Chapter 6
Eigenvalues and Eigenvectors
6.1 Introduction to Eigenvalues
Eigenvalues are the key to a system of n differential equations : dy=dt D ay becomes
d y=dt D Ay. Now A is a matrix and y is a vector .y1 .t/; : : : ; yn .t//. The vector
y changes with time. Here is a system of two equations with its 2 by 2 matrix A :
0 y1 0 D 4y1 C y2
y1
4 1
y1
is
D
:
(1)
y2
3 2
y2
y2 0 D 3y1 C 2y2
How to solve this coupled system, y 0 D Ay with y1 and y2 in both equations ? The
good way is to find solutions that “uncouple” the problem. We want y1 and y2 to grow
or decay in exactly the same way (with the same e t ) :
Look for
y1 .t/ D e t a
y.t/ D e t x
In vector notation this is
y2 .t/ D e t b
(2)
That vector x D .a; b/ is called an eigenvector. The growth rate is an eigenvalue. This
section will show how to find x and . Here I will jump to x and for the matrix in (1).
a
1
First eigenvector x =
D
and first eigenvalue D 5 in y D e 5t x
b
1
y1 D e 5t
5t
has
y2 D e
a
Second eigenvector x =
D
b
This y D e t x is a
second solution
y1 D
y2 D
y1 0 D 5 e 5t D 4y1 C
y2
5t
0
y2 D 5 e D 3y1 C 2y2
1
and second eigenvalue D 1 in y D e t x
3
et
3 et
325
has
y1 0 D
y2 0 D
e t D 4y1 C y2
3 e t D 3y1 C 2y2
326
Chapter 6. Eigenvalues and Eigenvectors
Those two x’s and ’s combine with any c1 , c2 to give the complete solution to y 0 D Ay :
5t e
et
1
5t 1
t
Complete solution y.t/ D c1 5t C c2
D c1 e
C c2 e
: (3)
e
3 et
1
3
This is exactly what we hope to achieve for other equations y 0 D Ay with constant A.
The solutions we want have the special form y.t/ D e t x. Substitute that solution
into y 0 D Ay, to see the equation Ax D x for an eigenvalue and its eigenvector x :
d t
.e x/ D A.e t x/
dt
is
e t x D Ae t x:
Divide both sides by e t :
Ax D x
Eigenvalue and eigenvector of A
(4)
Those eigenvalues (5 and 1 for this A) are a new way to see into the heart of a matrix.
This chapter enters a different part of linear algebra, based on Ax D x. The last page of
Chapter 6 has eigenvalue-eigenvector information about many different matrices.
Finding Eigenvalues from det.A
I/ D 0
Almost all vectors change direction, when they are multiplied by A. Certain very
exceptional vectors x are in the same direction as Ax. Those are the “eigenvectors.”
The vector Ax (in the same direction as x) is a number times the original x.
The eigenvalue tells whether the eigenvector x is stretched or shrunk or reversed
or left unchanged—when it is multiplied by A. We may find D 2 or 12 or 1 or 1.
The eigenvalue could be zero ! Ax D 0x puts this eigenvector x in the nullspace of A.
If A is the identity matrix, every vector has Ax D x. All vectors are eigenvectors of I .
Most 2 by 2 matrices have two eigenvector directions and two eigenvalues 1 and 2 .
To find the eigenvalues, write the equation Ax D x in the good form .A I /x D 0.
If .A I /x D 0, then A I is a singular matrix. Its determinant must be zero.
The determinant of A
I D
a
c
b
d
is
.a
/.d
/
bc D 0:
Our goal is to shift A by the right amount I , so that .A I /x D 0 has a solution.
Then x is the eigenvector, is the eigenvalue, and A I is not invertible. So we look
for numbers that make det.A I/ D 0. I will start with the matrix A in equation (1).
4 1
Example 1 For A D
, subtract from the diagonal and find the determinant :
3 2
4 1
det.A I/ D det
D 2 6 C 5 D . 5/. 1/:
(5)
3
2 I factored the quadratic, to see the two eigenvalues 1 D 5 and 2 D 1. The matrices
A 5I and A I are singular. We have found the ’s from det .A I / D 0.
327
6.1. Introduction to Eigenvalues
For each of the eigenvalues 5 and 1, we now find an eigenvector x :
.A
.A
5I / x D 0
is
1I / x D 0
is
1
3
1
3
3
3
1
1
x
x
D
D
0
0
0
0
and x D
and x D
1
1
1
3
Those were the vectors .a; b/ in our special solutions y D e t x. Both components of y
have the growth rate , so the differential equation was easily solved : y D e t x.
Two eigenvectors gave two solutions. Combinations c1 y 1 C c2 y 2 give all solutions.
Example 2
det.A
Find the eigenvalues and eigenvectors of the Markov matrix A D
I / D det
:8
:2
:3
:7
D 2
1
3
C D .
2
2
1/ :8
:2
1
2
:3
:7
.
:
I factored the quadratic into 1 times 12 , to see the two eigenvalues D 1 and 12 .
The eigenvectors x 1 and x 2 are in the nullspaces of A I and A 21 I .
.A
.A
I / x1
1
2 I / x2
" #
:6
x1 D
:4
" #
1
x2 D
1
D 0 is
D 0 is
Ax 1 D x 1
Ax 2 D 21 x 2
"
The first eigenvector is
The second eigenvector is
x1 D .:6; :4/
x2 D .1; 1/
#" #
:6
and Ax 1 D
D x 1 (Ax D x means that 1 D 1)
:4
"
#" # " #
:8 :3
1
:5
and Ax 2 D
D
(this is 12 x 2 so 2 D 21 ).
:2 :7
1
:5
:8 :3
:2 :7
If x 1 is multiplied again by A, we still get x 1 . Every power of A will give An x 1 D x 1 .
Multiplying x 2 by A gave 21 x 2 , and if we multiply again we get . 12 /2 times x 2 .
When A is squared, the eigenvectors x stay the same. A 2 x D A.x/ D .Ax/ D 2 x.
Notice 2 . This pattern keeps going, because the eigenvectors stay in their own directions.
They never get mixed. The eigenvectors of A100 are the same x 1 and x 2 . The eigenvalues
of A100 are 1100 D 1 and . 21 /100 D very small number.
We mention that this particular A is a Markov matrix. Its entries are positive and
every column adds to 1. Those facts guarantee that the largest eigenvalue must be D 1.
328
Chapter 6. Eigenvalues and Eigenvectors
D1
Ax 1 D x 1 D
" #
:6
A2 x 1 D .1/2 x 1
2
D1
:4
2 D :25
D :5
Ax 2 D 2 x 2 D
x2 D
"
"
:5
:5
1
1
2
2
A x 2 D .:5/ x 2 D
#
#
AD
:8
:2
:3
:7
"
:25
:25
#
Ax D x
An x D n x
Figure 6.1: The eigenvectors keep their directions. A2 has eigenvalues 12 and .:5/2 .
The eigenvector Ax1 D x 1 is the steady state—which all columns of Ak will approach.
Giant Markov matrices are the key to Google’s search algorithm. It ranks web pages.
Linear algebra has made Google one of the most valuable companies in the world.
Powers of a Matrix
When the eigenvalues of A are known, we immediately know the eigenvalues of all
powers Ak and shifts A C cI and all functions of A. Each eigenvector of A is also an
eigenvector of Ak and A 1 and A C cI :
If Ax D x then Ak x D k x and A
1
xD
1
x and .A C cI /x D . C c/x: (6)
Start again with A2 x, which is A times Ax D x. Then Ax is the same as Ax for any
number , and Ax is 2 x. We have proved that A2 x D 2 x.
For higher powers Ak x, continue multiplying Ax D x by A. Step by step you reach
k
A x D k x. For the eigenvalues of A 1 , first multiply by A 1 and then divide by :
Eigenvalues of A
We are assuming that A
1
are
1
1
Ax D x
x D A
1
x
A
1
xD
1
x
(7)
exists ! If A is invertible then will never be zero.
Invertible matrices have all ¤ 0. Singular matrices have the eigenvalue D 0.
The shift from A to A C cI just adds c to every eigenvalue (don’t change x) :
Shift of A
If Ax D x then .A C cI /x D Ax C cx D . C c/x:
(8)
As long as we keep the same eigenvector x, we can allow any function of A :
Functions of A
.A2 C 2A C 5I /x D .2 C 2 C 5/x
e A x D e x:
(9)
329
6.1. Introduction to Eigenvalues
I slipped in e A D I C A C 12 A2 C to show that infinite series produce matrices too.
Let me show you the powers of the Markov matrix A in Example 2. That starting matrix
is unrecognizable after a few steps.
"
#
"
# "
# "
#
:6000 :6000
:8 :3
:70 :45
:650 :525
:4000 :4000
:2 :7
:30 :55
:350 :475
(10)
A
A2
A3
A100
A100 was found by using D 1 and its eigenvector Œ:6; :4, not by multiplying 100 matrices.
The eigenvalues of A are 1 and 12 , so the eigenvalues of A100 are 1 and . 12 /100 . That last
number is extremely small, and we can’t see it in the first 30 digits of A100 .
How could you multiply A99 times another vector like v D .:8; :2/ ? This is not an
eigenvector, but v is a combination of eigenvectors. This is a key idea, to express any
vector v by using the eigenvectors.
" # " # " #
:8
:6
:2
Separate into eigenvectors
vD
D
C
:
(11)
v D x 1 C .:2/x 2
:2
:4
:2
Each eigenvector is multiplied by its eigenvalue, when we multiply the vector by A.
After 99 steps, x 1 is unchanged and x 2 is multiplied by . 12 /99 :
" #
" # 2 very 3
:8
:6
1
A99
is A99 .x 1 C :2x 2 / D x 1 C .:2/. /99 x 2 D
C 4 small 5 :
2
:2
:4
vector
This is the first column of A100 , because v D .:8; :2/ is the first column of A. The number
we originally wrote as :6000 was not exact. We left out .:2/. 21 /99 which wouldn’t show up
for 30 decimal places.
The eigenvector x 1 D .:6; :4/ is a “steady state” that doesn’t change (because 1 D 1/.
The eigenvector x 2 is a “decaying mode” that virtually disappears (because 2 D 1=2/.
The higher the power of A, the more closely its columns approach the steady state.
Bad News About AB and A C B
Normally the eigenvalues of A and B (separately) do not tell us the eigenvalues of AB.
We also don’t know about A C B. When A and B have different eigenvectors,
our reasoning fails. The good results for A2 are wrong for AB and A C B, when AB
is different from BA. The eigenvalues won’t come from A and B separately :
0 1
0 0
1 0
0 0
0 1
AD
BD
AB D
BA D
ACB D
0 0
1 0
0 0
0 1
1 0
All the eigenvalues of A and B are zero. But AB has an eigenvalue D 1, and A C B
has eigenvalues 1 and 1. But one rule holds : AB and BA have the same eigenvalues.
330
Chapter 6. Eigenvalues and Eigenvectors
Determinants
The determinant is a single number with amazing properties. It is zero when the matrix has
no inverse. That leads to the eigenvalue equation det.A I / D 0. When A is invertible,
the determinant of A 1 is 1=.det A/. Every entry in A 1 is a ratio of two determinants.
I want to summarize the algebra, leaving the details for my companion textbook
Introduction to Linear Algebra. The difficulty with det.A I / D 0 is that an n by n
determinant involves n Š terms. For n D 5 this is 120 terms—generally impossible to use.
For n D 3 there are six terms, three with plus signs and three with minus. Each of
those six terms includes one number from every row and every column :
2
1
6
44
7
3
2 3 1
7
5 65 4
8 9 7
2
5
Determinant from nŠ D 6 terms
8
C.1/.5/.9/
C C C
Three plus signs, three minus signs
.3/.5/.7/
C.2/.6/.7/ C.3/.4/.8/
.1/.6/.8/
.2/.4/.9/
That shows how to find the six terms. For this particular matrix the total must be det A D 0,
because the matrix happens to be singular : row 1 C row 3 equals 2.row 2/.
Let me start with five useful properties of determinants, for all square matrices.
1. Subtracting a multiple of one row from another row leaves det A unchanged.
2. The determinant reverses sign when two rows are exchanged.
3. If A is triangular then det A D product of diagonal entries.
4. The determinant of AB equals .det A/ times .det B/.
5. The determinant of AT equals the determinant of A.
By combining 1; 2; 3 you will see how the determinant comes from elimination :
The determinant equals ˙ (product of the pivots):
Property 1
Property 2
Property 3
(12)
says that A and U have the same determinant, unless rows are exchanged.
says that an odd number of exchanges would leave det A D det U .
says that det U is the product of the pivots on its main diagonal.
When elimination takes A to U , we find det A D ˙ (product of the pivots). This is how
all numerical software (like MATLAB or Python or Julia ) would compute det A.
Plus and minus signs play a big part in determinants. Half of the n Š terms have plus
signs, and half come with minus signs. For n D 3, one row exchange puts 3 5 7
or 1 6 8 or 2 4 9 on the main diagonal. A minus sign from one row exchange.
331
6.1. Introduction to Eigenvalues
Two row exchanges (an even number) take you back to (2) (6) (7) and (3) (4) (8). This
indicates how the 24 terms would go for n D 4, twelve terms with plus and twelve with
minus.
Even permutation matrices have det P D 1 and odd permutations have det P D 1.
Inverse of A If det A ¤ 0, you can solve Av D b and find A 1 using determinants :
Cramer’s Rule
v1 D
det B1
det A
v2 D
det B2
det A
vn D
det Bn
det A
(13)
The matrix Bj replaces the j th column of A by the vector b. Cramer’s Rule is expensive !
To find the columns of A 1 , we solve AA 1 D I . That is the Gauss-Jordan idea : For
each column b in I , solve Av D b to find a column v of A 1 .
In this special case, when b is a column of I , the numbers det Bj in Cramer’s Rule are
called cofactors. They reduce to determinants of size n 1, because b has so many zeros.
Every entry of A 1 is a cofactor of A divided by the determinant of A.
I will close with three examples, to introduce the “trace” of a matrix and to show
that real matrices can have imaginary (or complex) eigenvalues and eigenvectors.
"
#
2 1
Example 3 Find the eigenvalues and eigenvectors of S D
.
1 2
Solution You can see that x D .1; 1/ will be in the same direction as S x D .3; 3/.
Then x is an eigenvector of S with D 3. We want the matrix S I to be singular.
SD
"
2 1
1 2
#
det .S
"
#
2 1
I / D det
D 2
1
2 4 C 3 D 0:
Notice that 3 is the determinant of S (without ). And 4 is the sum 2 C 2 down the central
diagonal of S . The diagonal sum 4 is the “trace” of A. It equals 1 C 2 D 3 C 1.
Now factor 2 4 C 3 into . 3/. 1/. The matrix S I is singular (zero
determinant) for D 3 and D 1. Each eigenvalue has an eigenvector :
1 1
1
0
1 D 3 .S 3I / x 1 D
D
1 1
1
0
1 1
1
0
2 D 1 .S I / x 2 D
D
1 1
1
0
The eigenvalues 3 and 1 are real. The eigenvectors .1; 1/ and .1; 1/ are orthogonal.
Those properties always come together for symmetric matrices (Section 6.5).
Here is an antisymmetric matrix with AT D A. It rotates all real vectors by D 90ı .
Real vectors can’t be eigenvectors of a rotation matrix because it changes their direction.
332
Chapter 6. Eigenvalues and Eigenvectors
This real matrix has imaginary eigenvalues i , i and complex eigenvectors :
0
1
1
T
AD
D A
det.A I / D det
D 2 C 1 D 0:
1
0
1
Example 4
That determinant 2 C 1 is zero for D i and i . The eigenvectors are .1; i / and .1; i / :
0
1
1
i
1
0
1
1
i
1
D
Di
D
D i
1
0
i
1
i
1
0
i
1
i
Somehow those complex vectors x 1 and x 2 don’t get rotated (I don’t really know how).
Multiplying the eigenvalues .i /. i / gives det A D 1. Adding the eigenvalues gives
.i / C . i / D 0. This equals the sum 0 C 0 down the diagonal of A.
Product of eigenvalues = determinant
Sum of eigenvalues = “trace”
(14)
Those are true statements for all square matrices. The trace is the sum a11 C C ann
down the main diagonal of A. This sum and product are is especially valuable for 2
by 2 matrices, when the determinant 1 2 D ad bc and the trace 1 C 2 D a C d
completely determine 1 and 2 . Look now at rotation of a plane through any angle .
Example 5 Rotation comes from an orthogonal matrix Q. Then 1 D e i and 2 D e
cos QD
sin sin cos 1 D cos C i sin 2 D cos i sin i
:
1 C 2 D 2 cos D trace
1 2 D 1 D determinant
I multiplied .1 /.2 / to get cos2 C sin2 D 1. In polar form e i times e i is 1.
The eigenvectors of Q are .1; i / and .1; i / for all rotation angles .
Before ending this section, I need to tell you the truth. It is not easy to find eigenvalues
and eigenvectors of large matrices. The equation det.A I / D 0 is more or less limited
to 2 by 2 and 3 by 3. For larger matrices, we can gradually make them triangular without
changing the eigenvalues. For triangular matrices the eigenvalues are on the diagonal.
A good code to compute and x is free in LAPACK. The MATLAB command is eig .A/.
REVIEW OF THE KEY IDEAS
1. Ax D x says that eigenvectors x keep the same direction when multiplied by A.
2. Ax D x also says that det.A
3. The eigenvalues of A2 and A
I / D 0. This equation determines n eigenvalues.
1
are 2 and 1
, with the same eigenvectors as A.
4. Singular matrices have D 0. Triangular matrices have ’s on their diagonal.
333
6.1. Introduction to Eigenvalues
5. The sum down the main diagonal of A (the trace) is the sum of the eigenvalues.
6. The determinant is the product of the ’s. It is also ˙ (product of the pivots).
Problem Set 6.1
1
Example 2 has powers of this Markov matrix A :
:8 :3
:70 :45
AD
and A2 D
:2 :7
:30 :55
and A1 D
:6
:4
:6
:
:4
(a) A has eigenvalues 1 and 12 . Find the eigenvalues of A2 and A1 .
(b) What are the eigenvectors of A1 ? One eigenvector is in the nullspace.
(c) Check the determinant of A2 and A1 . Compare with .det A/2 and .det A/1 .
2
Find the eigenvalues and the eigenvectors of these two matrices :
1 4
2 4
AD
and A C I D
:
2 3
2 4
A C I has the
3
by 1.
Compute the eigenvalues and eigenvectors of A and also A 1 :
0 2
1=2 1
AD
and A 1 D
:
1 1
1=2 0
A 1 has the
has eigenvalues
4
eigenvectors as A. Its eigenvalues are
eigenvectors as A. When A has eigenvalues 1 and 2 , its inverse
. Check that 1 C 2 D trace of A D 0 C 1.
Compute the eigenvalues and eigenvectors of A and A2 :
1 3
7
2
AD
and A D
2 0
2
3
:
6
A2 has the same
as A. When A has eigenvalues 1 and 2 , the eigenvalues
of A2 are
. In this example, why is 21 C 22 D 13 ?
5
Find the eigenvalues of A and B (easy for triangular matrices) and A C B :
3 0
1 1
4 1
AD
and B D
and A C B D
:
1 1
0 3
1 4
Eigenvalues of A C B (are equal to) (might not be equal to) eigenvalues of A plus
eigenvalues of B.
334
6
Chapter 6. Eigenvalues and Eigenvectors
Find the eigenvalues of A and B and AB and BA :
1 0
1 2
1 2
3 2
AD
and B D
and AB D
and BA D
:
1 1
0 1
1 3
1 1
(a) Are the eigenvalues of AB equal to eigenvalues of A times eigenvalues of B ?
(b) Are the eigenvalues of AB equal to the eigenvalues of BA ? Yes !
7
8
Elimination produces a triangular matrix U . The eigenvalues of U are on its diagonal (why ?). They are not the eigenvalues of A. Give a 2 by 2 example of A and U .
(a) If you know that x is an eigenvector, the way to find is to
(b) If you know that is an eigenvalue, the way to find x is to
9
.
.
What do you do to the equation Ax D x, in order to prove (a), (b), and (c) ?
(a) 2 is an eigenvalue of A2 , as in Problem 4.
(b) 1
is an eigenvalue of A
1
, as in Problem 3.
(c) C 1 is an eigenvalue of A C I , as in Problem 2.
10
Find the eigenvalues and eigenvectors for both of these Markov matrices A and A1 .
Explain from those answers why A100 is close to A1 :
:6 :2
1=3 1=3
AD
and A1 D
:
:4 :8
2=3 2=3
11
A 3 by 3 matrix B has eigenvalues 0; 1; 2. This information allows you to find :
(a) the rank of B
12
13
(b) the eigenvalues of B 2
(c) the eigenvalues of .B 2 C I /
1
.
Find three eigenvectors for this matrix P . Projection matrices only have D 1 and 0.
Eigenvectors are in or orthogonal to the subspace that P projects onto.
2
3
:2 :4 0
Projection matrix P 2 D P D P T
P D 4 :4 :8 0 5 :
0 0 1
If two eigenvectors x and y share the same repeated eigenvalue , so do all their
combinations cx C d y. Find an eigenvector of P with no zero components.
From the unit vector u D 16 ; 61 ; 36 ; 65 construct the rank one projection matrix
P D uuT . This matrix has P 2 D P because uT u D 1.
(a) Explain why P u D .uuT /u equals u. Then u is an eigenvector with D 1.
(b) If v is perpendicular to u show that P v D 0. Then D 0.
(c) Find three independent eigenvectors of P all with eigenvalue D 0.
335
6.1. Introduction to Eigenvalues
14
Solve det.Q I / D 0 by the quadratic formula to reach D cos ˙ i sin :
cos sin QD
rotates the xy plane by the angle . No real ’s.
sin cos Find the eigenvectors of Q by solving .Q
I /x D 0. Use i 2 D 1.
15
Find three 2 by 2 matrices that have 1 D 2 D 0. The trace is zero and the
determinant is zero. A might not be the zero matrix but check that A2 is all zeros.
16
This matrix is singular with rank one. Find three ’s and three eigenvectors :
2 3
2
3
1 2 1 2
Rank one
A D 425 2 1 2 D 44 2 45:
1
2 1 2
17
When a C b D c C d show that .1; 1/ is an eigenvector and find both eigenvalues :
5 1
a b
Use the trace to find 2
AD
AD
:
2 4
c d
18
If A has 1 D 4 and 2 D 5 then det.A I / D . 4/. 5/ D 2 9 C 20.
Find three matrices that have trace a C d D 9 and determinant 20, so D 4 and 5.
19
Suppose Au D 0u and Av D 3v and Aw D 5w. The eigenvalues are 0; 3; 5.
(a) Give a basis for the nullspace of A and a basis for the column space.
(b) Find a particular solution to Ax D v C w. Find all solutions.
(c) Ax D u has no solution. If it did then
would be in the column space.
20
Choose the last row of A to produce .a/ eigenvalues 4 and 7 .b/ any 1 and 2 .
0 1
Companion matrix
AD
:
21
The eigenvalues of A equal the eigenvalues of A T . This is because det.A I /
equals det.AT I /. That is true because
. Show by an example that the
eigenvectors of A and AT are not the same.
22
Construct any 3 by 3 Markov matrix M : positive entries down each column add
to 1. Show that M T .1; 1; 1/ D .1; 1; 1/. By Problem 21, D 1 is also an eigenvalue
of M . Challenge : A 3 by 3 singular Markov matrix with trace 12 has what ’s ?
23
Suppose A and B have the same eigenvalues 1 ; : : :; n with the same independent
eigenvectors x 1 ; : : :; x n . Then A D B. Reason : Any vector v is a combination
c1 x 1 C C cn x n . What is Av ? What is Bv ?
336
Chapter 6. Eigenvalues and Eigenvectors
24
The block B has eigenvalues 1; 2 and C has eigenvalues 3; 4 and D has eigenvalues 5; 7. Find the eigenvalues of the 4 by 4 matrix A :
2
3
0 1 3 0
6 2 3 0 47
B C
7
AD
D6
4 0 0 6 15:
0 D
0 0 1 6
25
Find the rank and the four eigenvalues of A and C :
2
3
2
1 1 1 1
1
61 1 1 17
60
7
6
AD6
4 1 1 1 1 5 and C D 4 1
1 1 1 1
0
26
27
Subtract I from the previous A.
2
0 1 1
61 0 1
BDA I D6
41 1 0
1 1 1
0
1
0
1
1
0
1
0
Find the eigenvalues of B and
3
2
1
0
6 1
17
7 and
BD6
4 1
15
0
1
(Review) Find the eigenvalues of A, B, and C
2
3
2
1 2 3
0 0
A D 4 0 4 5 5 and B D 4 0 2
0 0 6
3 0
3
0
17
7:
05
1
B:
1
0
1
1
1
1
0
1
3
1
17
7:
15
0
:
3
1
05
0
2
2
and C D 4 2
2
3
2 2
2 25:
2 2
28
Every permutation matrix leaves x D .1; 1; : : :; 1/ unchanged. Then D 1. Find
two more ’s (possibly complex) for these permutations, from det.P I / D 0 :
2
3
2
3
0 1 0
0 0 1
P D 4 0 0 1 5 and P D 4 0 1 0 5 :
1 0 0
1 0 0
29
The determinant of A equals the product 1 2 n . Start with the polynomial
det.A I / separated into its n factors (always possible). Then set D 0 :
det.A
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I / D .1
/.2
/ .n
/
so
det A D
:
The sum of the diagonal entries (the trace) equals the sum of the eigenvalues :
a b
AD
has det.A I / D 2 .a C d / C ad bc D 0:
c d
p
The quadratic formula gives the eigenvalues D .aCd C
/=2 and D
Their sum is
. If A has 1 D 3 and 2 D 4 then det.A I/ D
.
.