MAT247H1/Unit 4
Unit 4 · Algebra II

Isometries: Orthogonal & Unitary Operators

The first structural class: operators that preserve length. It turns out preserving length forces preserving angle, forces $T^*T=I$, and forces the columns of the matrix to be an orthonormal basis. Four different-looking conditions, all the same condition.

By the end you can
Why it matters & where it comes from
Why it exists

These are the changes of coordinates that do not distort geometry. If you want to move to a nicer basis without changing lengths or angles — which is exactly what the spectral theorem does — the change-of-basis matrix must be unitary.

In the real world

Rotation matrices in graphics and robotics are orthogonal. Quantum time evolution is unitary, which is what conserves total probability. The discrete Fourier transform is unitary after normalising, which is why it preserves energy — Parseval's theorem in signal processing.

Story & open questions

The orthogonal group $O(n)$ and unitary group $U(n)$ are the first examples of Lie groups, the objects that organise modern physics. Their representation theory (MAT445H1) classifies the symmetries of quantum systems.

Four equivalent conditions

Definition

A linear $T:V\to V$ is an isometry if $\|Tv\|=\|v\|$ for all $v$. Over $\mathbb{R}$ such a $T$ is called orthogonal; over $\mathbb{C}$ it is called unitary.

Theorem Equivalences

For linear $T$ on a finite-dimensional inner product space, the following are equivalent:

(1) $\|Tv\|=\|v\|$ for all $v$ — lengths preserved.
(2) $\langle Tu,Tv\rangle=\langle u,v\rangle$ for all $u,v$ — inner products preserved.
(3) $T^*T=I$, equivalently $T^*=T^{-1}$.
(4) $T$ carries some (hence every) orthonormal basis to an orthonormal basis.
Proof The cycle
(2) ⇒ (1): take $u=v$.
(1) ⇒ (2): the polarization identity recovers the inner product from the norm, so preserving all norms preserves all inner products. Over $\mathbb{R}$, $\langle u,v\rangle=\tfrac14(\|u+v\|^2-\|u-v\|^2)$, and each norm on the right is preserved.
(2) ⇔ (3): condition (2) says $\langle u,v\rangle=\langle Tu,Tv\rangle=\langle u,T^*Tv\rangle$ for all $u,v$, that is $\langle u,(T^*T-I)v\rangle=0$ for all $u$, which forces $T^*T=I$.
(3) ⇔ (4): in an orthonormal basis $\{e_i\}$, the $(i,j)$ entry of $[T]^*[T]$ is $\langle Te_j,Te_i\rangle$. That matrix is $I$ exactly when the vectors $Te_i$ are orthonormal. ∎
Why finite dimension matters. $T^*T=I$ says $T$ is injective. In finite dimensions injective implies surjective, so $T$ is invertible and $TT^*=I$ too. In infinite dimensions the one-sided shift satisfies $T^*T=I$ but is not surjective — an isometry that is not unitary.

Orthogonal and unitary matrices

Definition

$A\in M_n(\mathbb{R})$ is orthogonal if $A^tA=I$; $A\in M_n(\mathbb{C})$ is unitary if $A^*A=I$. Equivalently the columns form an orthonormal basis — and since $A^*A=I$ gives $AA^*=I$, so do the rows.

Theorem Eigenvalues sit on the unit circle

If $T$ is unitary and $Tv=\lambda v$ with $v\ne0$, then $|\lambda|=1$.

$\|v\|=\|Tv\|=\|\lambda v\|=|\lambda|\,\|v\|$, and $\|v\|\ne0$.

Consequently $|\det T|=1$: the determinant is the product of the eigenvalues over $\mathbb{C}$. For a real orthogonal matrix $\det T=\pm1$.

Example All $2\times2$ real orthogonal matrices
The first column is a unit vector, so it is $(\cos\theta,\sin\theta)$ for some $\theta$. The second is a unit vector orthogonal to it, so it is $\pm(-\sin\theta,\cos\theta)$.
With $+$: $R_\theta =\left(\begin{smallmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{smallmatrix}\right)$, determinant $+1$ — a rotation.
With $-$: determinant $-1$, and the matrix is a reflection in the line at angle $\theta/2$. So every plane isometry fixing the origin is a rotation or a reflection, and the determinant tells you which.
Example A unitary that is not orthogonal

$U=\left(\begin{smallmatrix}0&i\\i&0\end{smallmatrix}\right)$ satisfies $U^*U=I$, so it is unitary. Its entries are not real, so it is not an orthogonal matrix; its eigenvalues are $\pm i$, both of modulus $1$.

Structure of the isometry groups

Theorem They form groups

Products and inverses of unitary operators are unitary, so they form a group $U(n)$ under composition; the real case gives $O(n)$. The elements of determinant $1$ form the subgroups $SU(n)$ and $SO(n)$.

If $S^*S=I$ and $T^*T=I$ then $(ST)^*(ST)=T^*S^*ST=T^*T=I$; and $(T^{-1})^*T^{-1}=(T^*)^*T^*=TT^*=I$.

Theorem Unitary change of basis

If $\beta$ and $\gamma$ are both orthonormal bases, the change-of-basis matrix from $\beta$ to $\gamma$ is unitary. This is why "orthogonally diagonalizable" in Unit 6 means $A=UDU^*$ rather than $A=PDP^{-1}$ — the conjugating matrix can be taken unitary, and then $P^{-1}$ costs nothing to compute.

Definition

$A$ and $B$ are unitarily equivalent if $B=U^*AU$ for some unitary $U$. This is stronger than similarity: it preserves the inner product structure as well as the operator. Unitarily equivalent matrices have the same eigenvalues and the same $\operatorname{tr}(A^*A)$.

Common pitfalls

• Calling a matrix orthogonal because its columns are pairwise orthogonal. They must also be unit vectors; "orthogonal matrix" really means orthonormal columns.
• Using $A^t$ instead of $A^*$ over $\mathbb{C}$. The transpose alone does not give the inverse of a unitary matrix.
• Concluding $\det T=1$. It is $\pm1$ in the real case and any unit complex number in general.
• Assuming an isometry is diagonalizable over $\mathbb{R}$. A rotation by $\theta\ne0,\pi$ has no real eigenvalue at all.

Practice

Original problems written for this guide in the style of Friedberg, Insel & Spence exercises. Write a full solution before opening the answer.

Practice 1 · Verify orthogonality

Is $\tfrac{1}{5}\left(\begin{smallmatrix}3&-4\\4&3\end{smallmatrix} \right)$ orthogonal?

Answer
Yes. Each column has norm $\tfrac15\sqrt{9+16}=1$ and their dot product is $\tfrac{1}{25}(-12+12)=0$. Determinant $\tfrac{1}{25}(9+16)=1$, so it is a rotation.
Practice 2 · Identify the isometry

Describe the map with matrix $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$.

Answer
Orthogonal with determinant $-1$: a reflection, in the line $y=x$. Its eigenvalues are $1$ (along $y=x$) and $-1$ (along $y=-x$).
Practice 3 · Eigenvalue modulus

Prove every eigenvalue of a unitary operator has $|\lambda|=1$.

Solution
From $Tv=\lambda v$ with $v\ne0$, $\|v\|=\|Tv\|=|\lambda|\|v\|$, so $|\lambda|=1$ after dividing by $\|v\|\ne0$.
Practice 4 · Products

Show the product of two unitary matrices is unitary.

Solution
$(UV)^*(UV) =V^*U^*UV=V^*IV=V^*V=I$.
Practice 5 · Preserving angles

Show an isometry maps orthogonal vectors to orthogonal vectors.

Solution
By characterisation (2), $\langle Tu,Tv\rangle=\langle u,v\rangle=0$ whenever $u\perp v$. Angle is preserved because both the inner product and the norms are.
Practice 6 · Reflection formula

For a unit vector $u$, show $T(v)=v-2\langle v,u\rangle u$ is orthogonal and describe it.

Solution
Compute $\|Tv\|^2=\|v\|^2-4\langle v,u\rangle^2+4\langle v,u\rangle^2\|u\|^2 =\|v\|^2$ using $\|u\|=1$. It fixes $u^\perp$ pointwise and sends $u$ to $-u$: reflection in the hyperplane $u^\perp$. These are the Householder reflections used in numerical QR.
Practice 7 · No real eigenvalues

Show $R_{\pi/2}$ has no real eigenvalue but is orthogonal.

Solution
Its matrix is $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ with characteristic polynomial $t^2+1$, which has no real root. It is still orthogonal — being an isometry says nothing about having eigenvectors over $\mathbb{R}$.
Practice 8 · Unitary invariant

Show $\operatorname{tr}(A^*A)$ is unchanged under $A\mapsto U^*AU$ for unitary $U$.

Solution
$(U^*AU)^*(U^*AU)=U^*A^*UU^*AU=U^*A^*AU$, and trace is invariant under similarity. So the sum of squared moduli of the entries is a unitary invariant — the Frobenius norm.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT247H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT247H1 calendar topics as the coverage spine, and pair lecture notes with a Friedberg-Insel-Spence-style linear algebra text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

Length preserved ⇒ inner products preserved ⇒ $T^*T=I$ ⇒ orthonormal columns. Each arrow is one short computation, and the chain is the whole unit.

Study routine. Given any concrete matrix, check unitarity by multiplying $A^*A$ rather than inspecting columns individually. It is faster and catches normalisation errors.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Matrix condition

State the matrix condition for unitarity.

Guide
$A^*A=I$, equivalently $A^{-1}=A^*$, equivalently the columns form an orthonormal basis.
Extension 2 · Determinant

What values can $\det$ take for a real orthogonal matrix?

Guide
Only $\pm1$. From $A^tA=I$, $(\det A)^2=1$. Determinant $+1$ means a rotation, $-1$ means orientation is reversed.
Extension 3 · Which implication needs polarization

Which step of the equivalence proof is not immediate?

Guide
Norms preserved ⇒ inner products preserved. It needs the polarization identity to recover $\langle u,v\rangle$ from norms alone.
Extension 4 · Unitary equivalence versus similarity

How do they differ?

Guide
Similarity allows any invertible $P$; unitary equivalence requires $P$ unitary, so it preserves the inner product too. Unitarily equivalent matrices share the Frobenius norm, which merely similar ones need not.