Linear algebra knowledge review, with proofs (2)

20 Mar, 2026

The following exercises are from Chapter 2 of Quantum Computation and Quantum Information by Isaac Chuang & Michael Nielsen.

Cauchy–Schwarz inequality: ${|⟨v\mid w⟩|}^2\le ⟨v\mid v⟩⟨w\mid w⟩$ for any two vectors $|v⟩,|w⟩$.

Proof 1: Construct $|z⟩=|v⟩-\lambda |w⟩$, where $\lambda$ is any scalar in $\mathbf{C}$.

Then $0\le ⟨z\mid z⟩=⟨v\mid v⟩-{\lambda }^{*}⟨w\mid v⟩-\lambda ⟨v\mid w⟩+{\left|\lambda \right|}^2⟨w\mid w⟩$.

Let $\lambda ={\displaystyle \frac{⟨w\mid v⟩}{⟨w\mid w⟩}}$,

we get $0\le ⟨v\mid v⟩-{\displaystyle \frac{⟨v\mid w⟩⟨w\mid v⟩}{⟨w\mid w⟩}}-{\displaystyle \frac{⟨w\mid v⟩⟨v\mid w⟩}{⟨w\mid w⟩}}+{\displaystyle \frac{{|⟨w\mid v⟩|}^2}{{|⟨w\mid w⟩|}^2}}⟨w\mid w⟩$

$\le ⟨v\mid v⟩-{\displaystyle \frac{{|⟨v\mid w⟩|}^2}{⟨w\mid w⟩}}$,

therefore ${|⟨v\mid w⟩|}^2\le ⟨v\mid v⟩⟨w\mid w⟩$. QED.

Proof 2 (from the book): Construct an orthonormal basis $|i⟩$ such that the first member is ${\displaystyle \frac{|w⟩}{\sqrt{⟨w\mid w⟩}}}$.

Using the completeness relation ${\displaystyle {\sum }_i|i⟩⟨i|=I}$,

we have ${\displaystyle ⟨v\mid v⟩⟨w\mid w⟩={\sum }_i⟨v\mid i⟩⟨i\mid v⟩⟨w\mid w⟩}$.

Substitute $|i⟩⟨i|$ with the first basis ${\displaystyle \frac{|w⟩}{\sqrt{⟨w\mid w⟩}}}{\displaystyle \frac{⟨w|}{\sqrt{⟨w\mid w⟩}}}$,

we have $⟨v\mid v⟩⟨w\mid w⟩\ge {\displaystyle \frac{⟨v\mid w⟩⟨w\mid v⟩}{⟨w\mid w⟩}}⟨w\mid w⟩=⟨v\mid w⟩⟨w\mid v⟩={|⟨v\mid w⟩|}^2$. QED.

Exercise 2.13: If $|w⟩$ and $|v⟩$ are any two vectors, show that ${(|w⟩⟨v|)}^{†}=|v⟩⟨w|$.

Proof: ${(|w⟩⟨v|)}^{†}={\left[{(|w⟩⟨v|)}^T\right]}^{*}={[⟨v\mid w⟩]}^{*}={(⟨v|)}^{*}{(|w⟩)}^{*}=|v⟩⟨w|$. QED.

Note: Transpose reverses order, while conjugation swaps bras and kets ($⟨v|\equiv |v{⟩}^{†}$).

Exercise 2.15: Show that ${\left(A^{†}\right)}^{†}=A$.

Proof: ${\left(A^{†}\right)}^{†}={\left\{{\left[{\left(A^T\right)}^{*}\right]}^T\right\}}^{*}$.

For each element, this means $A_{ij}\stackrel{T}{\to }{A}_{ji}\stackrel{*}{\to }{A}_{ji}^{*}\stackrel{T}{\to }{A}_{ij}^{*}\stackrel{*}{\to }{A}_{ij}$.

Therefore ${\left(A^{†}\right)}^{†}=A$. QED.

An operator $A$ that satisfies $A^{†}=A$ is known as a Hermitian or self-adjoint operator.

Projectors: Suppose $W$ is a $k$-dimensional vector subspace of $d$-dimensional vector space $V$. Using the Gram-Schmidt procedure it is possible to construct an orthonormal basis $|1⟩,\dots ,|d⟩$ for $V$ such that $|1⟩,\dots ,|k⟩$ is an orthonormal basis for $W$. By definition, the Hermitian operator ${\displaystyle P\equiv {\sum }_{i=1}^k|i⟩⟨i|}$ is the projector onto the subspace $W$.

The orthogonal complement of $P$ is the Hermitian operator $Q\equiv I-P$, also a projector.

Exercise 2.16: Show that any projector $P$ satisfies the equation $P^2=P$.

Proof: ${\displaystyle P^2={({\sum }_{i=1}^k|i⟩⟨i|)}^2={\sum }_{i=1}^k|i⟩⟨i\mid i⟩⟨i|={\sum }_{i=1}^k|i⟩⟨i|=P}$. QED.

An operator $A$ is said to be normal if $A{A}^{†}=A^{†}A$. Clearly, an operator which is Hermitian is also normal.

Spectral decomposition: Any normal operator $M$ on a vector space $V$ is diagonal with respect to some orthonormal basis for $V$. Conversely, any diagonalizable operator is normal.

Proof of the backward direction:

If $M$ is diagonalizable, then ${\displaystyle M={\sum }_i{\lambda }_i|i⟩⟨i|}$ for some orthonormal basis $|i⟩$.

${\displaystyle M^{†}={\sum }_i{\lambda }_i^{*}|i⟩⟨i|}$,

${\displaystyle M{M}^{†}={\sum }_i{\lambda }_i{\lambda }_i^{*}|i⟩⟨i|i⟩⟨i|={\sum }_i|{\lambda }_i{|}^2|i⟩⟨i|}$,

${\displaystyle M^{†}M={\sum }_i{\lambda }_i^{*}{\lambda }_i|i⟩⟨i|i⟩⟨i|={\sum }_i|{\lambda }_i{|}^2|i⟩⟨i|}$.

Then $M{M}^{†}=M^{†}M$, therefore $M$ is normal. QED.

Proof of the forward direction, by induction (from the book):

When dimension $d=1$, any operator on a 1D space is just a scalar:

$M=\lambda |1⟩⟨1|$.

When dimension $d>1$: Let $\lambda$ be an eigenvalue of $M$, $P$ the projector onto the $\lambda$-eigenspace, and $Q$ the projector onto the orthogonal complement.

Then $M=(P+Q)M(P+Q)=PMP+QMP+PMQ+QMQ$.

• By definition, the first term $PMP=\lambda P$.

• For the second term $QMP$:

  Let $|v⟩$ be any element of $P$, i.e., a vector in the $\lambda$-eigenspace. $M|v⟩=\lambda |v⟩$ is still in the $\lambda$-eigenspace. $Q$ projects it onto the orthogonal complement of $\lambda$-eigenspace, therefore $Q(M|v⟩)=0$. Therefore $QMP=0$.

• For the third term $PMQ$:

  Let $|v⟩$ be any element of $P$, then by definition $M{M}^{†}|v⟩=M^{†}M|v⟩=\lambda M^{†}|v⟩$ .

  Therefore $M^{†}|v⟩$ is an eigenvector of $M$ with eigenvalue $\lambda$, and is an element of $P$.

  Therefore by the same logic above, $Q{M}^{†}P=0$.

  By Hermitian conjugation of the equation, ${\left(Q{M}^{†}P\right)}^{†}=P^{†}M{Q}^{†}=PMQ=0$.

• For the fourth term $QMQ$:

  $QM=QMI=QM(P+Q)=QMP+QMQ=QMQ$,

  $Q{M}^{†}=Q{M}^{†}I=Q{M}^{†}(P+Q)=Q{M}^{†}P+Q{M}^{†}Q=Q{M}^{†}Q$.

  Note that $M{M}^{†}=M^{†}M$ (by normality), and $Q^2=Q$ (Exercise 2.16).

  $QMQQ{M}^{†}Q=QMQ{M}^{†}Q$

  $=QM{M}^{†}Q$

  $=Q{M}^{†}MQ$

  $=Q{M}^{†}QMQ$

  $=Q{M}^{†}QQMQ$, therefore $QMQ$ is normal.

  Since $QMQ$ is on a $<d$ dimensional space, by the inductive hypothesis, $QMQ$ is diagonal.

In conclusion, $M=\lambda P+QMQ$, where both $\lambda P$ and $QMQ$ are diagonal.

Therefore, $M$ is diagonal on the whole $d$-dimensional $V$. QED.

Exercise 2.17: Show that a normal matrix is Hermitian if and only if it has real eigenvalues.

Proof: A normal matrix $A$ can be diagonalized: ${\displaystyle A={\sum }_i{\lambda }_i|i⟩⟨i|}$.

Then ${\displaystyle A^{†}={\sum }_i{\lambda }_i^{*}|i⟩⟨i|}$.

• Forward: If $A=A^{†}$, then ${\displaystyle {\sum }_i{\lambda }_i|i⟩⟨i|={\sum }_i{\lambda }_i^{*}|i⟩⟨i|}$.

  Since $|i⟩⟨i|$ are linear independent, this implies ${\lambda }_i={\lambda }_i^{*}$ for all $i$, therefore ${\lambda }_i$ is real.

• Backward: If all ${\lambda }_i$ are real, then ${\lambda }_i={\lambda }_i^{*}$, so ${\displaystyle A^{†}={\sum }_i{\lambda }_i^{*}|i⟩⟨i|={\sum }_i{\lambda }_i|i⟩⟨i|=A}$, therefore $A$ is Hermitian.

QED.

Simultaneous diagonalization theorem: Suppose $A$ and $B$ are Hermitian operators. Then $[A,B]=0$ if and only if there exists an orthonormal basis such that both $A$ and $B$ are diagonal with respect to that basis. We say that $A$ and $B$ are simultaneously diagonalizable in this case.

Proof of the backward direction:

Let $|a,j⟩$ be an orthonormal basis for the eigenspace $V_a$ of $A$ with eigenvalue $a$.

Since $A,B$ are diagonal with respect to the same orthonormal basis, each $|a,j⟩$ is also an eigenvector of $B$ with some eigenvalue $b$.

Then $[A,B]|a,j⟩\equiv (AB-BA)|a,j⟩$

$=AB|a,j⟩-BA|a,j⟩$

$=Ab|a,j⟩-Ba|a,j⟩$

$=ab|a,j⟩-ba|a,j⟩=0$. QED.

Proof of the forward direction (from the book):

Let $|a,j⟩$ be an orthonormal basis for the eigenspace $V_a$ of $A$ with eigenvalue $a$.

We know that $AB|a,j⟩=BA|a,j⟩=aB|a,j⟩$,

therefore $B|a,j⟩$ is an element of the eigenspace $V_a$.

Let $P_a$ denote the projector onto the space $V_a$, define $B_a\equiv P_{a}B{P}_a$.

$B_a^{†}={\left(P_{a}B{P}_a\right)}^{†}=P_a^{†}{B}^{†}{P}_a^{†}=P_{a}B{P}_a=B_a$ ($P_a$ is projector ⇒ Hermitian; $B$ is Hermitian given in the theorem statement), therefore $B_a$ is Hermitian.

Therefore $B_a$ has a spectral decomposition in terms of an orthonormal set of eigenvectors which span the space $V_a$.

Let $|a,b,k⟩$ be this orthonormal set of eigenvectors, where indices $a,b$ label the eigenvalues of $A$ and $B_a$, and $k$ is an extra index to account for possible multiple linearly independent eigenvectors with the same eigenvalue $b$ that $B_a$ has.

$B|a,b,k⟩$ is an element of $V_a$, therefore $B|a,b,k⟩=P_{a}B|a,b,k⟩$.

Also, $P_a|a,b,k⟩=|a,b,k⟩$. Therefore $B|a,b,k⟩=P_{a}B|a,b,k⟩=P_{a}B{P}_a|a,b,k⟩=b|a,b,k⟩$.

Therefore $|a,b,k⟩$ is an eigenvector of $B$ with eigenvalue $b$.

Therefore $|a,b,k⟩$ is an orthonormal set of eigenvectors of both $A$ and $B$, and $A,B$ are diagonal with respect to that basis. QED.

#Mathematics