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.
- Prove the equivalence of the four characterisations of an isometry.
- Recognise orthogonal and unitary matrices and invert them by transposing.
- Show every eigenvalue of a unitary operator has modulus $1$.
- Classify the $2\times2$ real orthogonal matrices.
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.
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.
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
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.
For linear $T$ on a finite-dimensional inner product space, the following are equivalent:
Orthogonal and unitary matrices
$A\in M_n(\mathbb{R})$ 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.
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 $\det T=\pm1$.
$U=\left(\begin{smallmatrix}0&i\\i&0\end{smallmatrix}\right)$ 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
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 $(T^{-1})^*T^{-1}=(T^*)^*T^*=TT^*=I$.
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.
$A$ 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 $\operatorname{tr}(A^*A)$.
• 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.
Is $\tfrac{1}{5}\left(\begin{smallmatrix}3&-4\\4&3\end{smallmatrix} \right)$ orthogonal?
Answer
Describe the map with matrix $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$.
Answer
Prove every eigenvalue of a unitary operator has $|\lambda|=1$.
Solution
Show the product of two unitary matrices is unitary.
Solution
Show an isometry maps orthogonal vectors to orthogonal vectors.
Solution
For a unit vector $u$, show $T(v)=v-2\langle v,u\rangle u$ is orthogonal and describe it.
Solution
Show $R_{\pi/2}$ has no real eigenvalue but is orthogonal.
Solution
Show $\operatorname{tr}(A^*A)$ is unchanged under $A\mapsto U^*AU$ for unitary $U$.
Solution
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:
- Isometry characterisations. Prove all four conditions equivalent and know which implication uses polarization.
- Matrix criterion. Recognise orthogonal and unitary matrices by their columns and invert them by conjugate transpose.
- Spectral facts. Eigenvalues have modulus $1$ and $|\det|=1$.
- Classification in the plane. Rotations versus reflections by determinant.
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.
Length preserved ⇒ inner products preserved ⇒ $T^*T=I$ ⇒ orthonormal columns. Each arrow is one short computation, and the chain is the whole unit.
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.
State the matrix condition for unitarity.
Guide
What values can $\det$ take for a real orthogonal matrix?
Guide
Which step of the equivalence proof is not immediate?
Guide
How do they differ?