aHappy Thanksgiving! This note and the next one contain possibly the most relevant concepts and theorems in the context of quantum computing(QC) and quantum information(QI) theory. Not much argument of QC and QI is made here as a decent discussion of operators used in QC/QI research can not be justified in a short note. Therefore, current note still serves as a brief reference to important concepts and theorems. Because the difference between real and complex vector space does matter here, we will specify the type of vector space if a theorem is applicable only to complex (Cn${\displaystyle {\mathbb{C}}^n}$) or real vector spaces (Rn${\displaystyle {\mathbb{R}}^n}$).1. Self-adjoint (Hermitian) and Normal Operators 2. The Spectral Theorem 3. Positive Operators and Isometries 4. Polar Decomposition and Singular Value Decomposition 5. How the notes are made? 1. Self-adjoint (Hermitian) and Normal OperatorsWe start by making an analogy between operators and complex numbers. For each complex number z∈C${\displaystyle z\in \mathbb{C}}$, we have its conjugate ⏨z${\displaystyle \overline{z}}$ such that z⏨z=||z||2${\displaystyle z\overline{z}=||z|{|}^2}$. Also, z/||z||${\displaystyle z/||z||}$ is said to be the normalized complex number living on the unit circle on the complex plane. If each complex number is analogous to an operator, then we shall see in the later discussions that we can define "conjugate" and "normalized" operators, corresponding to the concept of ⏨z${\displaystyle \overline{z}}$ and z/||z||${\displaystyle z/||z||}$, perspectively. But why do we care this analogy? One of shallow reasons would be to make the study of operators complete. With analogous concepts defined we can manipulate operators in a way similar to vectors in the spaces of our interest."Adjoint" of a linear map is analogous to the concept of "conjugacy", and it is defined as the following:

Definition1Suppose v∈V${\displaystyle v\in V}$, w∈W${\displaystyle w\in W}$, and T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$, then the adjoint of T${\displaystyle T}$, T†∈L(W,V)${\displaystyle T^{†}\in \mathcal{L}(W,V)}$, is an operator satifying the following condition⟨Tv,w⟩=⟨v,T†w⟩$$⟨Tv,w⟩=⟨v,T^{†}w⟩$$for every v${\displaystyle v}$ and w${\displaystyle w}$.

To make sense out of the definition, we first notice that ⟨Tv,w⟩${\displaystyle ⟨Tv,w⟩}$ is a linear functional for all v∈V${\displaystyle v\in V}$. According to Riesz representation theorem, we can transform ⟨Tv,w⟩${\displaystyle ⟨Tv,w⟩}$ into a format of ⟨v,u⟩${\displaystyle ⟨v,u⟩}$ with u∈V${\displaystyle u\in V}$ being an unique vector. Replacing T†w${\displaystyle T^{†}w}$ with u${\displaystyle u}$, we recover the condition in Definition 1. A natural follow-up question would be: how do matrices of T${\displaystyle T}$ and T†${\displaystyle T^{†}}$ are related? The next theorem states that the two matrices are conjugate transposes to each other. To get the conjugate transpose of a matrix we simply transpose it and replace each entry with corresponding complex conjugate.

Theorem1let T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$ and let orthonormal bases of V${\displaystyle V}$ and W${\displaystyle W}$ be (e1,...,en)${\displaystyle (e_1,...,e_n)}$ and (f1,...,fm)${\displaystyle (f_1,...,f_m)}$, respectively. Then M(T)${\displaystyle \mathcal{M}(T)}$ with respect to the two bases is the conjugate transpose of M(T†)${\displaystyle \mathcal{M}\left(T^{†}\right)}$.

Proof: M(T)${\displaystyle \mathcal{M}(T)}$ maps ei${\displaystyle e_i}$ to ⟨Tei,f1⟩f1+⋯+⟨Tei,fm⟩fm${\displaystyle ⟨T{e}_i,f_1⟩f_1+\cdots +⟨T{e}_i,f_m⟩f_m}$ due to the definition of orthonormal basis. According to our previous discussion of matrix representation, the entry at j${\displaystyle j}$th row and i${\displaystyle i}$th column is ⟨Tei,fj⟩${\displaystyle ⟨T{e}_i,f_j⟩}$. On the other hand, M(T†)${\displaystyle \mathcal{M}\left(T^{†}\right)}$ maps fj${\displaystyle f_j}$ to ⟨T†fj,e1⟩e1+⋯+⟨T†fj,en⟩en${\displaystyle ⟨T^{†}{f}_j,e_1⟩e_1+\cdots +⟨T^{†}{f}_j,e_n⟩e_n}$. So the entry at i${\displaystyle i}$th row and j${\displaystyle j}$th column is ⟨T†fj,ei⟩${\displaystyle ⟨T^{†}{f}_j,e_i⟩}$. Because ⟨Tei,fj⟩=⟨ei,T†fj⟩=⏨⏨⏨⏨⟨T†fj,ei⟩${\displaystyle ⟨T{e}_i,f_j⟩=⟨e_i,T^{†}{f}_j⟩=\overline{⟨T^{†}{f}_j,e_i⟩}}$ for each entry of M(T)${\displaystyle \mathcal{M}(T)}$, we have M(T†)${\displaystyle \mathcal{M}\left(T^{†}\right)}$ being conjugate transpose of M(T)${\displaystyle \mathcal{M}(T)}$.□In the next box we list properties of the adjoint, and note that inner product is defined on the vector spaces mentioned below. We will prove (3) and (5) as they are common exercises given in introductory quantum mechanics class.

Proposition1 (1)(S+T)†=S†+T†, ∀S,T∈L(V,W)(2)(𝜆T)†=⏨𝜆T†(3)(T†)†=T(4)I†=I(5)(ST)†=T†S†, ∀S∈L(W,U) and ∀T∈L(V,W)$$\begin{array}{c}(S+T{)}^{†}=S^{†}+T^{†},\forall S,T\in \mathcal{L}(V,W)\\ (𝜆T{)}^{†}=\overline{𝜆}{T}^{†}\\ (T^{†}{)}^{†}=T\\ I^{†}=I\\ (ST{)}^{†}=T^{†}{S}^{†},\forall S\in \mathcal{L}(W,U)\text{ and }\forall T\in \mathcal{L}(V,W)\end{array}$$

Proof: To show (3), suppose T∈L(V,W)${\displaystyle T\in \mathcal{L}(V,W)}$, we have ⟨w,(T†)†v⟩=⟨T†w,v⟩=⟨w,Tv⟩${\displaystyle ⟨w,(T^{†}{)}^{†}v⟩=⟨T^{†}w,v⟩=⟨w,Tv⟩}$ because of Definition 1 for any v∈V${\displaystyle v\in V}$ and w∈W${\displaystyle w\in W}$. And the equality implies that (T†)†=T${\displaystyle (T^{†}{)}^{†}=T}$ as desired. To show (5) we have ⟨v,(ST)†u⟩=⟨(ST)v,u⟩=⟨Tv,S†u⟩=⟨v,T†S†u⟩$$⟨v,(ST{)}^{†}u⟩=⟨(ST)v,u⟩=⟨Tv,S^{†}u⟩=⟨v,T^{†}{S}^{†}u⟩$$which implies ${\displaystyle }$(ST)†=T†S†${\displaystyle (ST{)}^{†}=T^{†}{S}^{†}}$ for u∈U${\displaystyle u\in U}$ and v∈V${\displaystyle v\in V}$ as desired.□Theorem 1 and Proposition 1 apply to general linear maps, including operators. When T∈L(V)${\displaystyle T\in \mathcal{L}(V)}$, M(T)${\displaystyle \mathcal{M}(T)}$ and M(T†)${\displaystyle \mathcal{M}\left(T^{†}\right)}$ are two square matrices, making it possible to have M(T)=M(T†)${\displaystyle \mathcal{M}(T)=\mathcal{M}\left(T^{†}\right)}$. Recall that M${\displaystyle \mathcal{M}}$ is an isomorphism between L(V)${\displaystyle \mathcal{L}(V)}$ and Fn,n${\displaystyle {\mathcal{F}}^{n,n}}$, so this equality also implies T=T†${\displaystyle T=T^{†}}$, which gives the definition of self-adjoint operator.

Definition2Let T∈L(V)${\displaystyle T\in \mathcal{L}(V)}$, and it is a self-adjoint operator if T=T†${\displaystyle T=T^{†}}$. In the language of inner product, an operator is self-adjoint if and only if⟨Tv,w⟩=⟨v,Tw⟩$$⟨Tv,w⟩=⟨v,Tw⟩$$for every v,w∈V${\displaystyle v,w\in V}$.

From Definition 2, it should be easy to prove that self-adjoint operators have the following properties:

Proposition2Suppose operators T${\displaystyle T}$ and S${\displaystyle S}$ are self-adjoint, then the following equalities hold⟨(T+S)v,w⟩=⟨v,(T+S)w⟩⟨𝜆Tv,w⟩=⟨v,𝜆Tw⟩ for 𝜆∈R$$\begin{array}{c}⟨(T+S)v,w⟩=⟨v,(T+S)w⟩\\ ⟨𝜆Tv,w⟩=⟨v,𝜆Tw⟩\text{for}𝜆\in \mathbb{R}\end{array}$$

Going back our analogy at the beginning, the self-adjoint operators are analogous to real number in C.${\displaystyle \mathbb{C}.}$We note here that self-adjoint operators in the literature of physics are referred as Hermitian operators. In quantum mechanics, it is postulated that the energy of a physical system can be calculated by applying corresponding energy operator to wavefunction(a vector in Hilbert space). Such an operator is called as Hamiltonian operator, and it must have real eigenvaluse because the energy is a measurable quantity. All the Hamiltonian operators are postulated to be Hermitian because of the following proposition:

Proposition3Every eigenvalue of a self-adjoint (Hermitian) operator is real.

Proof: Let v∈V${\displaystyle v\in V}$ be an eigenvector of self-joint T∈M(V)${\displaystyle T\in \mathcal{M}(V)}$. Suppose Tv=𝜆v${\displaystyle Tv=𝜆v}$, then ⟨Tv,v⟩=𝜆||v||2=⟨v,Tv⟩=⏨𝜆||v||2$$⟨Tv,v⟩=𝜆||v|{|}^2=⟨v,Tv⟩=\overline{𝜆}||v|{|}^2$$which implies 𝜆=⏨𝜆${\displaystyle 𝜆=\overline{𝜆}}$, i.e., 𝜆${\displaystyle 𝜆}$ must be real number.To check if an operator is self-adjoint, the following theorem is of some use, which is only applicable to complex vector spaces.

Theorem2Suppose V${\displaystyle V}$ is a complex inner product space and T∈L(V)${\displaystyle T\in \mathcal{L}(V)}$. Then T${\displaystyle T}$ is self-adjoint if and only if⟨Tv,v⟩∈R$$⟨Tv,v⟩\in \mathbb{R}$$for every v∈V${\displaystyle v\in V}$.

For a vector space V${\displaystyle V}$, the self-adjoint operators compose of subspace of L(V)${\displaystyle \mathcal{L}(V)}$, and it is encompassed by another subspace of L(V)${\displaystyle \mathcal{L}(V)}$ made of normal operators whose definition is the following:

Definition3T∈L(V)${\displaystyle T\in \mathcal{L}(V)}$ is normal operator if TT†=T†T${\displaystyle T{T}^{†}=T^{†}T}$.

Apparently, self-adjoint operators must be normal. The following result provides a way to characterize normal operators:

Proposition4T∈L(V)${\displaystyle T\in \mathcal{L}(V)}$ is normal if and only if ||Tv||=||T†v||${\displaystyle ||Tv||=||T^{†}v||}$ for all v∈V${\displaystyle v\in V}$.

Being normal also gives interesting results regarding eigenvectors as shown below:

Theorem3Suppose T∈L(V)${\displaystyle T\in \mathcal{L}(V)}$ is normal and v∈V${\displaystyle v\in V}$ is an eigenvector of T${\displaystyle T}$ with eigenvalue 𝜆${\displaystyle 𝜆}$. Then v${\displaystyle v}$ is also an eigenvector of T†${\displaystyle T^{†}}$ with eigenvalue ⏨𝜆${\displaystyle \overline{𝜆}}$.

Proof: To see this, notice that T-𝜆I${\displaystyle T-𝜆I}$ is normal when T${\displaystyle T}$ is normal. Because Tv=𝜆v${\displaystyle Tv=𝜆v}$, we have||(T-𝜆I)v||=0=||(T-𝜆I)†v||=||(T†-⏨𝜆I)v||$$||(T-𝜆I)v||=0=||(T-𝜆I{)}^{†}v||=||(T^{†}-\overline{𝜆}I)v||$$which implies that v${\displaystyle v}$ is also an eigenvector of T†${\displaystyle T^{†}}$ with associated eigenvalue ⏨𝜆${\displaystyle \overline{𝜆}}$. □Also, eigenvectors of normal operators are orthogonal as shown in the following result.

Theorem4Suppose T∈L(V)${\displaystyle T\in \mathcal{L}(V)}$ is normal. Then eigenvectors of T${\displaystyle T}$ corresponding distinct eigenvalues are orthogonal.

Proof: Let v,u∈V${\displaystyle v,u\in V}$, and Tv=𝛼v, Tu=𝛽u${\displaystyle Tv=𝛼v,Tu=𝛽u}$. We have(𝛼-𝛽)⟨v,u⟩=⟨𝛼v,u⟩-⟨v,⏨𝛽u⟩=⟨Tv,u⟩-⟨v,T†u⟩=⟨Tv,u⟩-⟨Tv,u⟩=0$$(𝛼-𝛽)⟨v,u⟩=⟨𝛼v,u⟩-⟨v,\overline{𝛽}u⟩=⟨Tv,u⟩-⟨v,T^{†}u⟩=⟨Tv,u⟩-⟨Tv,u⟩=0$$where the second equality comes from Theorem 3. Because 𝛼≠𝛽${\displaystyle 𝛼\ne 𝛽}$, we must have ⟨v,u⟩=0${\displaystyle ⟨v,u⟩=0}$.□In previous notes, we already know that finite-dimensional complex spaces have eigenvalues, and we have not characterized eigenvalues of operators on real vector spaces. The following theorem provides a fact that as long as an operator is self-adjoint, no matter it is on real or complex vector space, it has eigenvalues.

Theorem5Suppose V≠{0}$V\ne \{0\}$ and T∈L(V)$T\in \mathcal{L}(V)$ is a self-adjoint operator. Then T$T$ has an eigenvalue.

2. The Spectral TheoremFor a given operator, the spectral theorem decides whether it is diagonalizable w.r.t orthonormal basis. On real vector spaces, such a condition is satisfied when the operator is self-adjoint as shown below.

Theorem6Suppose F=R$\mathcal{F}=\mathbb{R}$ and T∈L(V)$T\in \mathcal{L}(V)$. Then the following are equivalent:(a) T$T$ is self-adjoint.(b) V$V$ has an orthonormal basis consisting of eigenvectors of T$T$.(c) T$T$ has a diagonal matrix with respect to some orthonormal basis of V$V$.

We are not goint to prove the theorem but a proof might require the following lemma

Lemma1Suppose T∈L(V)$T\in \mathcal{L}(V)$ is self-adjoint and b,c∈R$b,c\in \mathbf{R}$ are such that b2<4c$b^2<4c$. ThenT2+bT+cI$$T^2+bT+cI$$is invertible.

Proof: we want to show that T2+bT+cI${\displaystyle T^2+bT+cI}$ is injective as V${\displaystyle V}$ is finite-dimensional. For nonzero v∈V${\displaystyle v\in V}$, we can expand the following⟨(T2+bT+cI)v,v⟩=c||v||2+||Tv||2+b⟨Tv,v⟩.$$⟨(T^2+bT+cI)v,v⟩=c||v|{|}^2+||Tv|{|}^2+b⟨Tv,v⟩.$$Because b⟨Tv,v⟩≤|b||⟨Tv,v⟩|≤|b|||Tv||||v||${\displaystyle b⟨Tv,v⟩\le |b||⟨Tv,v⟩|\le |b|||Tv||||v||}$, we must have⟨(T2+bT+cI)v,v⟩≥c||v||2+||Tv||2-|b|||Tv||||v||=(||v‖-|b|||v‖

2)2+(c-b2

4)||v‖2>0.$$⟨(T^2+bT+cI)v,v⟩\ge c||v|{|}^2+||Tv|{|}^2-|b|||Tv||||v||=(||v\Vert -\frac{|b|||v\Vert }{2}{)}^2+(c-\frac{b^2}{4})||v{\Vert }^2>0.$$Thus, null (T2+bT+cI)={0}${\displaystyle null(T^2+bT+cI)=\{0\}}$ as desired.□When F=C${\displaystyle \mathcal{F}=\mathbb{C}}$ and V${\displaystyle V}$ is complex vector space we have the complex spectral theorem.

Theorem7Suppose F=C$\mathcal{F}=\mathbb{C}$ and T∈L(V)$T\in \mathcal{L}(V)$. Then the following are equivalent:(a) T$T$ is normal.(b) V$V$ has an orthonormal basis consisting of eigenvectors of T$T$.(c) T$T$ has a diagonal matrix with respect to some orthonormal basis of V$V$.

3. Positive Operators and IsometriesWe have known that self-adjoint operators are composed of a subset of normal operators, but is there a famous subset to self-adjoint operators? Yes! And one of famous subset comprises positive operators defined as the following:

Definition4An operator T∈L(V)$T\in \mathcal{L}(V)$ is called positive if T$T$ is self-adjoint and⟨Tv,v⟩≥0$$⟨Tv,v⟩\ge 0$$for all v∈V$v\in V$.

In the definition above, we can drop the contraint of being self-adjoint for operators on complex spaces because of Theorem 2. Before we proceed to introduce the characterization of positive operators, we need to first define the concept of "square root" of a operator.

Definition5An operator R$R$ is called a square root of an operator T$T$ if R2=T$R^2=T$.

The set of propositions listed below are equivalent when T∈L(V)${\displaystyle T\in \mathcal{L}(V)}$ is positive operator, and we will prove (c) using the spectral theorem introduced above.

Theorem8Let T∈L(V)$T\in \mathcal{L}(V)$. Then the following are equivalent:(a) T$T$ is positive;(b) T$T$ is self-adjoint and all the eigenvalues of T$T$ are nonnegative;(c) T$T$ has a positive square root;(d) T$T$ has a self-adjoint square root;(e) there exists an operator R∈L(V)$R\in \mathcal{L}(V)$ such that T=R†R$T=R^{†}R$.

Proof for (c): Because (b) holds, we have an orthonormal basis made of eigenvectors of T${\displaystyle T}$ due to Theorem 6 and Theorem 7. Let the basis be e1,...en${\displaystyle e_1,...e_n}$, with corresponding eigenvalues being 𝜆1,....,𝜆n${\displaystyle {𝜆}_1,....,{𝜆}_n}$. Since 𝜆j${\displaystyle {𝜆}_j}$ is nonnegative. Let R${\displaystyle R}$ be a linear map from V${\displaystyle V}$ to V${\displaystyle V}$ such that Rej=𝜆jej${\displaystyle R{e}_j=\sqrt{{𝜆}_j}{e}_j}$, resulting⟨Rv,v⟩=<R(∑⟨v,ei⟩ei),(∑⟨v,ei⟩ei)>=∑|⟨v,ei⟩|2𝜆i >0.$$⟨Rv,v⟩=⟨R(\sum ⟨v,e_i⟩e_i),(\sum ⟨v,e_i⟩e_i)⟩=\sum |⟨v,e_i⟩{|}^2\sqrt{{𝜆}_i}>0.$$So R${\displaystyle R}$ is positive, and it is obvious that R2ej=𝜆jej=Tej${\displaystyle R^2{e}_j={𝜆}_j{e}_j=T{e}_j}$.□A positive operator can have infinite number of square roots, but only one of them is positive. In other words,

Proposition5Every positive operator has a unique positive square root.

We end this section by introducing another important class of operators, i.e. isometry, as defined below:

Definition6An operator S∈L(V)$S\in \mathcal{L}(V)$ is called an isometry if||Sv||=||v||$$||Sv||=||v||$$for all v∈V$v\in V$. In other words, an operator is an isometry if it preserves norms.

Like what we did for positive operator, the characterization of isometry is listed below. The equivalence of (a) and (c) [or (d)] shows that an operator is an isometry if and only if the list of columns of its matrix with respect to every [or some] basis is orthonormal.

Theorem9Suppose S∈L(V)$S\in \mathcal{L}(V)$. Then the following are equivalent:(a) S is an isometry;(b) ⟨Su,Sv⟩=⟨u,v⟩$⟨Su,Sv⟩=⟨u,v⟩$ for all u,v∈V$u,v\in V$(c) Se1,…,Sen$S{e}_1,\dots ,S{e}_n$ is orthonormal for every orthonormal list of vectors e1,…,en$e_1,\dots ,e_n$ in V$V$(d) there exists an orthonormal basis e1,…,en$e_1,\dots ,e_n$ of V$V$ such that Se1,…,Sen$S{e}_1,\dots ,S{e}_n$ is orthonormal;(e) S†S=I$S^{†}S=I$(f) SS†=I$S{S}^{†}=I$(g) S†$S^{†}$ is an isometry;(h) S$S$ is invertible and S-1=S†$S^{-1}=S^{†}$.

Isometry is closely related to unitary operators, the ones for describing behaviors of quantum gates. In Axler's book the author refers an isometry on a complex vector space as unitary operator, and orthogonal operator on a real vector sapce. Both isometry and unitary operators satisfying the condition of S†S=SS†=I${\displaystyle S^{†}S=S{S}^{†}=I}$. Accoding to literature of operators on Hilbert space, the major differences between the two are that a) unitary operators are defined on Hilbert space, H${\displaystyle H}$, and b) unitary operators are bounded linear operators. An operator S∈L(H)${\displaystyle S\in \mathcal{L}(H)}$ is bounded if and only if there exists some scalar M>0${\displaystyle M>0}$ such that for all v∈H${\displaystyle v\in H}$, ||Sv||≤M||v||${\displaystyle ||Sv||\le M||v||}$.(e) and (f) of Theorem 9 indicate that isometries are normal operators. By Theorem 7, we can show that eigenvalues of isometry have their absolute values being unity.

Theorem10Suppose V$V$ is a complex inner product space and S∈L(V)$S\in \mathcal{L}(V)$. Then the following are equivalent:(a) S$S$ is an isometry.(b) There is an orthonormal basis of V$V$ consisting of eigenvectors of S$S$ whose corresponding eigenvalues all have absolute value 1.

Proof: Suppose (a) holds, by Theorem 7 V${\displaystyle V}$ has an orthonormal basis made of eigenvector of S${\displaystyle S}$ as isometries are normal. Let e1,...,en${\displaystyle e_1,...,e_n}$ and 𝜆1,...,𝜆n${\displaystyle {𝜆}_1,...,{𝜆}_n}$ be the basis and associated eigenvalues, respectively. Because Sei=𝜆iei${\displaystyle S{e}_i={𝜆}_i{e}_i}$, we have⟨Sei,Sei⟩=⟨ei,ei⟩=|𝜆i|2=1$$⟨S{e}_i,S{e}_i⟩=⟨e_i,e_i⟩=|{𝜆}_i{|}^2=1$$due to (b) in Theorem 9.□4. Polar Decomposition and Singular Value DecompositionLet us review the analogy between operators and complex number proposed in Section 1. Each complex number z${\displaystyle z}$ can be written as z=(z

|z|)z⏨z.$$z=\left(\frac{z}{|z|}\right)\sqrt{z\overline{z}}.$$We would expect that on a vector space there should be two classes of operators analogous to z/|z|${\displaystyle z/|z|}$ and z⏨z${\displaystyle \sqrt{z\overline{z}}}$, respectively. Indeed, the theorem of polar decomposition tells us that for an arbitrary operator T${\displaystyle T}$ there is an isometry serving as z/|z|${\displaystyle z/|z|}$ and T†T${\displaystyle \sqrt{T^{†}T}}$ as z⏨z${\displaystyle \sqrt{z\overline{z}}}$.

Theorem11Suppose T∈L(V)$T\in \mathcal{L}(V)$. Then there exists an isometry S∈L(V)$S\in \mathcal{L}(V)$ such thatT=ST†T$$T=S\sqrt{T^{†}T}$$

It is worth noting that ${\displaystyle \sqrt{}}$ here refers to the unique positive square root of T†T${\displaystyle T^{†}T}$ as implied in Proposition 5.When V${\displaystyle V}$ in Theorem 11 is a complex space, then S${\displaystyle S}$ is a diagonal matrix with respect to an orthonormal basis. Because T†T${\displaystyle \sqrt{T^{†}T}}$ is positive, it is also a diagonal matrix with respect to an orthonormal basis. We emphasize here that these two bases may NOT be the same.Before we introduce the singular value decomposition, let us define what singular values are:

Definition7Suppose T∈L(V)$T\in \mathcal{L}(V)$. The singular values of T$T$ are the eigenvalues of T†T$\sqrt{T^{†}T}$, with each eigenvalue 𝜆$𝜆$ repeated dimE(𝜆,T†T)$\mathrm{d}\mathrm{i}\mathrm{m}E(𝜆,\sqrt{T^{†}T})$ times.

The singular values must be nonnegative, because T†T${\displaystyle T^{†}T}$ is positive for any operator, and T†T$\sqrt{T^{†}T}$ is the positive square root. With this, the singular value decomposition tells us that matrix of every operator is diagonal with respect to two distinct sets of bases, and diagonal values are singular values.

Theorem12Suppose T∈L(V)$T\in \mathcal{L}(V)$ has singular values s1,…,sn$s_1,\dots ,s_n$. Then there exist orthonormal bases e1,…,en$e_1,\dots ,e_n$ and f1,…,fn$f_1,\dots ,f_n$ of V$V$ such thatTv=s1<v,e1>f1+⋯+sn<v,en>fn$$Tv=s_1⟨v,e_1⟩f_1+\cdots +s_n⟨v,e_n⟩f_n$$for every v∈V$v\in V$.

Calculating singular values is not a trivial task in the study of computational linear algebra. To do so, people first calculate M(T†T)${\displaystyle \mathcal{M}\left(T^{†}T\right)}$ and its approximate eigenvalues. The nonnegative square roots of these eigenvalues are approximate singular values because of Theorem 8.5. How the notes are made?A supervisor of undergrad classes once asked me (then a Ph.D. candidate) the reason for making class notes. I am not a professor, and perhaps will never be one (not a big fan of teaching). So writing notes is not getting prepared for my future students. Rather, I do enjoy the time I spent on ranting things out of my own logic. The facts in these notes are the same as the ones in standard texts but the flow of logic must be mine. To prepare each note, I start by reading a chapter or two from a (non-)standard text, after which the author would review the content again and write down an outline for developing a note. The outline might be modified and even abandoned during the writting process, as details of a story might sound more intuitive if it is presented in a particular sequence. While writing the author definitely goes back to the text to assure himself nothing was missed, but the challenge is to avoid checking textbooks as much as possible. In order to relate the notes to quantum computing and quantum information, so extra arguments of quantum mechanics are made whenever the author finds it fit the big picture. As the final disclaimer, the posted notes were not examined with sufficient care so typos might not be uncommon. The author apologizes for that in advance just in case someone find them.