MAT247H1/Unit 6
Unit 6 · Algebra II

The Spectral Theorem

The high point of the course. Over $\mathbb{C}$, normal is exactly the right hypothesis: a normal operator has an orthonormal basis of eigenvectors, full stop. Over $\mathbb{R}$ the same is true for symmetric operators. Every diagonalization failure from MAT240H1 was a failure of this hypothesis.

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

MAT240H1 could diagonalize when it got lucky. This theorem replaces luck with a checkable hypothesis, and pays a bonus: the eigenbasis is orthonormal, so the change-of-basis matrix is unitary and its inverse is free.

In the real world

Principal component analysis is the spectral theorem applied to a covariance matrix. The normal modes of a vibrating structure are its eigenvectors. In quantum mechanics the theorem is the reason an observable has a complete set of states with real measured values.

Story & open questions

Cauchy proved the real symmetric case in 1829 while studying quadratic surfaces. Hilbert coined "spectrum" around 1904 for the infinite-dimensional analogue — before quantum mechanics gave the word its physical meaning, which he later called a lucky accident.

Schur's theorem

Theorem Schur

Let $T$ be an operator on a finite-dimensional complex inner product space. Then there is an orthonormal basis in which $[T]$ is upper triangular. Equivalently, every complex square matrix is unitarily equivalent to an upper triangular one: $A=UTU^*$ with $U$ unitary.

Proof Induction on dimension
Over $\mathbb{C}$ the characteristic polynomial of $T^*$ has a root, so $T^*$ has an eigenvector: $T^*w=\mu w$ with $\|w\|=1$.
Let $W=\operatorname{span}\{w\}$ and consider $W^\perp$, of dimension $n-1$. It is $T$-invariant: for $v\in W^\perp$, $\langle Tv,w\rangle=\langle v,T^*w\rangle=\langle v,\mu w\rangle =\overline{\mu}\langle v,w\rangle=0$.
By induction $T|_{W^\perp}$ has an orthonormal basis making it upper triangular. Appending $w$ gives an orthonormal basis of $V$ in which $[T]$ is upper triangular, since the first column has zeros below the diagonal by construction. ∎
Why start with $T^*$, not $T$. Taking an eigenvector of $T^*$ makes $W^\perp$ invariant under $T$, which is the direction the induction needs. Starting from an eigenvector of $T$ makes $W$ invariant instead, and $W^\perp$ need not be.

The complex spectral theorem

Theorem Spectral theorem, complex case

Let $T$ be an operator on a finite-dimensional complex inner product space. Then $T$ is normal if and only if $V$ has an orthonormal basis of eigenvectors of $T$. In matrix terms: $A$ is normal iff $A=UDU^*$ with $U$ unitary and $D$ diagonal.

Proof Schur plus normality
(⇐) If $[T]=D$ is diagonal in an orthonormal basis then $[T^*]=\overline{D}$, and diagonal matrices commute, so $T$ is normal.
(⇒) By Schur, take an orthonormal basis with $A=[T]$ upper triangular. Compare the $(1,1)$ entries of $A^*A$ and $AA^*$: the first is $|A_{11}|^2$, the second is $\sum_j|A_{1j}|^2$.
Normality forces these equal, so $A_{1j}=0$ for all $j>1$: the first row is zero off the diagonal. Repeating with the $(2,2)$ entries kills the second row, and so on. The triangular matrix is diagonal. ∎
Example Diagonalize a unitary

$U=\left(\begin{smallmatrix}0&i\\i&0\end{smallmatrix}\right)$ is unitary, hence normal.

$f(t)=t^2+1$, so the eigenvalues are $\pm i$ — on the unit circle, as Unit 4 requires.
Eigenvectors: $(1,1)$ for $i$ and $(1,-1)$ for $-i$. They are orthogonal, as normality guarantees. Normalising and using them as columns of $U_0$ gives $U_0^*UU_0 =\left(\begin{smallmatrix}i&0\\0&-i\end{smallmatrix}\right)$.

The real spectral theorem

Theorem Spectral theorem, real case

Let $T$ be an operator on a finite-dimensional real inner product space. Then $T$ is self-adjoint if and only if $V$ has an orthonormal basis of eigenvectors. In matrix terms: a real matrix is orthogonally diagonalizable iff it is symmetric, $A=QDQ^t$ with $Q$ orthogonal and $D$ real diagonal.

Proof Induction, using Unit 5
Unit 5 gave a real eigenvalue $\lambda$ with a real unit eigenvector $v$.
$\{v\}^\perp$ is $T$-invariant: for $w\perp v$, $\langle Tw,v\rangle=\langle w,Tv\rangle=\lambda\langle w,v\rangle=0$, using $T=T^*$.
The restriction of $T$ to $\{v\}^\perp$ is still self-adjoint, so induction gives an orthonormal eigenbasis there. Adjoining $v$ finishes. ∎
Normal is not enough over $\mathbb{R}$

The rotation $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ is normal — indeed orthogonal — but has no real eigenvalue, so no real eigenbasis exists. Over $\mathbb{R}$ the correct hypothesis is symmetry, not normality. The real classification of normal operators instead gives a block form with $2\times2$ rotation blocks.

Example Orthogonally diagonalize by hand

$A=\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$, symmetric.

$f(t)=(2-t)^2-1=t^2-4t+3$, so eigenvalues $3$ and $1$.
$E_3=\operatorname{span}\{(1,1)\}$, $E_1=\operatorname{span}\{(1,-1)\}$ — orthogonal, as the theorem promises.
Normalise into $Q=\tfrac{1}{\sqrt2}\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix} \right)$. Then $Q^tAQ=\left(\begin{smallmatrix}3&0\\0&1\end{smallmatrix} \right)$, and $Q^t$ is the inverse for free.

Spectral resolution

Definition

Let $T$ be normal (complex) or self-adjoint (real) with distinct eigenvalues $\lambda_1,\dots,\lambda_k$ and eigenspaces $E_1,\dots,E_k$. Writing $P_i$ for the orthogonal projection onto $E_i$, the spectral resolution is

$T=\sum_{i=1}^k\lambda_iP_i$, with $\sum_iP_i=I$, $P_iP_j=0$ for $i\ne j$, and $P_i^*=P_i=P_i^2$.

Equivalently $V=E_1\oplus\cdots\oplus E_k$, an orthogonal direct sum.

Theorem Functions of an operator

For any polynomial $p$, $p(T)=\sum_i p(\lambda_i)P_i$. More generally this defines $f(T)$ for any function $f$ on the spectrum — including $\sqrt{T}$ when the eigenvalues are non-negative, which Unit 7 needs.

Since $P_iP_j=0$ for $i\ne j$ and $P_i^2=P_i$, powers telescope: $T^m=\sum_i\lambda_i^mP_i$. Extend by linearity.

Example Resolve the last example

For $A=\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$ with eigenvalues $3,1$:

$P_3=\tfrac12\left(\begin{smallmatrix}1&1\\1&1 \end{smallmatrix}\right)$ and $P_1=\tfrac12\left(\begin{smallmatrix}1&-1\\-1&1\end{smallmatrix}\right)$. Check $P_3+P_1=I$ and $P_3P_1=0$. ✓
$3P_3+1\cdot P_1 =\tfrac12\left(\begin{smallmatrix}3+1&3-1\\3-1&3+1\end{smallmatrix}\right) =\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)=A$. ✓
And $\sqrt{A}=\sqrt3P_3+P_1$, whose square is $A$ — a square root obtained without solving any equations.
Common pitfalls

• Quoting "normal implies diagonalizable" over $\mathbb{R}$. Only symmetry works there.
• Forgetting to normalise the eigenvectors. $Q$ must be orthogonal, so its columns need norm $1$; otherwise $Q^t\ne Q^{-1}$.
• Assuming eigenvectors within a single eigenspace are automatically orthogonal. Vectors from distinct eigenvalues are; inside a repeated eigenspace you must run Gram–Schmidt yourself.
• Writing $A=QDQ^{-1}$ and stopping. The content of the theorem is that $Q$ can be taken orthogonal or unitary, so $Q^{-1}=Q^*$.

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 · Orthogonal diagonalization

Orthogonally diagonalize $\left(\begin{smallmatrix}3&1\\1&3\end{smallmatrix}\right)$.

Answer
Eigenvalues $4$ and $2$ with eigenvectors $(1,1)$ and $(1,-1)$. With $Q=\tfrac{1}{\sqrt2}\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix} \right)$, $Q^tAQ=\left(\begin{smallmatrix}4&0\\0&2\end{smallmatrix} \right)$.
Practice 2 · Repeated eigenvalue

Orthogonally diagonalize the $3\times3$ matrix with all entries $1$.

Solution
Eigenvalue $3$ with eigenvector $(1,1,1)$, and eigenvalue $0$ with the plane $x+y+z=0$. Run Gram–Schmidt inside that plane on $(1,-1,0)$ and $(1,0,-1)$ to get an orthonormal pair, then normalise $(1,1,1)$. The repeated eigenvalue is where the extra work lives.
Practice 3 · Which theorem applies

Can $\left(\begin{smallmatrix}0&-2\\2&0\end{smallmatrix}\right)$ be orthogonally diagonalized over $\mathbb{R}$?

Answer
No. It is normal but not symmetric, and its characteristic polynomial $t^2+4$ has no real root. Over $\mathbb{C}$ it is unitarily diagonalizable with eigenvalues $\pm2i$.
Practice 4 · Schur by hand

Find a unitary $U$ making $A=\left(\begin{smallmatrix}1&1\\0&2\end{smallmatrix}\right)$ upper triangular.

Solution
It already is upper triangular, so $U=I$ works. Note $A$ is not normal, so no unitary makes it diagonal — Schur is strictly weaker than the spectral theorem.
Practice 5 · Spectral resolution

Write the spectral resolution of $\left(\begin{smallmatrix}5&0\\0&5\end{smallmatrix}\right)$.

Solution
One eigenvalue $5$ with $E_5=\mathbb{R}^2$, so $P_1=I$ and $T=5I$. A single term — scalar operators are the degenerate case.
Practice 6 · Square root

Find a self-adjoint square root of $A=\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$.

Solution
From the resolution $A=3P_3+P_1$, take $\sqrt A=\sqrt3P_3+P_1 =\tfrac12\left(\begin{smallmatrix}\sqrt3+1&\sqrt3-1\\ \sqrt3-1&\sqrt3+1\end{smallmatrix}\right)$. Squaring returns $A$.
Practice 7 · Converse direction

Prove that if $A=UDU^*$ with $U$ unitary and $D$ diagonal then $A$ is normal.

Solution
$A^*=UD^*U^*$, so $A^*A=UD^*DU^*$ and $AA^*=UDD^*U^*$. Diagonal matrices commute, so $D^*D=DD^*$ and the two agree.
Practice 8 · Commuting operators

If $S$ and $T$ are commuting self-adjoint operators, show they are simultaneously orthogonally diagonalizable.

Solution
Each eigenspace $E_i$ of $T$ is $S$-invariant: if $Tv=\lambda v$ then $T(Sv)=S(Tv)=\lambda Sv$. Restrict $S$ to each $E_i$, where it is still self-adjoint, and apply the spectral theorem inside. Collecting the bases gives one orthonormal basis diagonalizing both.

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

Schur triangularizes any complex operator; normality then forces the triangle to be diagonal. Over $\mathbb{R}$, Unit 5's existence result plus invariance of $v^\perp$ runs the same induction for symmetric operators.

Study routine. After every orthogonal diagonalization, verify $Q^tQ=I$ and recompute $QDQ^t$. Both checks are quick and catch the normalisation slip that costs the most marks.

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 · Hypothesis by field

What hypothesis does the spectral theorem need over $\mathbb{C}$, and over $\mathbb{R}$?

Guide
Normal over $\mathbb{C}$; self-adjoint over $\mathbb{R}$. Normality alone is not enough over $\mathbb{R}$, as rotations show.
Extension 2 · Role of Schur

Why prove Schur's theorem first?

Guide
It supplies an orthonormal basis with $[T]$ upper triangular for any complex operator. Normality is then only needed to knock out the entries above the diagonal.
Extension 3 · Matrix statement

Write the theorem as a factorization.

Guide
$A=UDU^*$ with $U$ unitary, $D$ diagonal, over $\mathbb{C}$ for normal $A$; and $A=QDQ^t$ with $Q$ orthogonal, $D$ real diagonal, over $\mathbb{R}$ for symmetric $A$.
Extension 4 · Using the resolution

Given $T=\sum\lambda_iP_i$, what is $T^{10}$?

Guide
$\sum\lambda_i^{10}P_i$. The projections are orthogonal idempotents, so cross terms vanish and only the eigenvalues get raised to the power.
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