and quantum information theories, linear maps that map vectors to scaler field F${\displaystyle \mathcal{F}}$ play critical roles, as those maps are used to collapse wavefunctions into states with real physical properties. Because it is so special, we call linear map T∈L(V,F)${\displaystyle T\in \mathcal{L}(V,\mathcal{F})}$ linear functional. For an arbitrary vector space V${\displaystyle V}$, we call L(V,F)${\displaystyle \mathcal{L}(V,\mathcal{F})}$ its dual space, V′${\displaystyle V^{\prime }}$. With this, we can find that

Proposition7dim V′=dim V${\displaystyle dim{V}^{\prime }=dimV}$

Proof: To see this, we first note that the isomorphism between L(V,W)${\displaystyle \mathcal{L}(V,W)}$ and Fm,n${\displaystyle {\mathcal{F}}^{m,n}}$ results in dim L(V,W)=dim Fm,n${\displaystyle dim\mathcal{L}(V,W)=dim{\mathcal{F}}^{m,n}}$. Getting dim Fm,n${\displaystyle dim{\mathcal{F}}^{m,n}}$ is easy. We can think of its basis vector as m-by-n matrices with only one nonzero element at i${\displaystyle i}$th row and j${\displaystyle j}$the column. Thus, we have dim Fm,n=m×n=dimL(V,W)${\displaystyle dim{\mathcal{F}}^{m,n}=m\times n=dim\mathcal{L}(V,W)}$, and dimL(V,F)=n=dimV${\displaystyle dim\mathcal{L}(V,\mathcal{F})=n=dimV}$.□If v1,...,vn${\displaystyle v_1,...,v_n}$ is a basis of V${\displaystyle V}$, then we can define the dual basis of v1,...,vn${\displaystyle v_1,...,v_n}$ as

Definition10The dual basis of v1,...,vn∈V${\displaystyle v_1,...,v_n\in V}$ is the list 𝜙1,...,𝜙n${\displaystyle {𝜙}_1,...,{𝜙}_n}$ of elements of dual space V′${\displaystyle V^{\prime }}$, where each 𝜙j${\displaystyle {𝜙}_j}$ is a linear functional on V${\displaystyle V}$ such that𝜙j(vk)={a

1 if k=j

0 if k≠j

|$${𝜙}_j(v_k)=\{\begin{array}{c}1\text{if}k=j\\ 0\text{ if }k\ne j\end{array}$$

With this definition, we can proof the following

Proposition8Dual basis is a basis of the dual space

Proof: we just need to prove that 𝜙1,...,𝜙n${\displaystyle {𝜙}_1,...,{𝜙}_n}$ is linearly independent because every linearly independent list of lenth dimV′${\displaystyle dim{V}^{\prime }}$ is the basis of V′${\displaystyle V^{\prime }}$. Suppose a1,...,an∈F${\displaystyle a_1,...,a_n\in \mathcal{F}}$ gives thata1𝜙1+⋯+an𝜙n=0.$$a_1{𝜙}_1+\cdots +a_n{𝜙}_n=0.$$So (a1𝜙1+⋯+an𝜙n)(vj)=aj${\displaystyle (a_1{𝜙}_1+\cdots +a_n{𝜙}_n)(v_j)=a_j}$ for j=1,...,n${\displaystyle j=1,...,n}$, and the equation above implies that a1=a2=⋯=an=0${\displaystyle a_1=a_2=\cdots =a_n=0}$. Hence the list 𝜙1,...,𝜙n${\displaystyle {𝜙}_1,...,{𝜙}_n}$ is linearly independent □We can take a step further to define the dual map,

Definition11If T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$, then the dual map of T${\displaystyle T}$ is the linear map T′∈L(W′,V′)${\displaystyle T^{\prime }\in \mathcal{L}(W^{\prime },V^{\prime })}$ such that T′(𝜙)=𝜙∘T${\displaystyle T^{\prime }(𝜙)=𝜙\circ T}$ for 𝜙∈W′${\displaystyle 𝜙\in W^{\prime }}$.

𝜙∘T${\displaystyle 𝜙\circ T}$ means the composite application of two mapps. For arbitrary vector v∈V${\displaystyle v\in V}$, (𝜙∘T)${\displaystyle (𝜙\circ T)}$ first maps it to a vector Tv∈W${\displaystyle Tv\in W}$, and then apply 𝜙${\displaystyle 𝜙}$ to Tv${\displaystyle Tv}$ to map it to a scalar. The algebraic properties of dual maps listed below might be usefula

(S+T)′	=S′+T′ for all S,T∈L(V,W)
(𝜆T)′	=𝜆T′ for all 𝜆∈F
(ST)′	=T′S′ where T∈L(U,V) , S∈L(V,W)

$$\begin{array}{c}\begin{array}{rl}(S+T{)}^{\prime }& =S^{\prime }+T^{\prime }\text{for all}S,T\in \mathcal{L}(V,W)\\ (𝜆T{)}^{\prime }& =𝜆T^{\prime }\text{for all}𝜆\in \mathcal{F}\\ (ST{)}^{\prime }& =T^{\prime }{S}^{\prime }\text{where}T\in \mathcal{L}(U,V),S\in \mathcal{L}(V,W)\end{array}\\ \end{array}$$3.1. Null space and range of the dual of a linear mapWe can describe range T′${\displaystyle range{T}^{\prime }}$ and null T′${\displaystyle null{T}^{\prime }}$ in terms of range T${\displaystyle rangeT}$ and null T${\displaystyle nullT}$ by first defining the annihilator of subspaces.

Definition12For U⊂V${\displaystyle U\subset V}$, the annihilator of U${\displaystyle U}$, denoted as U0${\displaystyle U^0}$, is defined byU0={𝜙∈V′:𝜙(u)=0 ∀u∈U}$$U^0=\{𝜙\in V^{\prime }:𝜙(u)=0\forall u\in U\}$$

The dimension of the annihilator is related to dimV${\displaystyle dimV}$ through the folowing eqaution

Proposition9Suppose V${\displaystyle V}$ if finite-dimensional and U${\displaystyle U}$ is a subspace of V${\displaystyle V}$, then dimU+dimU0=dimV.${\displaystyle dimU+dim{U}^0=dimV.}$

Suppose V${\displaystyle V}$ and W${\displaystyle W}$ are finite-dimensional and T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$, then the following results hold. The author lists them here while have no clue how they are used in quantum-mechanical calculations.

a

nullT′	=(range T)0
dim nullT′	=dim null T+dimW-dimV
dim range T′	=dim range T
range T′	=(null T)0

$$\begin{array}{rl}null{T}^{\prime }& =(rangeT{)}^0\\ dimnull{T}^{\prime }& =dimnullT+dimW-dimV\\ dimrange{T}^{\prime }& =dimrangeT\\ range{T}^{\prime }& =(nullT{)}^0\end{array}$$

4. Things missed out in this noteLinear map is a rich branch of linear algebra. There are many interesting things left out here, among which products and quotients of space vectors are definitely worth diving. While quotients may not be pervasive in quantum computing literature, it seems to be a handy tool in proofs of quantum information theories.