MAT240H1/Unit 10

Unit 10 · Algebra I

Eigenvalues & Diagonalization

The finale: directions the map merely stretches. Find enough of them and the operator becomes diagonal — the simplest matrix it can ever have. Everything in the course converges here: determinants find the eigenvalues, null spaces find the eigenvectors, and change of basis cashes them in.

By the end you can

• Compute eigenvalues via the characteristic polynomial and eigenvectors via null spaces.
• Prove eigenvectors for distinct eigenvalues are independent.
• Decide diagonalizability and carry it out.
• State Cayley–Hamilton and verify it in examples.

Why it matters & where it comes from

Why it exists

Diagonalizing decouples a linear system into independent one-dimensional ones. Powers, exponentials, and long-run behaviour of a matrix all become trivial in an eigenbasis.

In the real world

Google's PageRank is an eigenvector; principal component analysis diagonalizes a covariance matrix; vibration modes, quantum energy levels, and Markov chain limits are all eigenvalue problems.

Story & open questions

"Eigen" is Hilbert's German for "own/characteristic." Cayley stated the Cayley–Hamilton theorem in 1858 after checking it on small matrices. When diagonalization fails, the Jordan canonical form (MAT247) finishes the classification.

Eigenvalues and eigenvectors

Definition

A nonzero $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ if $Av=\lambda v$. The eigenspace is $E_\lambda=N(A-\lambda I)$ — a subspace, consisting of all eigenvectors for $\lambda$ together with $0$.

Theorem How to find them

$\lambda$ is an eigenvalue of $A$ iff $\det(A-\lambda I)=0$: an eigenvector is a nonzero vector in $N(A-\lambda I)$, which exists exactly when $A-\lambda I$ is singular. So the eigenvalues are the roots of the characteristic polynomial $f(t)=\det(A-tI)$, and each eigenspace is computed by row reduction (Unit 7).

Example Full run

$A=\left(\begin{smallmatrix}2&1\\0&3\end{smallmatrix}\right)$.

$f(t)=\det\!\left(\begin{smallmatrix}2-t&1\\0&3-t \end{smallmatrix}\right)=(2-t)(3-t)$: eigenvalues $2$ and $3$.

$E_2=N\!\left(\begin{smallmatrix}0&1\\0&1\end{smallmatrix} \right)=\operatorname{span}\{(1,0)\}$; $E_3=N\!\left(\begin{smallmatrix}-1&1\\0&0\end{smallmatrix}\right) =\operatorname{span}\{(1,1)\}$ — check $A(1,1)=(3,3)$. ✓

Two independent eigenvectors in $\mathbb{R}^2$: with $P=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$, $P^{-1}AP=\left(\begin{smallmatrix}2&0\\0&3\end{smallmatrix}\right)$.

Independence and the diagonalizability test

Theorem Distinct eigenvalues

Eigenvectors belonging to distinct eigenvalues are linearly independent. Hence an $n\times n$ matrix with $n$ distinct eigenvalues is diagonalizable.

Proof (two eigenvalues; induction handles more)

Suppose $av_1+bv_2=0$ with $Av_i=\lambda_iv_i$, $\lambda_1\ne\lambda_2$. Apply $A$: $a\lambda_1v_1+b\lambda_2v_2=0$.

Subtract $\lambda_1$ times the first equation: $b(\lambda_2-\lambda_1)v_2=0$. Since $\lambda_2\ne\lambda_1$ and $v_2\ne0$, $b=0$; then $av_1=0$ gives $a=0$. ∎

Theorem Diagonalizability

$A$ is diagonalizable iff there is a basis of eigenvectors — equivalently, the eigenspace dimensions add up to $n$. Each eigenspace dimension (geometric multiplicity) is at least $1$ and at most the root's multiplicity in $f(t)$ (algebraic multiplicity); diagonalizable means every eigenvalue reaches that ceiling and $f$ splits over the field. Then $P$ (eigenvectors as columns) gives $P^{-1}AP=D$.

Example Two failure modes

Defective: $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ has only the eigenvalue $1$ with algebraic multiplicity $2$, but $E_1=N\!\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ is only one-dimensional — no second independent eigenvector, not diagonalizable.

Wrong field: the rotation $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ has $f(t)=t^2+1$ with no real roots — no real eigenvalues at all. Over $\mathbb{C}$ it has eigenvalues $\pm i$ and diagonalizes. Unit 9's field-dependence, live.

Why diagonal is the jackpot If $A=PDP^{-1}$ then $A^k=PD^kP^{-1}$, and powering a diagonal matrix means powering numbers. Linear recurrences, Markov chains, and systems of ODEs are solved by exactly this move.

Cayley–Hamilton

Theorem Cayley–Hamilton

Every square matrix satisfies its own characteristic polynomial: $f(A)=0$. Consequently the minimal polynomial (Unit 9) divides the characteristic polynomial.

Example Verify it

For $A=\left(\begin{smallmatrix}2&1\\0&3\end{smallmatrix}\right)$, $f(t)=t^2-5t+6$.

$A^2=\left(\begin{smallmatrix}4&5\\0&9\end{smallmatrix} \right)$, so $A^2-5A+6I=\left(\begin{smallmatrix}4-10+6&5-5\\0&9-15+6\end{smallmatrix} \right)=0$. ✓

Warning: "substitute $A$ into $\det(A-tI)$" is not a proof — $\det(A-AI)=\det 0=0$ is a scalar statement, not the matrix identity $f(A)=0$. The real proof is one of the highlights of the course.

Common pitfalls

• Eigenvectors are nonzero by definition; $E_\lambda$ contains $0$ but $0$ is never an eigenvector.
• Distinct eigenvalues guarantee diagonalizability, but repeated ones do not forbid it ($I$ has one eigenvalue and is already diagonal).
• Check geometric multiplicities — a repeated root with a thin eigenspace is exactly how diagonalization fails.
• Eigenvalues live in the field: real matrices can have complex eigenvalues, and "diagonalizable" must name the field.

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 · Full run

Find the eigenvalues and eigenvectors of $\left(\begin{smallmatrix}2&1\\0&3\end{smallmatrix}\right)$.

Answer

$\lambda=2$ with $(1,0)$; $\lambda=3$ with $(1,1)$. Diagonalizable, since the two are independent.

Practice 2 · Defective

Is $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ diagonalizable?

Answer

No: the only eigenvalue is $1$ and its eigenspace is $\operatorname{span}\{(1,0)\}$, dimension $1<2$.

Practice 3 · Symmetric swap

Diagonalize $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$.

Solution

$f(t)=t^2-1$: eigenvalues $1$ and $-1$, eigenvectors $(1,1)$ and $(1,-1)$. With those as the columns of $P$, $P^{-1}AP=\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)$.

Practice 4 · Independence

Prove: eigenvectors for two distinct eigenvalues are independent.

Solution

From $av_1+bv_2=0$, apply $A$ and subtract $\lambda_1$ times the original: $b(\lambda_2-\lambda_1)v_2=0$, so $b=0$, then $a=0$.

Practice 5 · Field matters

Show the rotation $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ has no real eigenvalues but two complex ones.

Solution

$f(t)=t^2+1$ has no real roots; over $\mathbb{C}$ the eigenvalues are $i$ and $-i$ (with eigenvectors $(1,-i)$ and $(1,i)$).

Practice 6 · Cayley–Hamilton

Verify Cayley–Hamilton for $A=\left(\begin{smallmatrix}2&1\\0&3\end{smallmatrix}\right)$.

Solution

$f(t)=t^2-5t+6$ and $A^2=\left(\begin{smallmatrix}4&5\\0&9\end{smallmatrix}\right)$; compute $A^2-5A+6I=0$ entry by entry.

Practice 7 · Triangular

Prove the eigenvalues of a triangular matrix are its diagonal entries.

Solution

$A-tI$ is triangular, so $\det(A-tI)=\prod_i(a_{ii}-t)$ (Unit 8) — the roots are exactly the diagonal entries.

Practice 8 · Projections again

Suppose $A=PDP^{-1}$ where $D$ is diagonal with only $0$s and $1$s on the diagonal. Prove $A^2=A$.

Solution

$D^2=D$ since $0^2=0$ and $1^2=1$; then $A^2=PD^2P^{-1}=PDP^{-1}=A$ — diagonalization turns matrix identities into scalar ones.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT240H1 topic map. It should cover the following items before you move on:

• Eigenvectors. Solve $Tv=\lambda v$ or $(A-\lambda I)v=0$ and distinguish eigenvalues from eigenvectors.
• Characteristic polynomial. Use det$(xI-A)$ or det$(A-\lambda I)$ consistently to find eigenvalues.
• Diagonalization. Use eigenbasis criteria and geometric/algebraic multiplicity.
• Cayley-Hamilton and minimal polynomial. Use polynomial identities in $A$ to connect characteristic and minimal polynomials.

Textbook/source alignment

Use the current UofT MAT240H1 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

The eigenvalue spine is: roots of the characteristic polynomial give candidate eigenvalues, eigenspaces give eigenvectors, and enough eigenvectors give diagonalization.

Study routine. For every diagonalization problem, compute eigenspace dimensions; factoring the characteristic polynomial is only the first half.

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 · Eigenvalue equation

What equation determines eigenvectors for eigenvalue $\lambda$?

Guide

Solve $(A-\lambda I)v=0$ for nonzero $v$.

Extension 2 · Diagonalizable criterion

When is an $n\times n$ matrix diagonalizable over $F$?

Guide

It is diagonalizable over $F$ iff $F^n$ has a basis of eigenvectors for the matrix, equivalently the sum of eigenspace dimensions is $n$.

Extension 3 · Cayley-Hamilton

State the Cayley-Hamilton theorem.

Guide

Every square matrix satisfies its own characteristic polynomial: if $p_A(t)=\det(tI-A)$, then $p_A(A)=0$.

Extension 4 · Minimal polynomial test

State the minimal-polynomial criterion for diagonalizability when it splits.

Guide

The operator is diagonalizable iff its minimal polynomial splits into distinct linear factors.

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