MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: eigenvalues and eigenvectors (MATH 133)

Q: How do I find the eigenvalues and eigenvectors of a 3 by 3 matrix by hand?

Form the matrix x times the identity minus A and look for a row or column with two zeros: expanding along it gives the characteristic polynomial already factored. Its roots are the eigenvalues. Then, for each eigenvalue, row reduce lambda times the identity minus A and read one basic eigenvector per free variable. Check each one by multiplying it by A.

Q: Why can't I read the eigenvalues off the row echelon form?

Because row operations change the eigenvalues. They preserve the null space of a matrix, not its characteristic polynomial. The matrix with rows 1, 2 and 3, 2 has eigenvalues 4 and minus 1, but one row operation turns its diagonal into 1 and minus 4. The eigenvalues can be read off the diagonal only when the original matrix is already triangular.

Q: Can zero be an eigenvalue?

Yes. Zero is an eigenvalue exactly when the matrix is not invertible, because then some non-zero vector is sent to the zero vector. What can never happen is the zero VECTOR being an eigenvector: it would satisfy the equation for every number at once, so the definition excludes it. Two different zeros, two different statements.

Q: Does every matrix have real eigenvalues?

No. The rotation of the plane by ninety degrees moves every line through the origin, so no non-zero vector stays on its line, and its characteristic polynomial x squared plus one has no real root. In MATH 133 the expected answer is exactly that: the matrix has no real eigenvalue and no eigenvector in the plane. Complex eigenvalues are not part of the course.

Q: How do I find a missing eigenvalue with the trace and the determinant?

The trace, the sum of the diagonal entries, equals the sum of the eigenvalues, and the determinant equals their product, each eigenvalue counted as often as it is a root. If two eigenvalues of a 3 by 3 are known, the third is the trace minus the other two. When zero is one of them the determinant is zero and tells you nothing, so use the trace.

This sheet is not a summary of section 3.3 of Nicholson: you already have the course notes. It answers one question only, what makes students lose marks on eigenvalues and eigenvectors in MATH 133 at McGill University, and which precise gesture avoids each loss.

The convention is the course's: $c_A(x) = \det(xI - A)$ . Every number below is done by hand, as on the exam, and every eigenvector quoted has been checked by the one multiplication that settles it: $Ax$ must be a multiple of $x$ .

The thread of the chapter

Every eigen question is a question about the single matrix $\lambda I - A$ : its DETERMINANT finds the eigenvalues, its NULL SPACE finds the eigenvectors, and the zero vector is never an answer. When the reduction of $\lambda I - A$ leaves no free variable, the $\lambda$ is wrong, not the vector.

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

Eigenvalue
Eigenvector
Characteristic polynomial
Eigenspace
Basic eigenvectors
Cofactor expansion
Block triangular matrix
Trace of a matrix
Invertible matrix theorem
Reflection and projection matrices
Rotation matrix

The essentials

Two questions, two tools, one matrix

• $x$ is an eigenvector of $A$ for the eigenvalue $\lambda$ when $Ax = \lambda x$ AND $x \neq 0$ . The number $\lambda = 0$ is allowed; the vector $x = 0$ never is.
• The NUMBERS: the eigenvalues are the real roots of $c_A(x) = \det(xI - A)$ , because $\lambda$ is an eigenvalue exactly when $\lambda I - A$ is singular.
• The DIRECTIONS: $E_\lambda(A)$ is the null space of $\lambda I - A$ . Its basic solutions, one per free variable, are the basic eigenvectors and form a basis of $E_\lambda(A)$ .
• A correct eigenvalue ALWAYS leaves at least one free variable in the reduction of $\lambda I - A$ . No free variable means a wrong $\lambda$ .
• An eigenvector is a direction that $A$ keeps on its LINE through the origin: stretched if $\lambda > 1$ , shrunk if $0 < \lambda < 1$ , reversed if $\lambda < 0$ , crushed to $0$ if $\lambda = 0$ .

For

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

u

is stretched by

5

and

w

is REVERSED, both stay on their dashed line and both are eigenvectors;

v

leaves its line, so it is not one.

Every question of the chapter is one of these two: find $\lambda$ (a determinant) or find $x$ (a row reduction). Writing which one you are doing on the first line of an answer is already half the method mark.

What you get for free once the eigenvalues are known

• $0$ is an eigenvalue of $A$ if and only if $A$ is NOT invertible. A visible dependency between the columns is an eigenvector for $0$ .
• When $c_A$ splits into real factors: $\text{tr}\,A = \lambda_1 + \dots + \lambda_n$ and $\det A = \lambda_1 \cdots \lambda_n$ , each eigenvalue counted as often as it is a root.
• $c_A(0) = \det(-A) = (-1)^n \det A$ , so for a $3 \times 3$ , $c_A(0) = -\det A$ .
• Triangular $A$ : the eigenvalues are the diagonal entries. Block triangular $\begin{pmatrix} B & C \\ 0 & D \end{pmatrix}$ : $c_A = c_B \, c_D$ .
• Powers, inverse, shifts, multiples and transpose: see the table below, each rule has a one-line proof starting from $Ax = \lambda x$ .

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.

Related matrices: which rules exist, and which do not

Read a line as: if $Ax = \lambda x$ with $x \neq 0$ , then the matrix of the first column has the eigenvector and the eigenvalue of the next two. The examples use $A = \begin{pmatrix} 5 & -6 \\ 3 & -4 \end{pmatrix}$ , with eigenvalues $2$ and $-1$ and eigenvectors $(2, 1)$ and $(1, 1)$ . The red lines are the rules students invent.

Matrix	Eigenvector	Eigenvalue
$A^k$	same $x$	$\lambda^k$
Example: $A^3$ has eigenvalues $8$ and $-1$ : $A^3(2, 1) = 8(2, 1)$ .
$A^{-1}$	same $x$	$\frac{1}{\lambda}$
Example: $A^{-1} = \frac{1}{2}\begin{pmatrix} 4 & -6 \\ 3 & -5 \end{pmatrix}$ and $A^{-1}(2, 1) = (1, \frac{1}{2})$ : eigenvalue $\frac{1}{2}$ , not $-2$ .
$A + cI$	same $x$	$\lambda + c$
Example: $A + 4I$ has eigenvalues $2 + 4 = 6$ and $-1 + 4 = 3$ .
$kA$	same $x$	$k\lambda$
Example: $3A$ has eigenvalues $6$ and $-3$ .
$A^T$	DIFFERENT	$\lambda$
Example: $A^T$ has eigenvalues $2$ and $-1$ , but $E_2(A^T) = \text{span}\{(1, -1)\}$ , not $(2, 1)$ .
$A + B$	no rule	$\lambda + \mu$ no such rule
Example: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ each have only $0$ , yet their sum has $1$ and $-1$ . What to do: Compute $c_{A+B}$ from scratch. The rule holds only when $B = cI$ .
echelon form of $A$	no rule	its diagonal no such rule
Example: $\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ has eigenvalues $4$ and $-1$ ; after $R_2 \to R_2 - 3R_1$ the diagonal reads $1$ and $-4$ . What to do: Form $xI - A$ first and take its determinant; reduce only $\lambda I - A$ , and only to find vectors.

Every line in blue is proved in one line from $Ax = \lambda x$ : apply the operation to both sides. If you cannot write that line, the rule does not exist, which is exactly what happens with $A + B$ .

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. Reading the eigenvalues off a row reduced matrix

the whole question, and every eigenvector computed after it

What not to write

“ $R_2 \to R_2 - 3R_1$ turns $\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ into $\begin{pmatrix} 1 & 2 \\ 0 & -4 \end{pmatrix}$ , so the eigenvalues are $1$ and $-4$ .”

What to write

“ $c_A(x) = \det\begin{pmatrix} x - 1 & -2 \\ -3 & x - 2 \end{pmatrix} = (x - 1)(x - 2) - 6 = x^2 - 3x - 4 = (x - 4)(x + 1)$ , so the eigenvalues are $4$ and $-1$ .”

Why: Row operations preserve the null space of $A$ , not its eigenvalues. The determinant has to be taken of $xI - A$ BEFORE any reduction; reduction comes afterwards, on $\lambda I - A$ , to find the vectors.

2. Giving the zero vector when the reduction has no free variable

2 marks, plus the eigenvalue that was actually missed

What not to write

“ $4I - A$ reduces to $I$ , so $E_4 = \{0\}$ and the eigenvector is $(0, 0)$ .”

What to write

“ $4I - A$ reduces to $I$ , so it is invertible and $4$ is NOT an eigenvalue: I go back to $c_A(x)$ and find the slip.”

Why: $\lambda I - A$ is singular by construction when $\lambda$ is a root of $c_A$ , so a reduction with three pivots on a $3 \times 3$ is an alarm, never an answer. For $\begin{pmatrix} 1 & 2 \\ -2 & 6 \end{pmatrix}$ , $\det(4I - A) = c_A(4) = -2 \neq 0$ , and the true eigenvalues are $2$ and $5$ .

3. Rejecting an eigenvector because its eigenvalue is negative

1 to 2 marks per missed eigenvector, and half the eigenvectors of any reflection

What not to write

“ $A(1, -1) = (-1, 1)$ points the other way, so $(1, -1)$ is not an eigenvector.”

What to write

“ $A(1, -1) = (-1, 1) = -1 \cdot (1, -1)$ : an eigenvector with eigenvalue $-1$ .”

Why: The definition asks for $Ax$ on the LINE of $x$ , not in its direction. A reversed vector is the eigenvector of $-1$ , the value every reflection has perpendicular to its mirror. The figure of the essentials shows $w$ reversed and kept.

4. Expanding a 3 by 3 along the wrong row, then fighting a cubic

10 minutes, and usually a sign error inside them

What not to write

“Along the first row: $c_A(x) = x^3 - 10x^2 + 31x - 30$ . I cannot factor it.”

What to write

“Row 2 of $xI - A$ is $(0,\ x - 5,\ 0)$ , so $c_A(x) = (x - 5)\left[(x - 4)(x - 1) + 2\right] = (x - 5)(x - 2)(x - 3)$ .”

Why: A row or a column of $xI - A$ with two zeros gives the polynomial ALREADY FACTORED. Scan for it before writing any cofactor: here $A = \begin{pmatrix} 4 & 1 & -2 \\ 0 & 5 & 0 \\ 1 & 1 & 1 \end{pmatrix}$ , and the first factor $(x - 5)$ comes out for free.

5. Giving one vector where the eigenspace needs two

1 mark, and the whole next chapter if the count is wrong

What not to write

“For $\lambda = 1$ the solutions are $(-t, s, t)$ , so the eigenvector is $(-1, 1, 1)$ .”

What to write

“Two free variables, so two basic eigenvectors: $E_1 = \text{span}\{(0, 1, 0), (-1, 0, 1)\}$ .”

Why: The question asks for a BASIS of the eigenspace: one basic eigenvector per free variable, obtained by setting one parameter to $1$ and the others to $0$ . $(-1, 1, 1)$ is an eigenvector, but alone it spans a line inside a plane. Worked in full in the example below.

6. Assuming A and its transpose share their eigenvectors

2 marks

What not to write

“ $A^T$ has the same characteristic polynomial, so $E_2(A^T) = \text{span}\{(2, 1)\}$ as for $A$ .”

What to write

“Same eigenvalues, but $2I - A^T = \begin{pmatrix} -3 & -3 \\ 6 & 6 \end{pmatrix}$ gives $E_2(A^T) = \text{span}\{(1, -1)\}$ .”

Same eigenvalues

2

and

-1

, four different lines: the eigenlines of

A

on the left, those of

A^T

on the right. Transposing keeps the numbers and moves the directions.

Why: $\det(xI - A^T) = \det\left((xI - A)^T\right) = \det(xI - A)$ proves the eigenvalues agree and nothing more: the null spaces of $\lambda I - A$ and $\lambda I - A^T$ are different. Here $A = \begin{pmatrix} 5 & -6 \\ 3 & -4 \end{pmatrix}$ and $A^T(2, 1) = (13, -16)$ .

7. Taking opposites instead of reciprocals for the inverse

1 mark for the answer and 1 for the missing proof

What not to write

“ $A$ has eigenvalues $2$ and $-1$ , so $A^{-1}$ has $-2$ and $1$ .”

What to write

“From $Ax = \lambda x$ and $\lambda \neq 0$ : $A^{-1}x = \frac{1}{\lambda}x$ , so $A^{-1}$ has $\frac{1}{2}$ and $-1$ , with the same eigenvectors.”

Why: Undoing a stretch by $2$ is a stretch by $\frac{1}{2}$ , not by $-2$ . The proof is one line, multiply by $A^{-1}$ and divide by $\lambda$ , and the division is where invertibility, that is $\lambda \neq 0$ , is used.

8. Mixing the two sign conventions for the characteristic polynomial

1 mark, and a wrong missing eigenvalue whenever the determinant is used to find one

What not to write

“ $c_A(0) = \det A$ , and $c_A(0) = (-\lambda_1)(-\lambda_2)(-\lambda_3)$ , so $\det A = -\lambda_1\lambda_2\lambda_3$ .”

What to write

“ $c_A(0) = \det(-A) = (-1)^3\det A = -\det A$ , and also $c_A(0) = -\lambda_1\lambda_2\lambda_3$ , so $\det A = \lambda_1\lambda_2\lambda_3$ .”

Why: $\det(A - xI) = (-1)^n\det(xI - A)$ : the two conventions agree for a $2 \times 2$ and are OPPOSITE for a $3 \times 3$ . The roots never change, the coefficients do. Pick Nicholson's $\det(xI - A)$ and keep it for the whole answer.

9. Inventing complex eigenvectors for a rotation

1 mark for the conclusion, and time lost outside the course

What not to write

“ $c_R(x) = x^2 + 1$ , so the eigenvalues of the rotation by $90^\circ$ are $\pm i$ and I look for the eigenvectors.”

What to write

“ $c_R(x) = x^2 + 1 \ge 1$ has no real root: the rotation has no real eigenvalue and no eigenvector in $\mathbb{R}^2$ , since a quarter turn moves every line.”

Why: In MATH 133 eigenvalues are real numbers attached to vectors of $\mathbb{R}^n$ . The expected answer is that observation, backed by the geometry of the decision tree figure below: no line through the origin survives a quarter turn.

Which method to choose

Where the eigenvalues come from, by the FORM of the matrix

Look at the shape of the matrix, or at what the statement already gives you, before writing any determinant

The reflection keeps

L

(

1

) and reverses its perpendicular (

-1

); the projection keeps

L

(

1

) and crushes the perpendicular to the dot (

0

); the quarter turn keeps no line at all.

If the matrix is triangular → the eigenvalues are the diagonal entries, no computation

Example: $\begin{pmatrix} 2 & 7 \\ 0 & -3 \end{pmatrix}$ has eigenvalues $2$ and $-3$
If the matrix is block triangular $\begin{pmatrix} B & C \\ 0 & D \end{pmatrix}$ → $c_A = c_B \, c_D$ , two small polynomials instead of one big one

Example: a $4 \times 4$ with $c_B = (x - 1)(x - 3)$ and $c_D = (x - 2)(x - 5)$
If a row or column of $xI - A$ holds two zeros → expand along it: the polynomial comes out already factored

Example: row $(0,\ x - 5,\ 0)$ gives the factor $(x - 5)$ at once
If a general $2 \times 2$ → $c_A(x) = x^2 - (\text{tr}\,A)x + \det A$

Example: $\begin{pmatrix} 1 & 2 \\ -2 & 6 \end{pmatrix}$ : $x^2 - 7x + 10 = (x - 2)(x - 5)$
If the matrix is described geometrically → read the lines the map sends onto themselves

Example: reflection $1$ and $-1$ , projection $1$ and $0$ , rotation by $90^\circ$ none
If some eigenvectors or eigenvalues are already given → multiply to get their eigenvalues, then the trace (or the determinant) for the last one

Example: eigenvalues $0$ and $5$ , trace $9$ : the third is $9 - 0 - 5 = 4$

If no branch applies, expand $\det(xI - A)$ along the row with the most zeros and test the divisors of the constant term: on an exam without a calculator the roots are always small integers. If none of them works, the polynomial is wrong.

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.

Finding a basis of an eigenspace

When to use it: Any question that says find the eigenvectors, find the eigenspaces, or give a basis of $E_\lambda(A)$

1 Write the eigenvalue you are using and form $\lambda I - A$ explicitly, entries written out: subtract $A$ from $\lambda I$ , so the off-diagonal entries change sign.
2 Row reduce, naming each operation ( $R_2 \to R_2 - 2R_1$ ). Check that at least one zero row appears; if not, stop and recheck the eigenvalue.
3 Write the general solution with one parameter per free variable, then split it into one vector per parameter: these are the basic eigenvectors.
4 Check each basic eigenvector by computing $Ax$ and comparing with $\lambda x$ , entry by entry.
5 Conclude with the span, never with a single vector unless there is a single free variable.

Concluding sentence

“For $\lambda = 1$ , the reduced form of $I - A$ gives $x_1 = -x_3$ with $x_2$ and $x_3$ free, so $E_1(A) = \text{span}\{(0, 1, 0), (-1, 0, 1)\}$ .”

The trap: Reducing $A$ instead of $\lambda I - A$ , or reducing $\lambda I - A$ with a sign error on the diagonal. The multiplication check of step 4 catches both in ten seconds.

Marking: Typically 1 mark for the matrix lambda I minus A written correctly, 1 for the reduction with named operations, 1 for the basis, and the check is what protects the other three.

Check before you hand in

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

One multiplication per eigenvector

Compute $Ax$ and compare with $\lambda x$ . It catches a wrong vector and a wrong eigenvalue at the same time, and it never needs the polynomial.

$A(1, 3, 1) = (5, 15, 5) = 5(1, 3, 1)$ : consistent. A result like $(5, 14, 5)$ means the reduction slipped.
The trace equals the sum

Add the diagonal of A and compare with the sum of your eigenvalues, repeated ones counted twice. Three seconds, and it catches most sign slips in the polynomial.

$A = \begin{pmatrix} 4 & 1 & -2 \\ 0 & 5 & 0 \\ 1 & 1 & 1 \end{pmatrix}$ : $4 + 5 + 1 = 10 = 2 + 3 + 5$ .
The determinant equals the product

Useful when the trace agrees by accident. If $\det A = 0$ , one eigenvalue must be $0$ ; if $0$ is on your list, $A$ must be singular.

$\det\begin{pmatrix} 4 & 0 & 3 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix} = 5 = 1 \cdot 1 \cdot 5$ .
A zero row must appear

When reducing $\lambda I - A$ , the absence of a zero row is not a result, it is a warning that $\lambda$ is not an eigenvalue.

$\det(4I - A) = -2$ for $A = \begin{pmatrix} 1 & 2 \\ -2 & 6 \end{pmatrix}$ : no zero row can appear, so $4$ was a wrong root.

The typical problem, taken apart

A 3 by 3 with a repeated eigenvalue, from the polynomial to the eigenspaces

Let $A = \begin{pmatrix} 4 & 0 & 3 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix}$ . Find the characteristic polynomial, the eigenvalues, and a basis of each eigenspace.

No calculator. Every step must be justified as on a MATH 133 final.

Step 1

$xI - A = \begin{pmatrix} x - 4 & 0 & -3 \\ -1 & x - 1 & -1 \\ -1 & 0 & x - 2 \end{pmatrix}$ . Its second COLUMN is $(0,\ x - 1,\ 0)$ : expand along it. The entry is in position $(2, 2)$ , sign $+$ .

Why

Scanning for zeros is the first mark of the question. A column works exactly like a row, and this one reduces a $3 \times 3$ determinant to a single $2 \times 2$ .

Step 2

$c_A(x) = (x - 1)\det\begin{pmatrix} x - 4 & -3 \\ -1 & x - 2 \end{pmatrix} = (x - 1)\left[(x - 4)(x - 2) - 3\right] = (x - 1)(x^2 - 6x + 5) = (x - 1)^2(x - 5)$ .

Why

The product of the off-diagonal entries is $(-3)(-1) = 3$ , and it is SUBTRACTED. The quadratic factors because $1 + 5 = 6$ and $1 \cdot 5 = 5$ , and it happens to share the root $1$ with the factor already out: the eigenvalue $1$ is a double root.

Step 3

$\lambda = 5$ : $5I - A = \begin{pmatrix} 1 & 0 & -3 \\ -1 & 4 & -1 \\ -1 & 0 & 3 \end{pmatrix}$ . $R_2 \to R_2 + R_1$ gives $(0, 4, -4)$ , $R_3 \to R_3 + R_1$ gives a zero row, $R_2 \to \frac{1}{4}R_2$ gives $(0, 1, -1)$ . So $x_1 = 3x_3$ , $x_2 = x_3$ : $E_5 = \text{span}\{(3, 1, 1)\}$ .

Why

The zero row is the confirmation that $5$ is right. One free variable, one basic eigenvector. Each operation is named because that is where the method marks are.

Step 4

$\lambda = 1$ : $I - A = \begin{pmatrix} -3 & 0 & -3 \\ -1 & 0 & -1 \\ -1 & 0 & -1 \end{pmatrix}$ . $R_1 \to -\frac{1}{3}R_1$ gives $(1, 0, 1)$ , then $R_2 \to R_2 + R_1$ and $R_3 \to R_3 + R_1$ give two zero rows. So $x_1 = -x_3$ with $x_2 = s$ and $x_3 = t$ free: $x = s(0, 1, 0) + t(-1, 0, 1)$ , and $E_1 = \text{span}\{(0, 1, 0), (-1, 0, 1)\}$ .

Why

Two free variables give TWO basic eigenvectors, one per parameter. The column of $x_2$ is entirely zero, so $x_2$ is free even though no equation mentions it: forgetting it is the classic way to lose $(0, 1, 0)$ .

Step 5

Checks: $A(3, 1, 1) = (15, 5, 5)$ , $A(0, 1, 0) = (0, 1, 0)$ , $A(-1, 0, 1) = (-1, 0, 1)$ . Trace $4 + 1 + 2 = 7 = 1 + 1 + 5$ and $\det A = 1 \cdot (8 - 3) = 5 = 1 \cdot 1 \cdot 5$ .

Why

Three multiplications and two sums confirm every number on the page. The repeated eigenvalue is counted twice in the trace and in the product, which is what makes both checks close.

The conclusion, written out

“ $c_A(x) = (x - 1)^2(x - 5)$ . The eigenvalues are $1$ (double root) and $5$ , with $E_1(A) = \text{span}\{(0, 1, 0), (-1, 0, 1)\}$ and $E_5(A) = \text{span}\{(3, 1, 1)\}$ .”

The classic mistake on this problem: Expanding along the first row, which gives $(x - 4)(x - 1)(x - 2) - 3(x - 1)$ and then an expanded cubic, $x^3 - 7x^2 + 11x - 5$ , to be factored by trial; or solving $I - A$ with only $x_1$ and $x_3$ in mind and reporting the single eigenvector $(-1, 0, 1)$ for $\lambda = 1$ .

Learn by heart

• $Ax = \lambda x$ with $x \neq 0$ . $\lambda = 0$ is allowed and means $A$ is not invertible.
• $c_A(x) = \det(xI - A)$ ; eigenvalues are its real roots; $E_\lambda(A)$ is the null space of $\lambda I - A$ .
• A correct $\lambda$ ALWAYS leaves a free variable in $\lambda I - A$ . One basic eigenvector per free variable.
• Expand $xI - A$ along the row or column with the most zeros. Triangular: the diagonal. Block triangular: $c_B \, c_D$ .
• $A^k$ : $\lambda^k$ . $A^{-1}$ : $\frac{1}{\lambda}$ . $A + cI$ : $\lambda + c$ . $kA$ : $k\lambda$ . Same eigenvectors. $A^T$ : same $\lambda$ , different vectors.
• $\text{tr}\,A =$ sum of the eigenvalues, $\det A =$ their product, repeated ones counted twice.
• Reflection: $1$ and $-1$ . Projection: $1$ and $0$ . Rotation by $90^\circ$ : no real eigenvalue.
• NEVER read eigenvalues off a row reduced matrix.

Frequently asked questions

How do I find the eigenvalues and eigenvectors of a 3 by 3 matrix by hand?

Form the matrix x times the identity minus A and look for a row or column with two zeros: expanding along it gives the characteristic polynomial already factored. Its roots are the eigenvalues. Then, for each eigenvalue, row reduce lambda times the identity minus A and read one basic eigenvector per free variable. Check each one by multiplying it by A.

Why can't I read the eigenvalues off the row echelon form?

Because row operations change the eigenvalues. They preserve the null space of a matrix, not its characteristic polynomial. The matrix with rows 1, 2 and 3, 2 has eigenvalues 4 and minus 1, but one row operation turns its diagonal into 1 and minus 4. The eigenvalues can be read off the diagonal only when the original matrix is already triangular.

Can zero be an eigenvalue?

Yes. Zero is an eigenvalue exactly when the matrix is not invertible, because then some non-zero vector is sent to the zero vector. What can never happen is the zero VECTOR being an eigenvector: it would satisfy the equation for every number at once, so the definition excludes it. Two different zeros, two different statements.

Does every matrix have real eigenvalues?

No. The rotation of the plane by ninety degrees moves every line through the origin, so no non-zero vector stays on its line, and its characteristic polynomial x squared plus one has no real root. In MATH 133 the expected answer is exactly that: the matrix has no real eigenvalue and no eigenvector in the plane. Complex eigenvalues are not part of the course.

How do I find a missing eigenvalue with the trace and the determinant?

The trace, the sum of the diagonal entries, equals the sum of the eigenvalues, and the determinant equals their product, each eigenvalue counted as often as it is a root. If two eigenvalues of a 3 by 3 are known, the third is the trace minus the other two. When zero is one of them the determinant is zero and tells you nothing, so use the trace.

Practise it

Corrected exercises: Eigenvalues and eigenvectors, 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 Linear transformations Next sheet Diagonalization

Looking for a MATH 133 tutor in Montreal?

Get in touch for a first session. Eigenvalues are where a first linear algebra course ties everything together: determinants, null spaces and invertibility, used on one matrix.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: eigenvalues and eigenvectors (MATH 133)

The essentials

Two questions, two tools, one matrix

What you get for free once the eigenvalues are known

The rules in table form

Related matrices: which rules exist, and which do not

The mistakes that cost marks

1. Reading the eigenvalues off a row reduced matrix

2. Giving the zero vector when the reduction has no free variable

3. Rejecting an eigenvector because its eigenvalue is negative

4. Expanding a 3 by 3 along the wrong row, then fighting a cubic

5. Giving one vector where the eigenspace needs two

6. Assuming A and its transpose share their eigenvectors

7. Taking opposites instead of reciprocals for the inverse

8. Mixing the two sign conventions for the characteristic polynomial

9. Inventing complex eigenvectors for a rotation

Which method to choose

Where the eigenvalues come from, by the FORM of the matrix

How the answer is expected to be written

Finding a basis of an eigenspace

Check before you hand in

The typical problem, taken apart

A 3 by 3 with a repeated eigenvalue, from the polynomial to the eigenspaces

Learn by heart

Frequently asked questions

How do I find the eigenvalues and eigenvectors of a 3 by 3 matrix by hand?

Why can't I read the eigenvalues off the row echelon form?

Can zero be an eigenvalue?

Does every matrix have real eigenvalues?

How do I find a missing eigenvalue with the trace and the determinant?

Corrected exercises: Eigenvalues and eigenvectors, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?