Complex inner products複素内積ふくそないせき and unitary matricesユニタリ行列ぎょうれつ - Basic Exercises

1Exercise plan

In complex calculations, the position of the complex conjugate複素共役ふくそきょうやく is essential. These exercises check what complex inner products複素内積ふくそないせき, conjugate transposes共役転置きょうやくてんち, Hermitian matricesエルミート行列ぎょうれつ, and unitary matricesユニタリ行列ぎょうれつ preserve or test.

For complex inner-product exercises, write conjugates explicitly. The test for orthogonality, norm preservation, and unitarity uses $A^{*}$A^*, the conjugate transpose, not just $A^T$A^T. This is the most common source of wrong answers when moving from real to complex vector spaces.

A unitary matrixユニタリ行列 can be checked by $U^{*}U=I$U^*U=I, by preservation of inner products, or by having orthonormal columns. These are equivalent tests, but each is useful in a different problem: column checks are quick for a concrete matrix, while inner-product preservation explains why lengths, angles, and projections are unchanged.

2Problem 1

Using the convention

for $u=(1,i{)}^T$u=(1,i)^T and $v=(i,1{)}^T$v=(i,1)^T, find $⟨u,v⟩$\langle u,v\rangle, $⟨v,u⟩$\langle v,u\rangle, and $\parallel u\parallel$\|u\|.

2.1Answer example

and

2.2Explanation

In this convention, the second variable is conjugated. Forgetting that conjugate changes orthogonality computations.

3Problem 2

For $A=\left(\begin{array}{cc}2& i\\ -i& 3\end{array}\right)$A=\begin{pmatrix}2&i\\-i&3\end{pmatrix}, find $A^{*}$A^* and decide whether $A$A is Hermitianエルミート行列ぎょうれつ.

3.1Answer example

Transpose first and conjugate each entry:

Thus $A^{*}=A$A^*=A, so $A$A is Hermitian.

3.2Explanation

In a Hermitian matrix, diagonal entries are real and off-diagonal mirror entries are complex conjugates.

4Problem 3

Confirm that $U=\left(\begin{array}{cc}1& 0\\ 0& i\end{array}\right)$U=\begin{pmatrix}1&0\\0&i\end{pmatrix} is unitary, and explain why $\parallel Ux\parallel =\parallel x\parallel$\|Ux\|=\|x\| for arbitrary $x=(x_1,x_2{)}^T$x=(x_1,x_2)^T.

4.1Answer example

so

Also $Ux=(x_1,i{x}_2{)}^T$Ux=(x_1,ix_2)^T, hence

4.2Explanation

A unitary matrix may change phase位相いそう, but it preserves lengths and inner products.

5Problem 4

For $y=(1,1+i{)}^T$y=(1,1+i)^T and $u=(1,i{)}^T$u=(1,i)^T, find the orthogonal projection直交射影ちょっこうしゃえい of $y$y onto $\mathrm{span}\left(u\right)$\operatorname{span}(u).

5.1Answer example

Under this convention,

Now

and $⟨u,u⟩=2$\langle u,u\rangle=2. Therefore

5.2Explanation

The denominator is positive real because $u\ne 0$u\ne0. The conjugate position in the projection coefficient matters.

6Preview review: eigenvalues, eigenvectors, diagonalization, and normal matrices

Problems 5 and later use eigenvalues, diagonalization, and normal matrices before the later exercise sets develop them fully. This review gives only the minimum needed to solve the problems here.

• If $Av=\lambda v$Av=\lambda v for some $v\ne 0$v\ne0, then $\lambda$\lambda is an eigenvalue固有値こゆうち and $v$v is an eigenvector固有ベクトル.
• If eigenvectors form a basis, then $A=PD{P}^{-1}$A=PDP^{-1}. This form is called diagonalization対角化たいかくか.
• If a unitary matrixユニタリ行列ぎょうれつ $U$U gives $U^{*}AU=D$U^*AU=D, then this is unitary diagonalizationユニタリ対角化たいかくか.
• A normal matrix正規行列せいきぎょうれつ is a matrix satisfying $A^{*}A=A{A}^{*}$A^*A=AA^*. Over finite-dimensional complex spaces, the later theorem says that normal matrices are exactly the matrices that can be unitarily diagonalized.

7Problem 5

Find the eigenvalues of $B=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$B=\begin{pmatrix}2&i\\-i&2\end{pmatrix} and check that they are real as expected for a Hermitian matrix.

7.1Answer example

Thus the eigenvalues are $1,3$1,3, both real.

7.2Explanation

Since $B^{*}=B$B^*=B, the matrix is Hermitian. Hermitian matrices may have complex entries, but their eigenvalues are real. This is one reason Hermitian matrices are used to represent observables in contexts such as quantum mechanics.

8Problem 6

Decide whether each statement is true and explain why.

1. A unitary matrix preserves the inner product.
2. A Hermitian matrix is always normal.
3. A unitary matrix is always Hermitian.

8.1Answer example

1 is true because $U^{*}U=I$U^*U=I implies $⟨Ux,Uy⟩=⟨x,y⟩$\langle Ux,Uy\rangle=\langle x,y\rangle. 2 is true because if $A^{*}=A$A^*=A, then $A^{*}A=A^2=A{A}^{*}$A^*A=A^2=AA^*. 3 is false: $\left(\begin{array}{cc}1& 0\\ 0& i\end{array}\right)$\begin{pmatrix}1&0\\0&i\end{pmatrix} is unitary but not Hermitian.

8.2Explanation

Unitary, Hermitian, and normal matrices are related but distinct. Separate preservation of inner products from diagonalization conditions.

9Problem 7

Over $\mathbb{C}$\mathbb C, confirm that

is normal and diagonalize it by a unitary matrix.

9.1Answer example

Since $A^{*}=-A$A^*=-A and $A^2=-I$A^2=-I,

Thus $A$A is normal. Its eigenvalues are $i,-i$i,-i. Eigenvectors may be chosen as $(1,i{)}^T$(1,i)^T and $(1,-i{)}^T$(1,-i)^T. Normalizing gives

with $U^{*}AU=D$U^*AU=D.

9.2Explanation

This matrix is not Hermitian, but it is normal. Normal matrices are the natural class for unitary diagonalization over complex spaces.

10Problem 8

Using the convention $⟨x,y⟩=\sum _k{x}_k\stackrel{\_}{y_k}$\langle x,y\rangle=\sum_kx_k\overline{y_k}, apply Gram-Schmidt to $x_1=(1,i{)}^T$x_1=(1,i)^T and $x_2=(1,0{)}^T$x_2=(1,0)^T.

10.1Answer example

Let $u_1=x_1=(1,i{)}^T$u_1=x_1=(1,i)^T. Since $⟨u_1,u_1⟩=2$\langle u_1,u_1\rangle=2,

10.2Explanation

The conjugate in the numerator is essential. Here $u_2$u_2 is orthogonal to $u_1$u_1.

11Problem 9

For

verify $⟨Ax,y⟩=⟨x,A^{*}y⟩$\langle Ax,y\rangle=\langle x,A^*y\rangle.

11.1Answer example

so

Also

so

11.2Explanation

The conjugate transpose is the adjoint随伴ずいはん relative to the complex inner product.

12Problem 10

Let $A=(1,i{)}^T$A=(1,i)^T and $b=(1,0{)}^T$b=(1,0)^T. Solve the complex least-squares normal equation and check residual orthogonality.

12.1Answer example

so

Thus $x=\frac{1}{2}$x=\frac12. Then

Finally,

12.2Explanation

The real least-squares statement "residual orthogonal to the column space" becomes $A^{*}r=0$A^*r=0 in the complex case.

Here the convention conjugates the second variable in the inner product, so the adjoint is $A^{*}=(1,-i)$A^*=(1,-i). The equation $A^{*}r=0$A^*r=0 says exactly that the residual has zero inner product with every vector in the one-dimensional column space列空間れつくうかん of $A$A.

13Problem 11

Confirm that

is unitary and verify $⟨Ux,Uy⟩=⟨x,y⟩$\langle Ux,Uy\rangle=\langle x,y\rangle for $x=(1,i{)}^T$x=(1,i)^T and $y=(1,0{)}^T$y=(1,0)^T.

13.1Answer example

The matrix $U$U is real symmetric and $U^{*}U=U^{T}U=I$U^*U=U^TU=I, so it is unitary. Also

Therefore

while $⟨x,y⟩=1$\langle x,y\rangle=1.

13.2Explanation

Unitary matrices preserve inner products and lengths even when they are not diagonal.

Although the components of $x$x and $y$y change after applying $U$U, the inner product does not. This is the defining geometric role of a unitary matrixユニタリ行列: it preserves lengths and angles in a complex inner product space.

14Related lectures

15Proof exercise: unitary matrices preserve inner products

15.1Problem

If $U^{*}U=I$U^*U=I, prove that for arbitrary complex vectors $x,y$x,y,

15.2Answer

With the convention used in this series, $⟨x,y⟩=y^{*}x$\langle x,y\rangle=y^*x, so

Taking $y=x$y=x gives $\parallel Ux\parallel =\parallel x\parallel$\|Ux\|=\|x\|.

15.3Explanation

A unitary matrix preserves the complex inner product, so it preserves lengths and angles.

16Supplementary theorem-check problem: normal matrices and unitary diagonalization

16.1Problem 12

Let $A$A be unitarily diagonalizable, so $U^{*}AU=D$U^*AU=D with $U$U unitary and $D$D diagonal. Prove that $A$A is normal. Then explain why the converse is the complex spectral theorem for normal matrices.

16.2Answer example

From $U^{*}AU=D$U^*AU=D, we have $A=UD{U}^{*}$A=UDU^* and $A^{*}=U{D}^{*}{U}^{*}$A^*=UD^*U^*. Hence

and

Since diagonal matrices commute with their conjugate transposes, $D^{*}D=D{D}^{*}$D^*D=DD^*. Thus $A^{*}A=A{A}^{*}$A^*A=AA^*, so $A$A is normal. The converse theorem says that every complex normal matrix admits such a unitary diagonalization.

16.3Explanation

Normality is exactly the condition that makes orthonormal eigenvector decompositions possible over $\mathbb{C}$\mathbb C. Hermitian and unitary matrices are important special cases of normal matrices.