MAT247H1/Unit 5
Unit 5 · Algebra II

Hermitian, Symmetric & Normal Operators

Unit 4 studied $T^*T=I$. Now the weaker condition $T^*T=TT^*$ — merely that $T$ commutes with its adjoint. That is exactly the hypothesis the spectral theorem needs, and self-adjoint operators are the special case where the eigenvalues come out real.

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

MAT240H1 left diagonalizability as a delicate condition that often fails. Adding one hypothesis about the adjoint makes it automatic — and gives an orthonormal eigenbasis, not just any basis. Normality is the exact boundary of that guarantee.

In the real world

Covariance matrices are symmetric, which is why principal component analysis always works. Quantum observables are self-adjoint precisely so that measured values — the eigenvalues — are real numbers. The adjacency matrix of an undirected graph is symmetric, and its spectrum governs connectivity and random walks.

Story & open questions

Hermite studied these forms in the 1850s. The name "normal" is Toeplitz's, from 1918, and reflects that this is the natural class rather than a contrived one. In infinite dimensions the spectral theorem for normal operators is a cornerstone of functional analysis (MAT436H1).

The definitions and the hierarchy

Definition

Let $T$ be a linear operator on a finite-dimensional inner product space.

$T$ is self-adjoint (Hermitian over $\mathbb{C}$, symmetric over $\mathbb{R}$) if $T^*=T$.
$T$ is skew-adjoint if $T^*=-T$.
$T$ is normal if $T^*T=TT^*$.

In matrix terms over an orthonormal basis: $A^*=A$, $A^*=-A$, and $A^*A=AA^*$.

Theorem Everything familiar is normal

Self-adjoint, skew-adjoint and unitary operators are all normal.

If $T^*=T$ then $T^*T=T^2=TT^*$. If $T^*=-T$ then $T^*T=-T^2=TT^*$. If $T^*T=I=TT^*$ then the two products are equal by definition. So normality is the common generalisation.

Example The hierarchy is strict
Normal, not self-adjoint: $A=\left(\begin{smallmatrix}i&0\\0&-i\end{smallmatrix}\right)$. Here $A^*=-A$, so $A^*A=AA^*=I$, but $A^*\ne A$.
Normal, not unitary: $2I$ satisfies $A^*A=AA^*=4I\ne I$.
Not normal: $A=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$. Then $A^*A=\left(\begin{smallmatrix}1&1\\1&2\end{smallmatrix}\right)$ but $AA^*=\left(\begin{smallmatrix}2&1\\1&1\end{smallmatrix}\right)$. This is the defective matrix from MAT240H1 — and its failure to be normal is exactly why it would not diagonalize.

Consequences of self-adjointness

Theorem Real eigenvalues

If $T$ is self-adjoint, every eigenvalue is real.

Let $Tv=\lambda v$ with $v\ne0$. Then $\lambda\|v\|^2=\langle Tv,v\rangle=\langle v,Tv\rangle =\overline{\lambda}\|v\|^2$, using $T^*=T$. Dividing by $\|v\|^2\ne0$ gives $\lambda=\overline{\lambda}$. ∎

Theorem $\langle Tv,v\rangle$ is real

$T$ is self-adjoint iff $\langle Tv,v\rangle\in\mathbb{R}$ for every $v$ — over $\mathbb{C}$. One direction is the computation above; the converse uses a polarization argument. Over $\mathbb{R}$ the statement is vacuous, and indeed the converse fails there: a rotation has $\langle Tv,v\rangle$ real but is not symmetric.

Theorem A real symmetric operator has a real eigenvalue

This is not automatic — the rotation in Unit 4 had none. Symmetry is what rescues it.

Pick an orthonormal basis and let $A=[T]$, a real symmetric matrix. Regard $A$ as acting on $\mathbb{C}^n$; the fundamental theorem of algebra gives an eigenvalue $\lambda\in\mathbb{C}$ with eigenvector $v\in\mathbb{C}^n$.
Since $A$ is real symmetric it is self-adjoint on $\mathbb{C}^n$, so by the previous theorem $\lambda$ is real. Then $A-\lambda I$ is a real singular matrix, so it has a nonzero real null vector. ∎
This is the bridge. Over $\mathbb{C}$ eigenvalues exist for free. Over $\mathbb{R}$ they do not, and this theorem is what lets the real spectral theorem get started. It is the one place the fundamental theorem of algebra enters the course.

Consequences of normality

Theorem Norms agree

$T$ is normal iff $\|Tv\|=\|T^*v\|$ for all $v$.

$\|Tv\|^2-\|T^*v\|^2=\langle Tv,Tv\rangle-\langle T^*v,T^*v\rangle =\langle (T^*T-TT^*)v,v\rangle$. If $T$ is normal this is $0$. Conversely, $T^*T-TT^*$ is self-adjoint and has $\langle Sv,v\rangle=0$ for all $v$, which forces $S=0$.

Theorem Shared eigenvectors

If $T$ is normal and $Tv=\lambda v$, then $T^*v=\overline{\lambda}v$.

Apply the previous theorem to $T-\lambda I$, which is also normal: $\|(T-\lambda I)v\|=\|(T-\lambda I)^*v\|=\|(T^*-\overline{\lambda}I)v\|$. The left side is $0$, so the right side is too.

Theorem Orthogonal eigenspaces

If $T$ is normal and $Tu=\lambda u$, $Tv=\mu v$ with $\lambda\ne\mu$, then $u\perp v$.

$\lambda\langle u,v\rangle=\langle Tu,v\rangle =\langle u,T^*v\rangle=\langle u,\overline{\mu}v\rangle=\mu\langle u,v\rangle$, where the third equality is the previous theorem.
So $(\lambda-\mu)\langle u,v\rangle=0$ and $\lambda\ne\mu$ forces $\langle u,v\rangle=0$. ∎

MAT240H1 proved such eigenvectors are independent. Normality upgrades that to orthogonal — which is precisely the extra strength the spectral theorem delivers.

Example Why non-normal fails

For $A=\left(\begin{smallmatrix}2&1\\0&3\end{smallmatrix}\right)$ from MAT240H1, the eigenvectors were $(1,0)$ and $(1,1)$ — independent but not orthogonal, since $\langle(1,0),(1,1)\rangle=1$. Consistent: $A$ is not normal.

Common pitfalls

• Assuming self-adjoint means real entries. A Hermitian matrix can have complex off-diagonal entries; it is the diagonal entries that must be real.
• Assuming normal implies self-adjoint. It does not — unitary operators are normal with non-real eigenvalues.
• Using $T^*v=\overline{\lambda}v$ without normality. For a general operator, an eigenvector of $T$ need not be one of $T^*$ at all.
• Forgetting the field when quoting "normal implies diagonalizable". Over $\mathbb{R}$ that is false; a rotation is normal and not diagonalizable.

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 · Classify

Is $A=\left(\begin{smallmatrix}2&1+i\\1-i&3\end{smallmatrix}\right)$ Hermitian?

Answer
Yes. The diagonal entries $2,3$ are real and the off-diagonal pair are conjugates: $\overline{1+i}=1-i$. So $A^*=A$, and its eigenvalues must be real.
Practice 2 · Normality test

Is $A=\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ normal over $\mathbb{R}$?

Answer
Yes: $A^t=-A$, so it is skew-symmetric and hence normal, with $A^tA=AA^t=I$. It is also orthogonal — the rotation by $\pi/2$.
Practice 3 · Real eigenvalues

Prove a Hermitian matrix has real eigenvalues.

Solution
From $Av=\lambda v$ with $v\ne0$, $\lambda\|v\|^2=\langle Av,v\rangle=\langle v,Av\rangle =\overline{\lambda}\|v\|^2$, so $\lambda=\overline{\lambda}$.
Practice 4 · Skew eigenvalues

Show a skew-adjoint operator has purely imaginary eigenvalues.

Solution
If $T^*=-T$ and $Tv=\lambda v$, then $\lambda\|v\|^2=\langle Tv,v\rangle =\langle v,T^*v\rangle=-\langle v,Tv\rangle=-\overline{\lambda}\|v\|^2$. So $\lambda=-\overline{\lambda}$, meaning $\lambda$ is purely imaginary (possibly $0$).
Practice 5 · Orthogonal eigenvectors

Prove eigenvectors of a normal operator for distinct eigenvalues are orthogonal.

Solution
Use $T^*v=\overline{\mu}v$ to get $\lambda\langle u,v\rangle=\langle Tu,v\rangle=\langle u,T^*v\rangle =\mu\langle u,v\rangle$, then divide by $\lambda-\mu\ne0$.
Practice 6 · Building normal operators

Show $T^*T$ is self-adjoint for any $T$, and its eigenvalues are non-negative.

Solution
$(T^*T)^*=T^*(T^*)^* =T^*T$. If $T^*Tv=\lambda v$ with $\|v\|=1$ then $\lambda=\langle T^*Tv,v\rangle=\|Tv\|^2\ge0$. This is the fact Unit 7 builds the singular value decomposition on.
Practice 7 · Sum of normals

Is the sum of two normal operators normal?

Solution
Not in general. Take $A=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$, both normal. Their sum $\left(\begin{smallmatrix}0&2\\0&0\end{smallmatrix} \right)$ satisfies $A^*A\ne AA^*$, so it is not normal. Normality is not preserved by addition.
Practice 8 · Decomposition

Show every operator splits as $T=A+iB$ with $A,B$ self-adjoint, and that $T$ is normal iff $A$ and $B$ commute.

Solution
Put $A=\tfrac12(T+T^*)$ and $B=\tfrac{1}{2i}(T-T^*)$; both are self-adjoint and $T=A+iB$. Expanding $T^*T-TT^*$ gives $2i(AB-BA)$, so normality is exactly commuting. This is the operator analogue of writing a complex number as real plus imaginary part.

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

Normality gives $T^*v=\overline{\lambda}v$, which gives orthogonal eigenspaces, which is the engine of the spectral theorem in Unit 6. Self-adjointness adds real eigenvalues and existence over $\mathbb{R}$.

Study routine. For every operator you meet, compute $T^*T$ and $TT^*$ before anything else. Whether they agree determines which theorems are available.

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 · Normal condition

State the defining condition for normality.

Guide
$T^*T=TT^*$: the operator commutes with its own adjoint.
Extension 2 · Spectrum by class

Where do the eigenvalues live for self-adjoint, skew-adjoint and unitary operators?

Guide
On the real axis, on the imaginary axis, and on the unit circle respectively. All three are normal, and the class shows up as a constraint on where the spectrum sits.
Extension 3 · Independence versus orthogonality

What does normality add to MAT240H1's eigenvector theorem?

Guide
MAT240H1 gave independence for distinct eigenvalues. Normality upgrades it to orthogonality, which is what makes the diagonalizing matrix unitary.
Extension 4 · Diagnosing a defect

An operator fails to diagonalize. What does that say about normality?

Guide
Over $\mathbb{C}$ it cannot be normal, since Unit 6 shows every complex normal operator diagonalizes. Over $\mathbb{R}$ the argument does not apply — a rotation is normal and still fails.
← Unit 04Isometries Unit 06 →The spectral theorem