Linear Algebra and Vector Geometry 201-SN4-RE • CEGEP, Montreal

Revision sheet: eigenvalues and diagonalization (201-SN4-RE)

This sheet does not repeat the definition of an eigenvalue: you have it in your notes. It deals with what decides the mark at the end of 201-SN4-RE (formerly 201-NYC): the diagonalizability criterion applied correctly, the order in which $P$ and $D$ are filled in, and the checks that catch a wrong eigenvalue in ten seconds.

It is written for CEGEP students in Science and in the Social Science mathematics profile (201-105). Once you have read it, the corrected exercise set on the same chapter puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

An eigenvector is a DIRECTION that the matrix merely stretches. The whole chapter is about finding these directions, and the most expensive mistake is to believe there are always enough of them to diagonalize.

This chapter is part of 201-SN4-RE, Linear Algebra and Vector Geometry

The essentials

What an eigenvalue means, and how to find it

• $\lambda$ is an EIGENVALUE of $A$ if there is a NON-ZERO vector $\vec{v}$ such that $A\vec{v}=\lambda\vec{v}$ . That $\vec{v}$ is an associated EIGENVECTOR.
• Geometrically, $A$ only STRETCHES $\vec{v}$ , without changing its direction. A negative $\lambda$ flips the vector, a zero $\lambda$ crushes it onto the origin.
• CHARACTERISTIC POLYNOMIAL: the eigenvalues are the roots of $\det(A-\lambda I)=0$ . Subtract $\lambda$ along the WHOLE diagonal, then compute the determinant.
• For a $2\times 2$ matrix it reads $\lambda^{2}-\operatorname{tr}(A)\lambda+\det(A)=0$ , where the trace is the sum of the diagonal entries.
• Two free checks: the SUM of the eigenvalues is the trace, and their PRODUCT is the determinant.

With

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

: the image of the eigenvector

\vec{v}

stays on its line, twice as long; the image of

\vec{w}

, which is not one, leaves its line.

These two checks are the best value in the chapter: they take five seconds and catch almost every calculation error in the characteristic polynomial.

Eigenspaces and the diagonalizability criterion

• $E_{\lambda}=\ker(A-\lambda I)$ is the EIGENSPACE: you get it by solving the homogeneous system $(A-\lambda I)\vec{v}=\vec{0}$ . It is always a subspace, of dimension at least $1$ .
• ALGEBRAIC MULTIPLICITY of $\lambda$ : its order as a root of the characteristic polynomial. GEOMETRIC MULTIPLICITY: $\dim E_{\lambda}$ .
• Always, geometric $\leq$ algebraic, and the inequality can be STRICT: that is exactly what prevents diagonalization.
• CRITERION: $A$ of order $n$ is diagonalizable if and only if the dimensions of the eigenspaces add up to $n$ , that is, if the two multiplicities agree for EVERY eigenvalue.
• Two easy cases where you conclude without calculation: $n$ DISTINCT eigenvalues, or a real SYMMETRIC matrix, which is always diagonalizable with orthogonal eigenvectors.

For

A=\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}

E_{5}

is the line of

(1, 1)

E_{2}

the line of

(1, -2)

, and each vector on them is stretched by its eigenvalue. Two lines: diagonalizable.

A simple eigenvalue never causes trouble: its geometric multiplicity is necessarily $1$ . Only REPEATED eigenvalues require computing the dimension of the eigenspace.

Diagonalizing, and putting it to use

• $A=PDP^{-1}$ , where the COLUMNS of $P$ are the eigenvectors and $D$ is diagonal with the eigenvalues in the SAME ORDER.
• Column $i$ of $P$ must be an eigenvector associated with the eigenvalue in position $i$ of $D$ . The order is free, but it must be the same on both sides.
• $A^{n}=PD^{n}P^{-1}$ , and $D^{n}$ is obtained by raising each diagonal entry to the power $n$ . That is the whole point of diagonalizing.
• A check that avoids computing $P^{-1}$ : $AP=PD$ . Two matrix products are enough to validate $P$ and $D$ together.
• A real symmetric matrix is diagonalized in an ORTHONORMAL basis: eigenvectors associated with different eigenvalues are orthogonal, and you only need to normalize them.

Same matrix: the directions of

X_{0}=(1, 0)

AX_{0}

A^{2}X_{0}

A^{3}X_{0}

(drawn at the same length) fold onto

E_{5}

: the eigenvalue with the largest absolute value imposes its direction.

For a sequence of vectors $X_{k+1}=AX_{k}$ , the eigenvalue with the largest absolute value ends up dominating: the limiting direction is that of its eigenvector, whatever the initial condition.

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. Mistaking the zero vector for an eigenvector

1 mark for rigour, and the question on the dimension of the eigenspace becomes inconsistent.

What not to write

“ $A\vec{0}=\lambda\vec{0}$ for every $\lambda$ , so $\vec{0}$ is an eigenvector.”

What to write

“An eigenvector is NON-ZERO by definition. Otherwise every real number would be an eigenvalue of every matrix, and the notion would have no content.”

Why: $\vec{0}$ does belong to $E_{\lambda}$ , which is a subspace, but it is not an EIGENVECTOR. The distinction shows in the write-up: “ $E_{\lambda}$ contains $\vec{0}$ ” is true, “ $\vec{0}$ is an eigenvector” is false.

2. Forming the matrix $A-\lambda I$ wrongly

the whole question, and the eigenvalues found have nothing to do with the matrix.

What not to write

“ $\det(A)-\lambda=0$ ”, or $\lambda$ subtracted from the top-left entry only.

What to write

“ $\det(A-\lambda I)=0$ : subtract $\lambda$ from ALL the diagonal entries, then compute the determinant of the resulting matrix.”

Why: $\lambda I$ is the diagonal matrix of $\lambda$ s, not the number $\lambda$ : you cannot subtract a number from a matrix. It is the same rule as for any matrix sum, entry by entry.

3. Giving an eigenspace reduced to a single vector

1 mark, and more if the question asked for the dimension of the eigenspace.

What not to write

“ $E_{4}=\{(1, 1)\}$ .”

What to write

“ $E_{4}=\operatorname{span}\{(1, 1)\}$ , that is, the whole LINE: every non-zero multiple of $(1, 1)$ is also an eigenvector associated with $4$ .”

E_{4}

is the whole LINE directed by

(1, 1)

(2, 2)

and

(-1, -1)

, in green, are eigenvectors for

4

as well. Write

E_{4}=\operatorname{span}\{(1, 1)\}

Why: If $A\vec{v}=\lambda\vec{v}$ , then $A(k\vec{v})=\lambda(k\vec{v})$ for every $k$ : the property survives multiplication. An eigenvector is never unique, only its DIRECTION is.

4. Assuming the geometric multiplicity equals the algebraic one

the whole question, with an invented matrix

P

whose columns are dependent.

What not to write

“ $\lambda=2$ is a double root, so $\dim E_{2}=2$ and the matrix is diagonalizable.”

What to write

“You have to COMPUTE $\dim E_{2}$ . For $A=\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ , the system $(A-2I)\vec{v}=\vec{0}$ gives $y=0$ : $E_{2}=\operatorname{span}\{(1, 0)\}$ , of dimension $1$ . The matrix is NOT diagonalizable.”

The only direction this matrix preserves is the horizontal one:

\vec{v_{2}}

tips toward it instead of staying on its line. One double eigenvalue, a single eigendirection.

Why: The inequality geometric $\leq$ algebraic can be strict, and precisely that case makes a matrix non-diagonalizable. A repeated eigenvalue always forces you to solve the system.

5. Filling $P$ and $D$ in different orders

the whole question, and the computation of

A^{n}

that follows gives nonsense.

What not to write

“ $P=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ with $\vec{v_{1}}$ , associated with $4$ , in the first column, and $D=\begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$ .”

What to write

“Column $i$ of $P$ must be an eigenvector for the eigenvalue in position $i$ of $D$ : here $D=\begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$ . The check $AP=PD$ confirms it at once.”

Why: The equality $AP=PD$ is read column by column: $A\vec{v_{i}}=\lambda_{i}\vec{v_{i}}$ . If the indices do not match, the equality fails for every column.

6. Reversing the order of $P$ and $P^{-1}$ in the power

the whole power question, and the matrix obtained does not even have the right eigenvalues.

What not to write

“ $A^{n}=P^{-1}D^{n}P$ .”

What to write

“ $A=PDP^{-1}$ , so $A^{n}=PD^{n}P^{-1}$ , in the SAME order as $A$ . The inner $P^{-1}P$ cancel and only the two outer factors remain.”

Why: The calculation $A^{2}=(PDP^{-1})(PDP^{-1})=PD(P^{-1}P)DP^{-1}=PD^{2}P^{-1}$ shows the order. Writing out this telescoping once is enough never to get it wrong again.

7. Concluding that there is no eigenvalue when the discriminant is negative

2 marks, and the geometric interpretation, often the next question, is lost.

What not to write

“The characteristic polynomial has a negative discriminant, so this matrix has no eigenvalues.”

What to write

“It has no REAL eigenvalue: no direction of the plane is preserved, which is the case for a rotation. Over $\mathbb{C}$ , it has two conjugate eigenvalues, and it is diagonalizable there.”

The rotation through

60^{\circ}

turns every vector through the same angle (image dashed): no arrow stays on its original line, so there is no real eigenvalue. Lengths, on the other hand, are preserved.

Why: The phrase “no eigenvalue” is wrong: you must say over which set. The geometric reading, on the other hand, is immediate and earns marks: a rotation through a non-zero angle leaves no line invariant.

8. Finding only the zero vector as a solution of the eigenvector system

the whole question, plus the time lost looking for the error somewhere other than where it is.

What not to write

“I solve $(A-\lambda I)\vec{v}=\vec{0}$ and get $\vec{v}=\vec{0}$ : there is no eigenvector for this $\lambda$ .”

What to write

“If $\lambda$ really is an eigenvalue, then $\det(A-\lambda I)=0$ , so the homogeneous system NECESSARILY has infinitely many solutions. Finding only $\vec{0}$ proves a calculation error, most often in $\lambda$ .”

Why: The eigenvalue was chosen precisely to make the matrix $A-\lambda I$ singular. A pivot must therefore disappear: if none does, the characteristic polynomial is wrong.

Which method to choose

Is this matrix diagonalizable?

Look at the characteristic polynomial and the shape of the matrix. Three branches out of five answer without computing a single eigenspace.

Eigendirections in order

2

: two lines (distinct eigenvalues), two perpendicular lines (symmetric), or a single one for a double eigenvalue: one is missing, no diagonalization.

If $n$ DISTINCT real eigenvalues for a matrix of order $n$ → YES, diagonalizable, with no further check

Example: $\lambda=4$ and $\lambda=2$ for a matrix of order $2$
If a real SYMMETRIC matrix → YES, diagonalizable, and even in an orthonormal basis

This is true even with repeated eigenvalues, and it spares you any dimension calculation.
If a REPEATED eigenvalue, non-symmetric matrix → you must COMPUTE $\dim E_{\lambda}$ and compare it with the algebraic multiplicity

Example: $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ has $\dim E_{2}=1$ for a multiplicity of $2$ , so no
If a NEGATIVE discriminant for a real matrix of order $2$ → not diagonalizable over $\mathbb{R}$ , for lack of a real eigenvalue; diagonalizable over $\mathbb{C}$

Geometrically, it is a rotation, possibly combined with a scaling.
If an already TRIANGULAR matrix → the eigenvalues are the diagonal entries, with no calculation; the multiplicities still have to be dealt with

A triple root $\lambda$ in order $3$ ? There is only one eigenspace, $E_{\lambda}$ : the matrix is diagonalizable if and only if $\dim E_{\lambda}=3$ , that is, if $A=\lambda I$ , with no calculation.

How to compute this expression once $A$ is diagonalized

Once $A=PDP^{-1}$ is written, most questions come down to a calculation on $D$ , which is diagonal and therefore immediate.

If you are asked for $A^{n}$ → $PD^{n}P^{-1}$ , with $D^{n}$ obtained by raising each diagonal entry to the power $n$

Example: $D=\operatorname{diag}(4, 2)$ gives $D^{3}=\operatorname{diag}(64, 8)$
If you are asked for $\det(A)$ or $\operatorname{tr}(A)$ → the product of the eigenvalues for the determinant, their sum for the trace

It also works as a check on the eigenvalues found, in the other direction.
If you are asked for the eigenvalues of $A^{-1}$ , of $A^{k}$ or of $A+cI$ → $\frac{1}{\lambda}$ , $\lambda^{k}$ , $\lambda+c$ , with the SAME eigenvectors

No new calculation: this is the most frequent theory question at the end of the chapter.
If you are asked for the behaviour of a sequence $X_{k+1}=AX_{k}$ → decompose $X_{0}$ in the eigenbasis, then multiply each component by $\lambda^{k}$

The eigenvalue with the largest absolute value dominates and sets the limiting direction.
If you are asked whether $A$ is invertible → check whether $0$ is an eigenvalue: $A$ is invertible if and only if no eigenvalue is zero

Never compute $P^{-1}$ unless you have to: the check $AP=PD$ validates the diagonalization without it, and many questions do not need it.

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.

Diagonalizing a matrix of order two

When to use it: The question asks you to diagonalize $A$ , or to find $P$ and $D$ such that $A=PDP^{-1}$ .

1 Write $A-\lambda I$ in full, then expand $\det(A-\lambda I)$ to get the characteristic polynomial.
2 Solve, give the eigenvalues, and check them with the trace and the determinant before going further.
3 For each eigenvalue, solve the homogeneous system $(A-\lambda I)\vec{v}=\vec{0}$ and give $E_{\lambda}$ in the form $\operatorname{span}\{\dots\}$ .
4 Justify diagonalizability: distinct eigenvalues, or dimensions adding up to the order.
5 Write $P$ with the eigenvectors AS COLUMNS and $D$ with the eigenvalues in the SAME order, then check $AP=PD$ .

Concluding sentence

“The eigenvalues are $4$ and $2$ , which are distinct, so $A$ is diagonalizable: with $P=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and $D=\begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$ , we have $A=PDP^{-1}$ .”

The trap: Forgetting to justify diagonalizability and jumping straight to $P$ and $D$ . The marking scheme always puts half a mark on this sentence, and it is the only one that mentions the criterion.

Marking: 1 mark for the characteristic polynomial, 1 mark for the eigenspaces, 0.5 mark for the justification, 0.5 mark for $P$ , $D$ and the check.

Computing a power by diagonalization

When to use it: The question asks for $A^{n}$ , or the state of a system after $n$ steps of an evolution $X_{k+1}=AX_{k}$ .

1 Diagonalize $A$ and restate $A=PDP^{-1}$ with $P$ and $D$ written out.
2 Write the telescoping $A^{n}=PD^{n}P^{-1}$ , justified by the cancellation of the inner $P^{-1}P$ .
3 Compute $D^{n}$ by raising each diagonal entry to the power $n$ , and say so.
4 Compute $P^{-1}$ , then carry out the two products in the order $P$ , then $D^{n}$ , then $P^{-1}$ .
5 Check on a small value of $n$ , for example $n=2$ , against the direct product $A\times A$ .

Concluding sentence

“Since $A=PDP^{-1}$ , we have $A^{n}=PD^{n}P^{-1}$ , hence $A^{3}=\begin{pmatrix} 36 & 28 \\ 28 & 36 \end{pmatrix}$ .”

The trap: Raising $P$ to the power $n$ along with $D$ . Only the diagonal matrix is raised to the power: $P$ and $P^{-1}$ stay unchanged, since they perform the change of basis.

Marking: 0.5 mark for writing $A^{n}=PD^{n}P^{-1}$ , 0.5 mark for $D^{n}$ , 1 mark for $P^{-1}$ , 1 mark for the final product.

Check before you hand in

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

Check with the trace and the determinant

The sum of the eigenvalues must equal the trace, their product the determinant. Two additions and one multiplication validate the whole characteristic polynomial.

$4+2=6$ and $4\times 2=8$ : the trace of $\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ is indeed $6$ and its determinant $8$ .
Multiply the matrix by each eigenvector found

Compute $A\vec{v}$ and check that you get exactly $\lambda\vec{v}$ . It is the only check that tests the eigenvalue and the vector at the same time.

$\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 4 \\ 4 \end{pmatrix}=4\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ : everything is consistent.
Check $AP=PD$ instead of computing $P^{-1}$

Two matrix products validate $P$ and $D$ together, order included, without inverting anything. It is three times faster than checking with $P^{-1}$ .

If $AP$ and $PD$ differ in one column, that column of $P$ does not match the eigenvalue placed opposite it in $D$ .
Check that a pivot has disappeared

In the system $(A-\lambda I)\vec{v}=\vec{0}$ , one row must become proportional to another. If the two rows stay independent, the eigenvalue is wrong.

For $\lambda=4$ on $\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ , you get $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$ : the two rows are opposite, the expected sign.
Test a power at $n=2$

Compare the formula $PD^{2}P^{-1}$ with the direct product $A\times A$ , which costs only eight multiplications in order two. Agreement validates the whole diagonalization.

$A^{2}=\begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$ by both methods: the formula is right, and $A^{n}$ can be written without fear.

The typical problem, taken apart

Diagonalizing a symmetric matrix, then computing its cube

Let $A=\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ .

1) Find the eigenvalues and the eigenspaces of $A$ .

2) Write $P$ and $D$ such that $A=PDP^{-1}$ , and check.

3) Deduce $A^{3}$ .

The data:

A

sends the unit square (in blue) onto the parallelogram built on its columns

(3, 1)

and

(1, 3)

. Question 1 asks which directions this change merely stretches.

Step 1

$\operatorname{tr}(A)=3+3=6$ and $\det(A)=9-1=8$ , so the characteristic polynomial is $\lambda^{2}-6\lambda+8=0$ .

Why

The trace-determinant form avoids writing $A-\lambda I$ in full in order $2$ , and it gives right away the two checks used in the next step.

Step 2

$\Delta=36-32=4$ , so $\lambda_{1}=\dfrac{6+2}{2}=4$ and $\lambda_{2}=\dfrac{6-2}{2}=2$ . Check: $4+2=6$ and $4\times 2=8$ .

Why

The check takes five seconds and catches any sign error in the polynomial. Two DISTINCT eigenvalues are already enough to guarantee diagonalizability.

Step 3

$E_{4}$ : $(A-4I)\vec{v}=\vec{0}$ gives $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\vec{v}=\vec{0}$ , that is $-x+y=0$ , so $E_{4}=\operatorname{span}\{(1, 1)\}$ .

Why

The two rows are opposite: a pivot has indeed disappeared, which confirms that $4$ is a genuine eigenvalue. The subspace is a LINE, not an isolated vector.

Step 4

$E_{2}$ : $(A-2I)\vec{v}=\vec{0}$ gives $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\vec{v}=\vec{0}$ , that is $x+y=0$ , so $E_{2}=\operatorname{span}\{(1, -1)\}$ .

Why

The two eigenvectors are orthogonal, which was guaranteed in advance: $A$ is symmetric. It is a free check on both subspaces at once.

Step 5

$A$ has two distinct eigenvalues in order $2$ , so it is diagonalizable: $P=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and $D=\begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$ .

Why

The order is free but it must be the SAME on both sides: $\vec{v_{1}}$ in the first column of $P$ forces $4$ in the first position of $D$ .

Step 6

Check: $AP=\begin{pmatrix} 4 & 2 \\ 4 & -2 \end{pmatrix}$ and $PD=\begin{pmatrix} 4 & 2 \\ 4 & -2 \end{pmatrix}$ : the two agree.

Why

Two products, no inversion: it is the cheapest check, and it tests the order along with the eigenvectors.

Step 7

$\det(P)=-2$ , so $P^{-1}=\dfrac{1}{-2}\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix}=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{pmatrix}$ .

Why

Apply the order-two recipe: swap the diagonal entries, change the sign of the off-diagonal ones, divide by the determinant. Here $P$ is symmetric, so $P^{-1}$ is too.

Step 8

$A^{3}=PD^{3}P^{-1}$ with $D^{3}=\begin{pmatrix} 64 & 0 \\ 0 & 8 \end{pmatrix}$ , which gives $A^{3}=\begin{pmatrix} 36 & 28 \\ 28 & 36 \end{pmatrix}$ .

Why

Only $D$ is raised to the power; $P$ and $P^{-1}$ stay as they are. Direct check: $A^{2}=\begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$ , then $A^{2}A$ gives the same result.

The conclusion, written out

$A$ has eigenvalues $4$ and $2$ , with $E_{4}=\operatorname{span}\{(1, 1)\}$ and $E_{2}=\operatorname{span}\{(1, -1)\}$ ; it diagonalizes as $A=PDP^{-1}$ and $A^{3}=\begin{pmatrix} 36 & 28 \\ 28 & 36 \end{pmatrix}$ .

The classic mistake on this problem: Writing $D=\begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$ while keeping $(1, 1)$ in the first column of $P$ . The check $AP=PD$ then fails on both columns, but only if you do it.

Learn by heart

• $A\vec{v}=\lambda\vec{v}$ with $\vec{v}\neq\vec{0}$ : the zero vector is never an eigenvector.
• Eigenvalues: roots of $\det(A-\lambda I)=0$ , with $\lambda$ subtracted along the WHOLE diagonal.
• In order $2$ : $\lambda^{2}-\operatorname{tr}(A)\lambda+\det(A)=0$ . Sum $=$ trace, product $=$ determinant.
• $E_{\lambda}=\ker(A-\lambda I)$ is a SUBSPACE, so at least a line, never a single vector.
• Geometric $\leq$ algebraic, and equality for EVERY eigenvalue is the diagonalizability criterion.
• $n$ distinct eigenvalues, or a real symmetric matrix: diagonalizable with no further check.
• $A=PDP^{-1}$ : column $i$ of $P$ and position $i$ of $D$ must match.
• $A^{n}=PD^{n}P^{-1}$ , and only $D$ is raised to the power.
• Check with $AP=PD$ , which avoids computing $P^{-1}$ .
• Negative discriminant: no REAL eigenvalue, so no direction preserved, as for a rotation.

Frequently asked questions

How do I find the eigenvalues of a two by two matrix?

Write the characteristic polynomial as lambda squared, minus the trace times lambda, plus the determinant, set equal to zero. The trace is the sum of the two diagonal entries. Then solve this quadratic equation, and check that the sum of the roots equals the trace and their product the determinant.

Why is the zero vector not an eigenvector?

Because the definition explicitly excludes it. If it were allowed, any number would be an eigenvalue of any matrix, since a matrix always sends the zero vector to itself. The notion would lose all content. The zero vector still belongs to the eigenspace, which is a subspace.

When is a matrix not diagonalizable?

When it has a repeated eigenvalue whose eigenspace has a dimension strictly smaller than its multiplicity. In other words, there are not enough eigendirections to form a basis. Over the real numbers, this is also the case for a matrix whose characteristic polynomial has a negative discriminant, such as a rotation.

What is diagonalization used for in practice?

It turns a hard calculation into an immediate one. Raising a matrix to a high power becomes a simple calculation on the diagonal entries, which lets you study an evolution after a large number of steps, for example two coupled populations, without doing the multiplications one by one.

How do I check a diagonalization without computing the inverse?

Compare two products: the original matrix times the change-of-basis matrix, and the change-of-basis matrix times the diagonal matrix. They must be identical. This check tests the eigenvectors, the eigenvalues and the order in which they are placed all at once, and it costs only two products.

Is a symmetric matrix always diagonalizable?

Yes, as long as it is real. It is a result of the course, valid even when an eigenvalue is repeated. Moreover, eigenvectors associated with different eigenvalues are automatically perpendicular to each other, which lets you choose an orthonormal basis of eigenvectors.

Practise it

Corrected exercises: Eigenvalues and diagonalization, 201-SN4-RE

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.

16 corrected exercises
160 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-sn4-eigenvalues-diagonalization. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Get in touch for a first session. We go over the moves that cost marks on the linear algebra exam, from the characteristic polynomial to justifying diagonalizability, then put them to the test on exercises at the real level of CEGEP evaluations.

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Revision sheet: eigenvalues and diagonalization (201-SN4-RE)

The essentials

What an eigenvalue means, and how to find it

Eigenspaces and the diagonalizability criterion

Diagonalizing, and putting it to use

The mistakes that cost marks

1. Mistaking the zero vector for an eigenvector

2. Forming the matrix A−λIA-\lambda IA−λI wrongly

3. Giving an eigenspace reduced to a single vector

4. Assuming the geometric multiplicity equals the algebraic one

5. Filling PPP and DDD in different orders

6. Reversing the order of PPP and P−1P^{-1}P−1 in the power

7. Concluding that there is no eigenvalue when the discriminant is negative

8. Finding only the zero vector as a solution of the eigenvector system

Which method to choose

Is this matrix diagonalizable?

How to compute this expression once AAA is diagonalized

How the answer is expected to be written

Diagonalizing a matrix of order two

Computing a power by diagonalization

Check before you hand in

The typical problem, taken apart

Diagonalizing a symmetric matrix, then computing its cube

Learn by heart

Frequently asked questions

How do I find the eigenvalues of a two by two matrix?

Why is the zero vector not an eigenvector?

When is a matrix not diagonalizable?

What is diagonalization used for in practice?

How do I check a diagonalization without computing the inverse?

Is a symmetric matrix always diagonalizable?

Corrected exercises: Eigenvalues and diagonalization, 201-SN4-RE

See also

Looking for a 201-SN4-RE tutor in Montreal?

2. Forming the matrix $A-\lambda I$ wrongly

5. Filling $P$ and $D$ in different orders

6. Reversing the order of $P$ and $P^{-1}$ in the power

How to compute this expression once $A$ is diagonalized