aTOC1. Brief Intro 2. Vector Space, A Short Definition 3. Subspaces 3.1. Sum of subspaces 4. Prove Theorems 5. Juice for Me 1. Brief IntroIf you wander in the world of quantum computing(QC) long enough, it becomes painful to follow the arguments of quantum algorithms with your vague and shaky knowledge base of linear algebra. Inspired by this, the author takes an initiative to write a series of notes on linear algebra before we undertake the endeavor of QC rigorously. By writing the notes with QC in mind, the author hopes to emphasize the key parts of linear algebra that will repeatedly be used in QC applications. The objectives of this note are:• Introduce the concepts of vector space, subspace, (direct) sum of subspaces• Some nice proof exercises for knowledge testing 2. Vector Space, A Short DefinitionIn quantum computing, we mostly care about complex numbers and/or real numbers. Complex and real numbers live in the sets of C${\displaystyle \mathbb{C}}$ and R${\displaystyle \mathbb{R}}$, respectively. These two sets are called Fields in abstract algebra as +/-/×/÷${\displaystyle +/-/\times /÷}$are defined between the elements from those fields. Before diving into vector spaces, we first define the concept of tuple. Unlike set, tuple in mathematics allows repetition, and the sequence of the elements in a tuple matters. A tuple containing n elements is called n-tuple. In most literature, we repesent a n-tuple as (1)(x1,x2,x3,...xn)$$(x_1,x_2,x_3,...x_n)$$ We say that a n-tuple lives in the vector space Rn${\displaystyle {\mathbb{R}}^n}$ when its n elements are all real numbers. Conversely, the vector space Rn${\displaystyle {\mathbb{R}}^n}$ is a set of n-tuples that satisfying some criteria, for which we will discuss in detail later. Using the example of Rn${\displaystyle {\mathbb{R}}^n}$, we can define the vector space as (2)Rn={(x1,x2,x3,...,xn):xi∈R,i=1,2,3,...,n}$${\mathbb{R}}^n=\{(x_1,x_2,x_3,...,x_n):x_i\in \mathbb{R},i=1,2,3,...,n\}$$ where constraints on xi${\displaystyle x_i}$ are listed after semicolon(i.e., xi∈R${\displaystyle x_i\in \mathbb{R}}$). Similarly, we can define Cn${\displaystyle {\mathbb{C}}^n}$ as (3)Cn={(x1,x2,x3,...,xn):xi∈C,i=1,2,3,...,n}.$${\mathbb{C}}^n=\{(x_1,x_2,x_3,...,x_n):x_i\in \mathbb{C},i=1,2,3,...,n\}.$$ To simplify our following discussion, we use F${\displaystyle \mathcal{F}}$ to represent real or complex number field. By doing so, we indicate that the following discussions hold for both fields. For two elements in Fn${\displaystyle {\mathcal{F}}^n}$, we define the addition of the two elements as (4)(x1,x2,x3,...,xn)+(y1,y2,y3,...,yn)=(x1+y1,x2+y2,x3+y3,...,xn+yn)$$(x_1,x_2,x_3,...,x_n)+(y_1,y_2,y_3,...,y_n)=(x_1+y_1,x_2+y_2,x_3+y_3,...,x_n+y_n)$$where xi,yi∈F${\displaystyle x_i,y_i\in \mathcal{F}}$. Let x=(x1,x2,x3,...,xn)${\displaystyle x=(x_1,x_2,x_3,...,x_n)}$ and y=(y1,y2,y3,...,yn)${\displaystyle y=(y_1,y_2,y_3,...,y_n)}$, it's easy to see that (5)x+y=y+x.$$x+y=y+x.$$The equation above is sometimes referred as commutativity of addition. With the definition of addition, it is also easy to show that for x,y,z∈Fn${\displaystyle x,y,z\in {\mathcal{F}}^n}$, (6)x+(y+z)=(x+y)+z,$$x+(y+z)=(x+y)+z,$$that is, associativity of addition. Elements in F${\displaystyle \mathcal{F}}$ are called scalars, a fancy way to call real or complex numbers, on which we can perform multiplication operation, i.e., (7)a×b=b×a for a,b∈F$$a\times b=b\times a\text{for}a,b\in \mathcal{F}$$ In vector space Fn${\displaystyle {\mathcal{F}}^n}$, we can define scalar multiplication as (8)a×x=(ax1,ax2,ax3,...,axn) for a∈F and x∈Fn$$a\times x=(a{x}_1,a{x}_2,a{x}_3,...,a{x}_n)\text{for}a\in \mathcal{F}\text{and}x\in {\mathcal{F}}^n$$Defining scalar multiplication and addition on Fn${\displaystyle {\mathcal{F}}^n}$ allow x∈Fn${\displaystyle x\in {\mathcal{F}}^n}$ to hold interesting properties. Before we list those properties, let's define the concept of null in Fn${\displaystyle {\mathcal{F}}^n}$. When n=1${\displaystyle n=1}$, we know null means the scalar "0". For n>1${\displaystyle n>1}$, we have similarly, (9) 0=(0,0,0,...,0)⏠⏣⏣⏣⏡⏣⏣⏣⏢n$$0=\underset{n}{\underbrace{(0,0,0,...,0)}}$$0${\displaystyle 0}$ is also referred as additive identity, as we shall now see the reason. The properties of any elementu,v,w∈Fn${\displaystyle u,v,w\in {\mathcal{F}}^n}$ are (10)a

additive identity:	u+0=u;
additive inverse:	∀u∈Fn,∃w∈Fn such that u+w=0;
commutativity of addition:	u+v=v+u
associativity of addition:	u+(v+w)=(u+v)+w
multiplicative identity:	1×u=u;
distributivity of scalar multiplication:	a(u+v)=au+av for a∈F.

$$\begin{array}{rl}\text{additive identity: }& u+0=u;\\ \text{additive inverse:}& \forall u\in {\mathcal{F}}^n,\exists w\in {\mathcal{F}}^n\text{such that}u+w=0;\\ \text{commutativity of addition:}& u+v=v+u\\ \text{associativity of addition:}& u+(v+w)=(u+v)+w\\ \text{multiplicative identity:}& 1\times u=u;\\ \text{distributivity of scalar multiplication:}& a(u+v)=au+av\text{for}a\in \mathcal{F}.\end{array}$$ Note that the additive inverse of u${\displaystyle u}$ is -u=(-1)×u${\displaystyle -u=(-1)\times u}$, a scalar multiplication. It should be clear now that all the properties of elements in vector spaces are derived from the definition of addition and scalar multiplication. With this observation, we now give the formal defintion of vector space(with constraints on its elements specified):

Definition1A vector space is a set V (e.g.Fn${\displaystyle {\mathcal{F}}^n}$) equipped with addition and scalar multiplication of its elements such that properties in Eqn.(10) are satisfied.

When writing out the properties in Eqn.(10), we need to specify that the scalar a∈F${\displaystyle a\in \mathcal{F}}$. Because of this, we say that the vector space V${\displaystyle V}$ is defined over F${\displaystyle \mathcal{F}}$. As an example, R3${\displaystyle {\mathbb{R}}^3}$ is a vector space defined over R${\displaystyle \mathbb{R}}$ as R3={(x1,x2,x3):xi∈R for i=1,2,3}.$${\mathbb{R}}^3=\{(x_1,x_2,x_3):x_i\in \mathbb{R}\text{ for }i=1,2,3\}.$$We should be fairly familiar with R3${\displaystyle {\mathbb{R}}^3}$ geometrically, as we use 3-tuples in R3${\displaystyle {\mathbb{R}}^3}$ to discribe points or vectors in 3D space. The geometrical representation of vector space is delibrately ignored here as the definitions are independent of the geometries. When the dimension of vector spaces is higher than 3, it's hard for us to visualize it using 3D geometries, but we can still perform algbra on them without any changes in our definitions of addition and scalar multiplication. 3. Subspaces

Definition2A subspace W${\displaystyle W}$ of a vector space V${\displaystyle V}$ is a subset of V${\displaystyle V}$, but is a vector space itself.

For example, {(x1,x2,0):x1,x2∈F}${\displaystyle \{(x_1,x_2,0):x_1,x_2\in \mathcal{F}\}}$ is a subspace of F3${\displaystyle {\mathcal{F}}^3}$(verify this as an exercise). A subset U${\displaystyle U}$ if vector space V${\displaystyle V}$ over F${\displaystyle \mathcal{F}}$ is a subspace if and only if U${\displaystyle U}$ satisfies the following conditions: (11)a

additive identity:	0∈U
closed under addition:	u+v∈U for u,v∈U
closed under scalar multiplication:	au∈U for a∈F, and u∈U

$$\begin{array}{rl}\text{additive identity: }& 0\in U\\ \text{closed under addition: }& u+v\in U\text{for }u,v\in U\\ \text{closed under scalar multiplication: }& au\in U\text{for }a\in \mathcal{F},\text{ and }u\in U\end{array}$$3.1. Sum of subspacesIn applications of QC in composite quantum system that has multiple loosely coupled subsystems, We use elements in different subspaceS of Hilbert space(a vector space) to describe quantum states of the subsystems. Thus, it is useful to learn the notion of the sum of subspaces.

Definition3Suppose U1,...,Um${\displaystyle U_1,...,U_m}$ are subspaces of a vector space V${\displaystyle V}$. The sum of the subspaces is denoted as U1+U2+...+Um${\displaystyle U_1+U_2+...+U_m}$, with the following definition: (12)U1+U2+...+Um={u1+u2+⋯+um:u1∈U1,...,um∈Um}$$U_1+U_2+...+U_m=\{u_1+u_2+\cdots +u_m:u_1\in U_1,...,u_m\in U_m\}$$

From the definition above, every element of U1+U2+...+Um${\displaystyle U_1+U_2+...+U_m}$ can be written as u1+u2+⋯+um.$$u_1+u_2+\cdots +u_m.$$If each element can be written in terms of {ui}${\displaystyle \{u_i\}}$ in only one way, then the sum U1+U2+...+Um${\displaystyle U_1+U_2+...+U_m}$ is said to be direct sum, denoted as U1⊕U2⊕...⊕Um${\displaystyle U_1\oplus U_2\oplus ...\oplus U_m}$.

Condition for U1+U2+...+Um${\displaystyle U_1+U_2+...+U_m}$ to be direct sum:U1+U2+...+Um${\displaystyle U_1+U_2+...+U_m}$ is a direct sum if and only if the only way to write additive identity 0${\displaystyle 0}$ as a sum u1+u2+⋯+um${\displaystyle u_1+u_2+\cdots +u_m}$, is by taking each ui=0${\displaystyle u_i=0}$.

Proof: (⇒${\displaystyle \Rightarrow }$) suppose U1+U2+...+Um${\displaystyle U_1+U_2+...+U_m}$ is a direct sum, then the definition of direct sum implies that 0${\displaystyle 0}$ is obtained by setting ui=0${\displaystyle u_i=0}$ for i=1,2,3,...,m${\displaystyle i=1,2,3,...,m}$. (⇐${\displaystyle \Leftarrow }$) Suppose the only way to write 0${\displaystyle 0}$ is by taking each ui=0${\displaystyle u_i=0}$. We can show U1+U2+...+Um${\displaystyle U_1+U_2+...+U_m}$ is a direct sum by first writting a element of the sum asv=u1+u2+⋯+um$$v=u_1+u_2+\cdots +u_m$$Suppose we also have another representation for v${\displaystyle v}$ asv=v1+v2+⋯+vm$$v=v_1+v_2+\cdots +v_m$$We then have0=(u1-v1)+(u2-v2)+⋯+(um-vm)$$0=(u_1-v_1)+(u_2-v_2)+\cdots +(u_m-v_m)$$Because ui-vi∈Ui${\displaystyle u_i-v_i\in U_i}$, and we supposed element from Ui${\displaystyle U_i}$ in 0${\displaystyle 0}$ can only be 0${\displaystyle 0}$, we have ui=vi${\displaystyle u_i=v_i}$, and v${\displaystyle v}$ has an unique representation.□

Theorem1Suppose U${\displaystyle U}$ and W${\displaystyle W}$ are subspaces of V${\displaystyle V}$. The U+W${\displaystyle U+W}$ is a direct sum if and only if U∩W={0}${\displaystyle U\cap W=\{0\}}$.

Proof:(⇒${\displaystyle \Rightarrow }$)First suppose that U+W${\displaystyle U+W}$ is a direct sum. By the definition, if v∈U${\displaystyle v\in U}$, and 0=v+(-v)${\displaystyle 0=v+(-v)}$, we have -v∈W${\displaystyle -v\in W}$. By the uniqueness of 0${\displaystyle 0}$ in a direct sum, we must have v=0${\displaystyle v=0}$. So U∩W={0}.${\displaystyle U\cap W=\{0\}.}$(⇐${\displaystyle \Leftarrow }$) Suppose U∩W={0}${\displaystyle U\cap W=\{0\}}$, We need to prove U+W${\displaystyle U+W}$ is a direct sum. Let v∈V${\displaystyle v\in V}$ and w∈W${\displaystyle w\in W}$, and v+w=0${\displaystyle v+w=0}$. Notice that v⏦∈W=-w⏠⏣⏣⏡⏣⏣⏢(-1)w∈W$$\underset{\in W}{\underbrace{v}}=\underset{(-1)w}{\underbrace{-w}}\in W$$So v∈U∩W${\displaystyle v\in U\cap W}$, and hence, v=0=w${\displaystyle v=0=w}$. By the condition for U1+U2+...+Um${\displaystyle U_1+U_2+...+U_m}$ to be direct sum, we have U+W${\displaystyle U+W}$ is a direct sum.4. Prove Theorems The following theorems are arranged in a ascending order of the difficulty of their proofs.

Theorem2The intersection of every collection of subspaces of V${\displaystyle V}$ is a subspace of V${\displaystyle V}$

Hint: Starting from two subspaces,U1${\displaystyle U_1}$ and U2${\displaystyle U_2}$. Let x,y∈U1∩U2${\displaystyle x,y\in U_1\cap U_2}$⇒x,y∈U1 and x+y∈U1${\displaystyle \Rightarrow x,y\in U_{1}andx+y\in U_1}$, similarly, x+y∈U2${\displaystyle x+y\in U_2}$. So x+y=U1∩U2${\displaystyle x+y=U_1\cap U_2}$(closed under addition). Similar procedure for proving "closed under scalar multiplication".

Theorem3The union of two subspaces of V${\displaystyle V}$ is a subspace of V${\displaystyle V}$ if and only if one of the subpsaces is contained in the other.

Hint: prove by contradiction. Let v1∈U1${\displaystyle v_1\in U_1}$and v2∈U2${\displaystyle v_2\in U_2}$, we can have v1+v2∈U1 or U2 or U1∪U2${\displaystyle v_1+v_2\in U_{1}or{U}_{2}or{U}_1\cup U_2}$

Theorem4The union of three subspaces of V is a subspace of V${\displaystyle V}$ if and only if one othe subspace contains the other two.

Hint: this is quite challenging. You need to construct two scalars, a${\displaystyle a}$ and b∈F${\displaystyle b\in \mathcal{F}}$, such that a-b=1${\displaystyle a-b=1}$. You might find a quite nice solution at Math StackExchange. 5. Juice for Me Writing note like this is enjoyable for me but time-consuming at the same time. The only time I can find for this is my weekends. Thus, my next note might take a while to brew, stay tuned.