MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: diagonalization and its applications (MATH 133)

Q: How do I know if a matrix is diagonalizable without finding P?

Factor the characteristic polynomial. If the eigenvalues are all distinct, the matrix is diagonalizable and you can stop. For each repeated eigenvalue, row reduce A minus lambda I and count the free variables: that number is the dimension of the eigenspace. The matrix is diagonalizable exactly when every such dimension equals the number of times the eigenvalue is repeated.

Q: Does a repeated eigenvalue mean the matrix is not diagonalizable?

No. A repeated eigenvalue is only the place where diagonalization can fail. The identity matrix has a single repeated eigenvalue and is already diagonal. What decides is the dimension of the eigenspace of that eigenvalue: if it equals the multiplicity of the root, the matrix is diagonalizable; if it is smaller, it is not.

Q: In which order do I put the eigenvalues in D?

In the same order as the eigenvectors in the columns of P. The first diagonal entry of D is the eigenvalue of the first column of P, and so on. Any order is allowed, but the two lists must move together: swapping two columns of P without swapping the matching entries of D gives a wrong answer even when every number in it is correct.

Q: Why does diagonalization make powers of a matrix easy?

Because A equals P times D times the inverse of P, and in a product of such factors every inverse of P cancels the P next to it. What remains is P times D to the n times the inverse of P, and a diagonal matrix is raised to a power entry by entry. Geometrically, each eigenvector axis is simply stretched by its eigenvalue, n times over.

Q: How do I find the steady state of a two-state Markov chain?

Write the transition matrix with columns adding up to one, so that one is an eigenvalue. Find an eigenvector for the eigenvalue one, then scale it so that its entries add up to the total, or to one for proportions. The other eigenvalue is smaller than one in size, so its part of the solution dies out and every starting point with the same total ends at that steady state.

This sheet is not a summary of the diagonalization chapter: you already have Nicholson and your lecture notes. It answers one question only, what makes students lose marks on diagonalization and its applications in MATH 133 at McGill University, and which precise gesture avoids each loss.

Everything below is done by hand, as the final requires: characteristic polynomials that factor, ranks read off a row reduction whose operations are named, and a check $AP = PD$ that costs two or three matrix-vector products. Orthogonal diagonalization, complex eigenvalues and the Jordan form are not in this course and do not appear here.

The thread of the chapter

Diagonalizable is a COUNTING question: are there $n$ independent eigenvectors or not? A repeated eigenvalue is not a verdict, the dimension of its eigenspace is, and it costs one rank of $A - \lambda I$ . Once $P$ is built, powers, long-run behaviour, Markov chains and recurrences are all read in the eigenvector axes, where $A$ only stretches.

This chapter is part of MATH 133, Linear Algebra and Geometry (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Diagonalizable matrix
Characteristic polynomial
Algebraic multiplicity
Geometric multiplicity
Eigenspace
Powers of a matrix
Similar matrices
Linear dynamical system
Dominant eigenvalue
Markov chain
Steady-state vector
Linear recurrence relation

The essentials

Diagonalizable means: enough independent eigenvectors

• $A$ ( $n \times n$ ) is diagonalizable when $P^{-1}AP = D$ for some invertible $P$ and diagonal $D$ ; equivalently $A = PDP^{-1}$ , with $P$ on the LEFT.
• Criterion: $A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors. They are the COLUMNS of $P$ , and the $i$ -th diagonal entry of $D$ is the eigenvalue of the $i$ -th column.
• Algebraic multiplicity of $\lambda$ : its multiplicity as a root of $\det(A - \lambda I)$ . Geometric multiplicity: $\dim E_\lambda = n - \text{rank}(A - \lambda I)$ . Always $1 \le \dim E_\lambda \le$ algebraic multiplicity.
• Diagonalizable if and only if $\dim E_\lambda$ equals the algebraic multiplicity for EVERY eigenvalue. A simple eigenvalue always passes, so only the repeated ones need the test.
• $n$ distinct eigenvalues: diagonalizable, nothing else to check. The converse is false: $2I$ has one eigenvalue and is already diagonal.

Same polynomial

(\lambda - 2)^2

: for

2I

every direction is an eigenvector; for

M = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}

only the horizontal line is, so no

P

exists.

The polynomial gives the eigenvalues; it never gives the verdict. Two matrices with the same polynomial can land on opposite sides, and the only thing that separates them is the rank of $A - \lambda I$ at the repeated eigenvalue.

What diagonalization buys, once $P$ is built

• Powers: $A^n = PD^nP^{-1}$ , and ONLY $D$ is raised entry by entry: $D^n = \text{diag}(\lambda_1^n, \dots, \lambda_n^n)$ .
• Dynamical system $v_{k+1} = Av_k$ : write $v_0 = c_1u_1 + \dots + c_nu_n$ in eigenvectors, then $v_k = c_1\lambda_1^ku_1 + \dots + c_n\lambda_n^ku_n$ . The coefficients $c_i$ are $P^{-1}v_0$ .
• Long run: the eigenvalue of largest absolute value sets the growth factor, and its eigenvector sets the eventual PROPORTIONS, provided its coefficient $c_1$ is not zero.
• Markov chain: the columns of $T$ add up to $1$ , so $1$ is an eigenvalue; the steady state is its eigenvector scaled to the total.
• Recurrence $x_{k+2} = ax_{k+1} + bx_k$ : set $V_k = (x_k, x_{k+1})$ , so $V_{k+1} = \begin{pmatrix} 0 & 1 \\ b & a \end{pmatrix}V_k$ , and diagonalize.
• Similar matrices $B = P^{-1}AP$ share determinant, trace, rank, characteristic polynomial and eigenvalues. Sharing all of them does NOT prove similarity.

Every application is the same move: rewrite the starting vector in the eigenvector axes, let each coordinate be multiplied by its own eigenvalue, then come back.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Similarity: what survives a change of basis, and what does not

Read this grid to DISPROVE similarity: one quantity that differs settles the question. A red case is never an answer, it is an order to go through a diagonal matrix instead.

Quantity	For $B = P^{-1}AP$	Shared?
$\det$	$\det B = \det A$	yes
Example: $\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ and $\text{diag}(4, -1)$ : both determinants equal $-4$ .
trace	$\text{tr}\,B = \text{tr}\,A$	yes
Example: $1 + 2 = 3 = 4 + (-1)$ for the same pair.
$\text{rank}(A - \lambda I)$	$P^{-1}(A - \lambda I)P$	yes
Example: $\text{rank}(M - 2I) = 1$ but $\text{rank}(2I - 2I) = 0$ : $M = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ and $2I$ are not similar.
eigenvectors	$u$ for $B$ gives $Pu$ for $A$	moved not shared
Example: $B = D = \text{diag}(4, -1)$ has eigenvector $(1, 0)$ ; $A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ has $P(1,0) = (2, 3)$ instead. Same form, other result: Same eigenvalue $4$ for both matrices, two different eigenvectors: $A(1,0) = (1, 3)$ is not a multiple of $(1, 0)$ . What to do: Transport them: an eigenvector $u$ of $B = P^{-1}AP$ gives the eigenvector $Pu$ of $A$ .
same polynomial	so similar?	no rule that does not exist
Example: $M = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ and $2I$ both give $(\lambda - 2)^2$ , but $P^{-1}(2I)P = 2I$ for every $P$ , never $M$ . What to do: To PROVE similarity, show both matrices are similar to the same diagonal $D$ , eigenvalues in the same order.

The last two rows are the whole exam question: invariants can only say no.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Declaring a double eigenvalue defective without computing a rank

the whole question, often 4 or 5 marks

What not to write

“ $\lambda = 3$ is a double root of $(3-\lambda)^2(1-\lambda)$ , so $A$ is not diagonalizable.”

What to write

“ $\lambda = 3$ is a double root; $A - 3I$ reduces to one non-zero row, so $\dim E_3 = 3 - 1 = 2$ , equal to the algebraic multiplicity, and $A$ IS diagonalizable.”

Why: A repeated root is where the count CAN fail, not where it does. The matrices $\begin{pmatrix} 3 & 0 & 0 \\ 1 & 2 & 1 \\ -1 & 1 & 2 \end{pmatrix}$ and $\begin{pmatrix} 3 & 0 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$ share that polynomial and differ by one sign: ranks $1$ and $2$ for $A - 3I$ , opposite verdicts.

2. Taking one eigenvector per eigenvalue, whatever its multiplicity

3 marks: a $3 \times 2$ matrix has no inverse

What not to write

“One eigenvector for $\lambda = 3$ and one for $\lambda = 1$ , so $P = \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 0 & -1 \end{pmatrix}$ .”

What to write

“ $\dim E_3 = 2$ , so $\lambda = 3$ contributes TWO columns: $P = \begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & -1 \end{pmatrix}$ and $D = \text{diag}(3, 3, 1)$ .”

Why: $P$ must be square and invertible, so it needs $n$ columns. A double eigenvalue that passes the rank test hands you a basis of TWO vectors, and both go into $P$ , next to two copies of the eigenvalue in $D$ .

3. Listing the eigenvalues in D in a different order from the columns of P

2 marks, although every number on the line is right

What not to write

“ $P = \begin{pmatrix} 1 & 2 \\ -1 & 3 \end{pmatrix}$ and $D = \text{diag}(4, -1)$ .”

What to write

“The first column $(1, -1)$ belongs to $-1$ , so $P = \begin{pmatrix} 1 & 2 \\ -1 & 3 \end{pmatrix}$ goes with $D = \text{diag}(-1, 4)$ .”

Why: $AP = PD$ multiplies the $i$ -th column of $P$ by the $i$ -th entry of $D$ . For $\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ , swap one list and you must swap the other, or $AP \neq PD$ .

4. Putting the eigenvectors in the rows of P

the whole construction, and $P^{-1}AP$ comes out non-diagonal

What not to write

“The eigenvectors are $(2, 3)$ and $(1, -1)$ , so $P = \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}$ .”

What to write

“The eigenvectors are the COLUMNS: $P = \begin{pmatrix} 2 & 1 \\ 3 & -1 \end{pmatrix}$ .”

Why: The product $AP$ applies $A$ to the columns of $P$ and never reads its rows. With the rows version, $P^{-1}AP = \frac{1}{5}\begin{pmatrix} 28 & 22 \\ -12 & -13 \end{pmatrix}$ , which is the check that should have raised the alarm.

5. Writing A as the inverse of P times D times P

every entry of $A^n$, typically 3 marks

What not to write

“From $P^{-1}AP = D$ I get $A = P^{-1}DP$ , so $A^n = P^{-1}D^nP$ .”

What to write

“Multiplying by $P$ on the left and by $P^{-1}$ on the right: $A = PDP^{-1}$ and $A^n = PD^nP^{-1}$ .”

Why: With $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $D = \text{diag}(1, 2)$ , $PDP^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ but $P^{-1}DP = \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}$ . The check at $n = 1$ catches it at once: the formula must give back $A$ .

6. Raising the entries of A to the power n

the whole power question

What not to write

“ $A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$ , so $A^2 = \begin{pmatrix} 16 & 4 \\ 1 & 1 \end{pmatrix}$ .”

What to write

“ $A^2 = AA = \begin{pmatrix} 14 & -10 \\ 5 & -1 \end{pmatrix}$ ; only the DIAGONAL matrix $D$ is raised entry by entry.”

Why: Entry-by-entry powers are correct for a diagonal matrix and for nothing else, which is precisely why we go through $D$ . The formula $A^n = PD^nP^{-1}$ gives $\begin{pmatrix} 46 & -38 \\ 19 & -11 \end{pmatrix}$ at $n = 3$ , and direct multiplication agrees.

7. Quoting the eigenvector as the steady state of a Markov chain

2 marks, and an answer that loses $495$ bikes

What not to write

“The eigenvector for $\lambda = 1$ is $(3, 2)$ , so in the long run there are $3$ bikes at Hilltop and $2$ at Riverside.”

What to write

“The eigenvector $(3, 2)$ scaled to the $500$ bikes gives the steady state $(300, 200)$ , that is $60$ and $40$ per cent.”

Why: An eigenvector is only a direction; the steady state is the point of that line that has the right TOTAL, because a Markov matrix conserves it. Divide by the sum of the entries, then multiply by the total.

8. Reading a stable ratio as a stable population

the interpretation mark, usually 2 marks

What not to write

“The ratio of the two classes tends to $2 : 1$ , so the population stabilises.”

What to write

“The PROPORTIONS settle along the dominant eigenvector; the SIZES are multiplied by $\lambda_1 = 0.8$ each step, so the population dies out in the ratio $2 : 1$ .”

Points

x_k = 8(0.8)^k(2, 1) + 2(0.4)^k(-1, 1)

: they line up on the dashed line of ratio

2 : 1

while heading to the origin. Stable ratio, vanishing population.

Why: Two different limits are hiding in one sentence. The direction of $v_k$ always tends to the dominant eigenvector; its length follows $|\lambda_1|^k$ , and only $\lambda_1 = 1$ gives a population that stabilises.

Which method to choose

Is $A$ diagonalizable? Decide on the form

Look at the eigenvalues first, then spend a rank test only on the repeated ones

If the matrix is triangular → its eigenvalues are the diagonal entries, no determinant to expand

Example: $\begin{pmatrix} 2 & a & 1 \\ 0 & 2 & b \\ 0 & 0 & -1 \end{pmatrix}$ has $2, 2, -1$
If $n$ distinct eigenvalues → diagonalizable, stop: distinct eigenvalues give independent eigenvectors

Example: $\lambda^2 - 3\lambda - 4 = (\lambda - 4)(\lambda + 1)$ : yes
If an eigenvalue $\lambda$ repeated $m$ times → row reduce $A - \lambda I$ ; it passes exactly when $n - \text{rank} = m$

Example: $n = 3$ , $m = 2$ : rank $1$ passes, rank $2$ fails
If a $2 \times 2$ with a double eigenvalue $\lambda$ → diagonalizable only if $A = \lambda I$ already

Example: $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ : no; $\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ : yes
If the characteristic polynomial has no real root → no real eigenvector, so no real $P$ ; say so and stop, complex eigenvalues are not in MATH 133

Example: a rotation by $90^{\circ}$ : $\lambda^2 + 1$ , no real root

Notice what never decides: the size of the entries, the look of the matrix, or the fact that a root is repeated. A triangular parameter can matter or not depending on WHERE it sits: in $\begin{pmatrix} 2 & a & 1 \\ 0 & 2 & b \\ 0 & 0 & -1 \end{pmatrix}$ , only $a = 0$ counts.

Long-run behaviour of $v_{k+1} = Av_k$

Order the eigenvalues by absolute value and read the largest one, $\lambda_1$

The same starting size under three dominant eigenvalues:

1.2

explodes,

1

stays level,

0.8

dies out. The step from one regime to the next is where

\lambda_1

crosses

1

If $|\lambda_1| > 1$ → the system grows by the factor $\lambda_1$ per step, in the proportions of $u_1$

Example: $\lambda_1 = 1.5$ , $u_1 = (2, 1)$ : $+50$ per cent a year, two juveniles per adult
If $\lambda_1 = 1$ , all the others smaller in absolute value → convergence to $c_1u_1$ , a steady state

Example: Markov chain with $\lambda = 1, 0.5$ : $(300, 200)$
If $|\lambda_1| < 1$ → everything tends to $0$ , in the proportions of $u_1$

Example: $\lambda = 0.9$ and $0.7$ : extinction at $10$ per cent a year
If $v_0$ has no component along $u_1$ , that is $c_1 = 0$ → the next eigenvalue takes over

Example: $\text{diag}(2, 0.5)$ with $v_0 = (0, 1)$ : $v_k \to 0$

The eigenvalue decides the SIZE, the eigenvector decides the SHAPE. An answer that gives one without the other has answered half the question.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Diagonalize a matrix: the answer a marker can grade

When to use it: “Diagonalize $A$ ”, “find an invertible $P$ and a diagonal $D$ ”, or “is $A$ diagonalizable?”

1 Write $\det(A - \lambda I)$ , say along which row or column you expand and why (the one with the most zeros), and FACTOR it: list the eigenvalues with their algebraic multiplicities.
2 For each eigenvalue, row reduce $A - \lambda I$ naming every operation ( $R_3 \to R_3 + R_1$ ), then write its rank and $\dim E_\lambda = n - \text{rank}$ , with a basis of $E_\lambda$ .
3 Compare each $\dim E_\lambda$ with the algebraic multiplicity. If one falls short, stop and write the verdict with the eigenvalue that fails.
4 Build $P$ from the basis vectors AS COLUMNS and $D$ with the eigenvalues in the same order, a repeated eigenvalue appearing as many times as it has columns.
5 Check $AP = PD$ column by column and state $\det P \neq 0$ .

Concluding sentence

“The eigenvalues are $3$ , of algebraic multiplicity $2$ , and $1$ . Since $\dim E_3 = 2$ and $\dim E_1 = 1$ , $A$ has $3$ linearly independent eigenvectors, so it is diagonalizable, with $P = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \end{pmatrix}$ and $D = \text{diag}(3, 3, 1)$ , so that $P^{-1}AP = D$ .”

The trap: Writing the verdict before the rank of $A - \lambda I$ has been computed for the repeated eigenvalue.

Marking: Typically 2 marks for the factored polynomial, 3 for the eigenspaces with their dimensions, 1 for the verdict with its reason, 2 for $P$ and $D$ in matching order and 2 for the check. A correct $P$ with no reasoning is worth about a quarter of the question.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

$AP = PD$ , column by column

Multiply $A$ by each column of $P$ : you must get that same column times the matching diagonal entry of $D$ . Two or three matrix-vector products, and no inverse to compute.

$A(2, 3) = (8, 12) = 4(2, 3)$ for $A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ : correct. A result such as $(8, 11)$ says the eigenvector is wrong.
Trace and determinant against $D$

The diagonal of $D$ must add up to the trace of $A$ and multiply to $\det A$ . Five seconds, and it catches a wrong eigenvalue before any eigenvector is computed.

$1 + 2 = 3 = 4 + (-1)$ and $2 - 6 = -4 = 4 \times (-1)$ . A trace that disagrees means a slip in the polynomial.
The power formula at $n = 0$ and $n = 1$

Any formula for $A^n$ must give $I$ at $n = 0$ and $A$ itself at $n = 1$ . A $P^{-1}$ on the wrong side fails the second test at once.

$A^n = \begin{pmatrix} 2 \cdot 3^n - 2^n & 2^{n+1} - 2 \cdot 3^n \\ 3^n - 2^n & 2^{n+1} - 3^n \end{pmatrix}$ gives $I$ , then $\begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$ : consistent.
A steady state, three conditions

Entries non-negative, adding up to the total you started with, and $Tx = x$ exactly. A steady state that fails the total is an unscaled eigenvector.

$(300, 200)$ : $0.8 \times 300 + 0.3 \times 200 = 300$ and $0.2 \times 300 + 0.7 \times 200 = 200$ .

The typical problem, taken apart

A $3 \times 3$ with a double eigenvalue, from polynomial to check

Decide whether $A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -2 & 3 \end{pmatrix}$ is diagonalizable. If it is, give an invertible $P$ and a diagonal $D$ with $P^{-1}AP = D$ .

Step 1

Expand along ROW 2, which holds two zeros: $\det(A - \lambda I) = (3 - \lambda)\det\begin{pmatrix} 1-\lambda & 0 \\ 2 & 3-\lambda \end{pmatrix} = (3 - \lambda)^2(1 - \lambda)$ . Eigenvalues: $3$ with algebraic multiplicity $2$ , and $1$ .

Why

The choice of row is worth saying: it turns a $3 \times 3$ expansion into one $2 \times 2$ that is already triangular. The polynomial comes out factored, which is the only form in which multiplicities can be read.

Step 2

$A - 3I = \begin{pmatrix} -2 & 2 & 0 \\ 0 & 0 & 0 \\ 2 & -2 & 0 \end{pmatrix}$ ; $R_3 \to R_3 + R_1$ gives a second zero row, so the rank is $1$ and $\dim E_3 = 3 - 1 = 2$ . The equation $x = y$ leaves $y$ and $z$ free: $E_3 = \text{span}\{(1, 1, 0), (0, 0, 1)\}$ .

Why

This is the step that decides. The double root keeps a plane of eigenvectors, so the verdict is yes, and it is a rank that says so, not the polynomial.

Step 3

$A - I = \begin{pmatrix} 0 & 2 & 0 \\ 0 & 2 & 0 \\ 2 & -2 & 2 \end{pmatrix}$ ; $R_2 \to R_2 - R_1$ kills row $2$ , so $y = 0$ and $2x - 2y + 2z = 0$ gives $x = -z$ : $E_1 = \text{span}\{(1, 0, -1)\}$ .

Why

A simple eigenvalue can never fail, so this step is not a test: it only fills the last column of $P$ .

Step 4

$P = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \end{pmatrix}$ and $D = \text{diag}(3, 3, 1)$ , the two columns of $E_3$ first because the two $3$ come first in $D$ .

Why

Columns, not rows, and the same order in both lists: two independent marks on a typical grid, lost by students who had every number right.

Step 5

Check: $A(1, 1, 0) = (3, 3, 0)$ , $A(0, 0, 1) = (0, 0, 3)$ , $A(1, 0, -1) = (1, 0, -1)$ , so $AP = PD$ ; and $\det P = 1$ , expanding along column $2$ , whose only non-zero entry is the $1$ in row $3$ . Trace: $1 + 3 + 3 = 7 = 3 + 3 + 1$ .

Why

Three products and one determinant replace computing $P^{-1}$ . The trace check costs nothing and confirms the eigenvalues independently of the eigenvectors.

The conclusion, written out

“ $A$ is diagonalizable: $\dim E_3 = 2$ equals the algebraic multiplicity of $3$ , and $P^{-1}AP = D$ with $P = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \end{pmatrix}$ and $D = \text{diag}(3, 3, 1)$ .”

The classic mistake on this problem: Stopping at step 1 with “double root, so not diagonalizable”, or keeping only $(1, 1, 0)$ for $\lambda = 3$ : both lose the question. The mirror slip is $D = \text{diag}(1, 3, 3)$ with this $P$ , for which $AP \neq PD$ .

Learn by heart

• Diagonalizable $\iff$ $n$ independent eigenvectors $\iff$ $\dim E_\lambda =$ algebraic multiplicity for EVERY eigenvalue.
• $\dim E_\lambda = n - \text{rank}(A - \lambda I)$ , and $1 \le \dim E_\lambda \le$ algebraic multiplicity.
• $n$ distinct eigenvalues: diagonalizable. A repeated eigenvalue: compute ONE rank before any verdict.
• $P^{-1}AP = D \iff A = PDP^{-1}$ : eigenvectors in the COLUMNS of $P$ , eigenvalues in $D$ in the SAME order.
• $A^n = PD^nP^{-1}$ , and only $D$ is raised entry by entry.
• $v_k = c_1\lambda_1^ku_1 + \dots + c_n\lambda_n^ku_n$ : the dominant $\lambda_1$ sets the size, $u_1$ sets the proportions.
• Similar matrices share $\det$ , trace, rank, polynomial and eigenvalues; sharing them proves NOTHING.

Frequently asked questions

How do I know if a matrix is diagonalizable without finding P?

Factor the characteristic polynomial. If the eigenvalues are all distinct, the matrix is diagonalizable and you can stop. For each repeated eigenvalue, row reduce A minus lambda I and count the free variables: that number is the dimension of the eigenspace. The matrix is diagonalizable exactly when every such dimension equals the number of times the eigenvalue is repeated.

Does a repeated eigenvalue mean the matrix is not diagonalizable?

No. A repeated eigenvalue is only the place where diagonalization can fail. The identity matrix has a single repeated eigenvalue and is already diagonal. What decides is the dimension of the eigenspace of that eigenvalue: if it equals the multiplicity of the root, the matrix is diagonalizable; if it is smaller, it is not.

In which order do I put the eigenvalues in D?

In the same order as the eigenvectors in the columns of P. The first diagonal entry of D is the eigenvalue of the first column of P, and so on. Any order is allowed, but the two lists must move together: swapping two columns of P without swapping the matching entries of D gives a wrong answer even when every number in it is correct.

Why does diagonalization make powers of a matrix easy?

Because A equals P times D times the inverse of P, and in a product of such factors every inverse of P cancels the P next to it. What remains is P times D to the n times the inverse of P, and a diagonal matrix is raised to a power entry by entry. Geometrically, each eigenvector axis is simply stretched by its eigenvalue, n times over.

How do I find the steady state of a two-state Markov chain?

Write the transition matrix with columns adding up to one, so that one is an eigenvalue. Find an eigenvector for the eigenvalue one, then scale it so that its entries add up to the total, or to one for proportions. The other eigenvalue is smaller than one in size, so its part of the solution dies out and every starting point with the same total ends at that steady state.

Practise it

Corrected exercises: Diagonalization and its applications, MATH 133 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Eigenvalues and eigenvectors

Looking for a MATH 133 tutor in Montreal?

Get in touch for a first session. Diagonalization is where the whole course comes together: a determinant, a row reduction and a rank, then one change of axes that makes everything else easy.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: diagonalization and its applications (MATH 133)

The essentials

Diagonalizable means: enough independent eigenvectors

What diagonalization buys, once PPP is built

The rules in table form

Similarity: what survives a change of basis, and what does not

The mistakes that cost marks

1. Declaring a double eigenvalue defective without computing a rank

2. Taking one eigenvector per eigenvalue, whatever its multiplicity

3. Listing the eigenvalues in D in a different order from the columns of P

4. Putting the eigenvectors in the rows of P

5. Writing A as the inverse of P times D times P

6. Raising the entries of A to the power n

7. Quoting the eigenvector as the steady state of a Markov chain

8. Reading a stable ratio as a stable population

Which method to choose

Is AAA diagonalizable? Decide on the form

Long-run behaviour of vk+1=Avkv_{k+1} = Av_kvk+1​=Avk​

How the answer is expected to be written

Diagonalize a matrix: the answer a marker can grade

Check before you hand in

The typical problem, taken apart

A 3×33 \times 33×3 with a double eigenvalue, from polynomial to check

Learn by heart

Frequently asked questions

How do I know if a matrix is diagonalizable without finding P?

Does a repeated eigenvalue mean the matrix is not diagonalizable?

In which order do I put the eigenvalues in D?

Why does diagonalization make powers of a matrix easy?

How do I find the steady state of a two-state Markov chain?

Corrected exercises: Diagonalization and its applications, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?

What diagonalization buys, once $P$ is built

Is $A$ diagonalizable? Decide on the form

Long-run behaviour of $v_{k+1} = Av_k$

A $3 \times 3$ with a double eigenvalue, from polynomial to check