Linear algebra knowledge review, with proofs (1)

20 Mar, 2026

The following exercises are from Chapter 1-2 of Quantum Computation and Quantum Information by Isaac Chuang & Michael Nielsen. A lot of linear algebra knowledge, but then again, reviewing them again really made me appreciate the elegance of maths and proofs. So I should probably continue onwards in the next few days.

Rants: Don't like how limited MathML can be — they don't even support commands from the physics package. Had to manually replace those \ket{v} with |v\rangle for them to render properly. And they are still rendered pretty badly. Anyway!

Important notations:

• Complex conjugate: $z^{*}$, and $A^{*}$.
• Transpose: $A^T$.
• Hermitian conjugate / adjoint: $A^{†}\equiv {\left(A^T\right)}^{*}$, and $⟨v|\equiv |v{⟩}^{†}$.
  • This operator is anti-linear: ${\displaystyle {\left({\sum }_i{a}_i{A}_i\right)}^{†}={\sum }_i{a}_i^{*}{A}_i^{†}}$.
• Inner product: $⟨\phi \mid \psi ⟩$, and $⟨\phi \mid A\mid \psi ⟩$.
  • Positive definiteness: $⟨v\mid v⟩\ge 0$, with equality if and only if $|v⟩=0$.
  • Well-defined: conjugate symmetry, linearity in the second argument, and positive definiteness.
• Outer product: $|\phi ⟩⟨\psi |$.
• Tensor product: $|\phi ⟩\otimes |\psi ⟩$, abbreviated as $|\phi ⟩|\psi ⟩$.
  • Basic properties: commutative, associative, and distributive.
  • Well-defined: If $A,B$ are linear operators on $V,W$ respectively, then $A\otimes B$ is a well-defined linear operator on $V\otimes W$.
  • Well-defined inner product of tensor products: For any element $|v_i⟩\otimes |w_i⟩$ in $V\otimes W$, i.e., $v_i\in V$, $w_i\in W$, and another element $|v_j⟩\otimes |w_j⟩$ in $V\otimes W$, i.e., $v_j\in V$, $w_j\in W$, define $⟨v_i\otimes w_i\mid v_j\otimes w_j⟩\equiv ⟨v_i\mid v_j⟩⟨w_i\mid w_j⟩$, which satisfies the three rules of a well-defined inner product.
  • $|\psi {⟩}^{\otimes k}\equiv \underset{k\text{ times}}{\underbrace{|\psi ⟩\otimes \cdots \otimes |\psi ⟩}}$.
• Trace: ${\displaystyle tr(A)\equiv {\sum }_i{A}_{ii}}$.
  • Basic properties: cyclic, and linearity.
  • ${\displaystyle tr(A|\psi ⟩⟨\psi |)={\sum }_i⟨i\mid A\mid \psi ⟩⟨\psi \mid i⟩=⟨\psi \mid A\mid \psi ⟩}$.
• Commutator: $[A,B]\equiv AB-BA$.
• Anti-commutator: $\{A,B\}\equiv AB+BA$.
  • ${\displaystyle \frac{[A,B]+\{A,B\}}{2}}={\displaystyle \frac{AB-BA+AB+BA}{2}}=AB$.
  • $[A,B{]}^{†}=(AB-BA{)}^{†}=B^{†}{A}^{†}-A^{†}{B}^{†}=(BA-AB)=[B,A{]}^{†}$.
  • $[A,B]=AB-BA=-(BA-AB)=-[B,A]$.

Exercise 2.47: Suppose $A$ and $B$ are Hermitian. Show that $i[A,B]$ is Hermitian.

Proof: $(i[A,B]{)}^{†}=(iAB-iBA{)}^{†}$

$=-i{B}^{†}{A}^{†}+i{A}^{†}{B}^{†}$

$=-iBA+iAB$

$=iAB-iBA=i[A,B]$. QED.

Gram–Schmidt procedure: a procedure to produce an orthonormal basis set. Suppose $|w_1⟩,\dots ,|w_d⟩$ is a basis set, then define

$|v_1⟩\equiv {\displaystyle \frac{|w_1⟩}{\Vert |w_1⟩\Vert }}$,

$|v_{k+1}⟩\equiv {\displaystyle \frac{|w_{k+1}⟩-{\sum }_{i=1}^k⟨v_i\mid w_{k+1}⟩|v_i⟩}{\Vert |w_{k+1}⟩-{\sum }_{i=1}^k⟨v_i\mid w_{k+1}⟩|v_i⟩\Vert }}$, where $1\le k\le d-1$.

Proof by induction:

When $k=1$, $|v_1⟩\equiv {\displaystyle \frac{|w_1⟩}{\Vert |w_1⟩\Vert }}$ is normalized, therefore is a unit vector with $⟨v_1\mid v_1⟩=1$. $\{|v_1⟩\}$ is orthonormal.

When $k>1$, assume $\{|v_1⟩,\dots ,|v_k⟩\}$ are orthonormal.

Let ${\displaystyle |v_{k+1}^{\prime }⟩\equiv |w_{k+1}⟩-{\sum }_{i=1}^k⟨v_i\mid w_{k+1}⟩|v_i⟩}$,

For $j\le k$ we have ${\displaystyle ⟨v_j\mid v_{k+1}^{\prime }⟩=⟨v_j\mid w_{k+1}⟩-{\sum }_{i=1}^k⟨v_i\mid w_{k+1}⟩\underset{{\delta }_{ij}}{\underbrace{⟨v_j\mid v_i⟩}}}$

${\displaystyle =⟨v_j\mid w_{k+1}⟩-⟨v_j\mid w_{k+1}⟩=0}$.

By the way $\Vert |v_{k+1}^{\prime }⟩\Vert \ne 0$. By contradiction, if $\Vert |v_{k+1}^{\prime }⟩\Vert =0$, then ${\displaystyle |w_{k+1}⟩={\sum }_{i=1}^k⟨v_i\mid w_{k+1}⟩|v_i⟩}$ meaning $|w_{k+1}⟩$ is in $span\{|v_1⟩,\dots ,|v_k⟩\}=span\{|w_1⟩,\dots ,|w_k⟩\}$, contradicting that $\{|w_1⟩,\dots ,|w_k⟩,|w_{k+1}⟩\}$ is a basis set.

Therefore, by normalization we get $|v_{k+1}⟩\equiv {\displaystyle \frac{|v_{k+1}^{\prime }⟩}{\Vert |v_{k+1}^{\prime }⟩\Vert }}$ which is a unit vector that satisfies $⟨v_j\mid v_{k+1}⟩={\delta }_{j,k+1}$. $\{|v_1⟩,\dots ,|v_k⟩,|v_{k+1}⟩\}$ are orthonormal. QED.

Pauli matrices:

${\sigma }_0\equiv I\equiv \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$,

${\sigma }_1\equiv {\sigma }_x\equiv X\equiv \left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$,

${\sigma }_2\equiv {\sigma }_y\equiv Y\equiv \left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]$,

${\sigma }_3\equiv {\sigma }_z\equiv Z\equiv \left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$.

Exercise 2.9: (Pauli operators and the outer product) The Pauli matrices can be considered as operators with respect to an orthonormal basis $|0⟩,|1⟩$ for a two-dimensional Hilbert space. Express each of the Pauli operators in the outer product notation.

With the knowledge of

$|0⟩⟨0|=\left[\begin{array}{c}1\\ 0\end{array}\right]\left[\begin{array}{cc}1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$,

$|0⟩⟨1|=\left[\begin{array}{c}1\\ 0\end{array}\right]\left[\begin{array}{cc}0& 1\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$,

$|1⟩⟨0|=\left[\begin{array}{c}0\\ 1\end{array}\right]\left[\begin{array}{cc}1& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$,

$|1⟩⟨1|=\left[\begin{array}{c}0\\ 1\end{array}\right]\left[\begin{array}{cc}0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$,

We can write

$I=|0⟩⟨0|+|1⟩⟨1|$ ,

$X=|0⟩⟨1|+|1⟩⟨0|$,

$Y=-i|0⟩⟨1|+i|1⟩⟨0|$,

$Z=|0⟩⟨0|-|1⟩⟨1|$ .

Exercise 2.10: Suppose $|v_i⟩$ is an orthonormal basis for an inner product space $V$. What is the matrix representation for the operator $|v_j⟩⟨v_k|$, with respect to the $|v_i⟩$ basis?

Similar to Exercise 2.9, we can see that $|v_j⟩⟨v_k|$ is an array whose element at row $j$, column $k$ is 1 and 0 otherwise.

Commutation relations for the Pauli matrices:

$XY=(|0⟩⟨1|+|1⟩⟨0|)(-i|0⟩⟨1|+i|1⟩⟨0|)$

$=i|0⟩⟨1\mid 1⟩⟨0|-i|1⟩⟨0\mid 0⟩⟨1|$

$=i|0⟩⟨0|-i|1⟩⟨1|=iZ$,

$YX=(-i|0⟩⟨1|+i|1⟩⟨0|)(|0⟩⟨1|+|1⟩⟨0|)$

$=-i|0⟩⟨1\mid 1⟩⟨0|+i|1⟩⟨0\mid 0⟩⟨1|$

$=-i|0⟩⟨0|+i|1⟩⟨1|=-iZ$,

Therefore $[X,Y]=XY-YX=2iZ$.

Similarly, $[Y,Z]=YZ-ZY=2iX$,

$[Z,X]=ZX-XZ=2iY$.

Anti-commutation relations for the Pauli matrices:

$\{X,Y\}=XY+YX=0$,

Similarly, $\{Y,Z\}=\{Z,X\}=0$.

$X^2=XX=(|0⟩⟨1|+|1⟩⟨0|)(|0⟩⟨1|+|1⟩⟨0|)$

$=|0⟩⟨0|+|1⟩⟨1|=I$.

Similarly, $Y^2=Z^2=I$.

A diagonal representation for an operator $A$ is a representation $A={\sum }_i{\lambda }_i|i⟩⟨i|$ where the vectors $|i⟩$ form an orthonormal set of eigenvectors for $A$ with corresponding eigenvalues ${\lambda }_i$.

Common one-qubit states:

$|+⟩={\displaystyle \frac{1}{\sqrt{2}}}|0⟩+{\displaystyle \frac{1}{\sqrt{2}}}|1⟩={\displaystyle \frac{1}{\sqrt{2}}}[\begin{array}{c}1\\ 1\end{array}]$,

$|-⟩={\displaystyle \frac{1}{\sqrt{2}}}|0⟩-{\displaystyle \frac{1}{\sqrt{2}}}|1⟩={\displaystyle \frac{1}{\sqrt{2}}}[\begin{array}{c}1\\ -1\end{array}]$,

$|+i⟩={\displaystyle \frac{1}{\sqrt{2}}}|0⟩+{\displaystyle \frac{i}{\sqrt{2}}}|1⟩={\displaystyle \frac{1}{\sqrt{2}}}[\begin{array}{c}1\\ i\end{array}]$,

$|-i⟩={\displaystyle \frac{1}{\sqrt{2}}}|0⟩-{\displaystyle \frac{i}{\sqrt{2}}}|1⟩={\displaystyle \frac{1}{\sqrt{2}}}[\begin{array}{c}1\\ -i\end{array}]$.

Exercise 2.11: (Eigendecomposition of the Pauli matrices) Find the eigenvectors, eigenvalues, and diagonal representations of the Pauli matrices $X,Y$ , and $Z$.

For Pauli $X$:

• Eigenvalues: $det(X-\lambda I)=0$

  $⟹det\left[\begin{array}{cc}-\lambda & 1\\ 1& -\lambda \end{array}\right]=0$

  $⟹{\lambda }^2-1=0⟹\lambda =\pm 1$.

• Eigenvectors: When $\lambda =1$, $(X-I)|v⟩=0$

  $⟹|v⟩={\displaystyle \frac{1}{\sqrt{2}}}[\begin{array}{c}1\\ 1\end{array}]\equiv |+⟩$.

  When $\lambda =-1$, eigenvector is $|-⟩$.

• Diagonal representation: $X=|+⟩⟨+|-|-⟩⟨-|$.

For Pauli $Y$:

• Eigenvalues: $\lambda =\pm 1$.

• Eigenvectors: When $\lambda =1$, eigenvector is $|+i⟩$.

  When $\lambda =-1$, eigenvector is $|-i⟩$.

• Diagonal representation: $Y=|+i⟩⟨+i|-|-i⟩⟨-i|$.

For Pauli $Z$:

• Eigenvalues: $\lambda =\pm 1$.

• Eigenvectors: When $\lambda =1$, eigenvector is $|0⟩$.

  When $\lambda =-1$, eigenvector is $|1⟩$.

• Diagonal representation: $Z=|0⟩⟨0|-|1⟩⟨1|$.

#Math notes