3.3 & 5.5 Eigenvalues and
Diagonalization
Definitions
A is n x n.  is an eigenvalue of A if
AX = X has non zero solutions X (called eigenvectors)
If  is an eigenvalue of A, the set
E = E(A) = {X | X in n, AX = X}
is a vector space called the eigenspace associated w/ 
(i.e. E is all eigenvectors corresponding to  & 0 vector)
 is eigenvalue if E has at least one non-zero vector.
Can also write AX = X as (In - A)X = 0
Example
Show that  = -3 is an eigenvalue of A, and find the
eigenspace E-3.
8
16 
 5
A 

4


 4
1
4


 11 

8
Write out (In - A)X = 0 and solve.
Get:
 1
 
X s 1 
 

 0 

2 
 
t 0
 

 1 

So it is an eigenvalue since there is a
non-zero solution. Eigenspace is:
E3
1   2 
   
 span  1 , 0 
   
   
 0   1 
Discussion
Now we have (In - A)X = 0, and  is an eigenvalue iff there
exists a nonzero solution X.
Recall that a matrix U is invertible iff UX = 0 implies X = 0.
So, since we are looking for a nonzero solution above,
(In-A) cannot be invertible for  to be an eigenvalue.
So det (In-A) =0.
Definition
The characteristic polynomial of the n x n matrix A is:
cA(x) = det(xI - A)
Theorem 1
A (n x n). The eigenvalues of A are the real roots of the
characteristic polynomial of A --the real numbers  satisfying:
cA() = det(In - A) = 0
The eigenspace E = {X | (I - A)X = 0}
consists of all solutions to a system of n linear equations in n
variables.
The eigenvectors corresponding to  are the nonzero vectors
in the eigenspace.
Summary
So there are two issues: finding eigenvalues, and finding
eigenspaces (and eigenvectors).
Finding the eigenvalues can be difficult - won’t do much here.
Spend more time dealing with eigenspaces.
Example
Find the characteristic polynomial, eigenvalues, and
eigenspaces of A:
3 
1  2

A  2


1
6
2

6

 1 

Set up cA(x) = det (xI - A)
Eigenvalues will be the roots of the polynomial as those will
give us where det is 0.
Then use those  to find eigenspace: X such that ( I-A)X=0
Example
If A is a triangular matrix, show that the eigenvalues of A are
the entries on the main diagonal.
Proof: cA(x) = det (xI - A) = det ( a triangular matrix) =
product of entries on main diagonal of (xI - A).
The matrix showing entries on main diagonal is:
x  a 11





x  a 22
...





x  a nn 
det = (x-a11)(x-a22)…(x-ann)
So eigenvalues are{a11,a22,…,ann}
Example
Show that A and AT have the same characteristic polynomial
and thus the same eigenvalues.
Proof: From chapter 3, we know that a matrix and its
transpose will have the same determinant.
c
(x )  det( xI  A )  det(( xI )  A )
T
A
T
T
 det( xI  A )  det( xI  A )  c A ( x )
T
T
Theorem 2
If A is a real symmetric matrix, each root of the characteristic
polynomial cA(x) is real. (to be proven later)
Show this is true for a (2 x 2):
a
A  
b
b 

c 
x  a
c A ( x )  det 
  b
 b 
2

(
x

a
)(
x

c
)

b

x  c 
 x  x (a  c )  (ac  b )
2
2
Recall that we can determine the nature of the roots from the
discriminant: (b2-4ac) = (a+c)2-4(ac+b2) = a2+c2+2ac-4ac+4b2
=a2-2ac+c2+4b2 = (a-c)2 + 4b2 which is always pos so real roots.
Similar Matrices
A, B (n x n) are similar (we say A~B) if B = P-1AP
holds for some invertible matrix.
P is not unique.
Example
Find P-1AP in the following case, then compute An.
1
P  
1
5 
6
, A  
2 
2
 5 

1 
We are able to find a similar matrix B.
Then P-1AP=B.
So A = PBP-1
So A2=(PBP-1)(PBP-1)=PB2P-1
Generally An=PBnP-1
Life is made easy is B is diagonal since we just raise entries to n.
Interesting Fact
Similar Matrices will have the same determinant.
Proof:
P-1AP = D
det(D) = det (P-1AP) = (detP-1)(detA)(detP) =
(1/detP)(detA)(det P) = det A.•
Example
Show that A and B are not similar.
1
A  
2
2 
 1
, B  
1 
 1
1

1
Just need to show that they do not have the same determinant.
Trace
The trace of a square matrix A (tr A) is the sum of the entries
on the main diagonal of A.
Theorem 3
A,B (n x n), k is a scalar:
1. tr(A + B) = tr A + tr B and tr(kA) = k tr A
2. tr (AB) = tr (BA)
Proof:

a b

1. (homework)


a b
2.
AB 
n
1j
j 1
j1
n
2j





j2
j 1
...
 n

tr ( AB )    a ij b ji  

i  1  j  1
n






n

a
b
 nj jn 
j 1



  b ji a ij   tr ( BA )
j 1
i 1
n
n
Theorem 4
If A~B, they have the same determinant, the same rank, the same
trace, the same characteristic polynomial, and the same
eigenvalues. (similarity invariants)
Proof: Already shown that they have the same determinant.
Rank: Have B = P-1AP
rank (B) = rank (P-1AP) = rank(AP)=rankA since P is invertible
(and using cor 4 of thm 4 in 5.5)
tr B = tr (P-1AP) = tr[(AP)P-1] = tr (A) (uses thm 3)
Theorem 4 - cont
Characteristic polynomial
cB(x) = det (xI - B) = det(xI - P-1AP)=det(P-1xIP - P-1AP)
(since xI = P-1xIP -- since xI is diagonal)
= det [P-1(xI - A)P]=(1/detP)(det(xI-A))(det P) = det(xI-A) = cA(x)
Eigenvalues: all matrices with the same characteristic poly will
have the same eigenvalues since the eigenvalues are the roots of
the characteristic polynomial.
•
Fact
The invariants do not imply similarity.
Ex. I  1 0 , A  1 2 
0
1 
0
1 
Have same det,tr,rank,characteristic poly, eigenvalues, but are
not similar since P-1IP = I ≠ A
Theorem 5
A,B,C (n x n). Then:
1. A~A for all A.
2. If A ~ B, then B~A
3. If A ~ B and B ~ C, then A~C.
Proof of 2 (others follow):
A~B  B = P-1AP
Let Q = P-1, then B = QAQ-1, so A= Q-1BQ
Which means that B ~ A.
Use of thm 5
Proving similarity is not always easy. But if we can find a simple
(often diagonal) matrix to which both A and B are both similar,
then: A~D and B~D means D~B by (2)
and A~B then by (3)
Definitions
A (n x n) is diagonalizable if it is similar to a real diagonal
matrix (i.e. there exists real and invertible P such that P-1AP
is a diagonal matrix).
The convenient thing about this is that P-1AP will have the
same eigenvalues as P, and that they are just the entries on
the main diagonal. This makes life very easy!
Furthermore, since we said that P would be real, the
eigenvalues will be real.
Definitions
A (n x n) is diagonalizable if it is similar to a real diagonal
matrix (i.e. there exists real and invertible P such that P-1AP
is a diagonal matrix).
The convenient thing about this is that P-1AP will have the
same eigenvalues as P, and that they are just the entries on
the main diagonal. This makes life very easy!
Furthermore, since we said that P would be real, the
eigenvalues will be real.
Method (continued)
We write P-1AP = D as AP = PD, so that:
AX1
A X
1
X2
AX
X n   X1
...
2
...
X2
AX n    1 X 1
...
 1

0

X n  0

...
 0

 2X2
...
0
0
...
2
0
...
0
3
...
...
...
...
0
0
...
0 

0

0 

... 
 n 

 nXn
AX i   i X i , i  1, 2 ,..., n
This just means that the columns of P are the eigenvectors of
A corresponding to the eigenvalues 1, 2,…, n (in order).
Method (continued)
Since P must be invertible, we know that eigenvectors will have
to be LI.
Conversely, we know that if the columns X1,X2,…,Xn are LI,
then P is invertible, and we can use the reverse of the argument
we just used to show that AP = PD where D is diagonal. So
P-1AP = D. This gives us:
Theorem 1: An n x n matrix is diagonalizable iff it has n LI
eigenvectors (which are then a basis of n.
Diagonalization Algorithm
A (n x n)
1) Find the eigenvalues of A.
2) If possible, find n LI eigenvectors X1,X2,…,Xn
3) Form P = [X1 X2 … Xn]
4) P-1AP is diagonal, the diagonal entries are eigenvalues
corresponding to X1,X2,…,Xn (in order)
Example
Determine whether A is diagonalizable, and if it is, find P
such that P-1AP is diagonal.
1 0 0  Here, c (x) = (x-1)2(x-2) so  =1,  =2.
A
1
2

A  1


0
2
0


1 

1
E1=span{X1,X2}, E2=span{X3}
 1
 1
0 
 
 
 
X1  0 , X2  1 , X3  1
 
 
 



 1 

 0 

0 

Find
P-1,
show
P-1AP=D
These are LI so:
1

D 0


0
0
1
0
 1

P  0


 1
0 

0

2 

1
1
0
0 

1

0 

If not Diagonalizable
It may happen that you the eigenvalues do not give n LI
eigenvectors. (You may get fewer than n eigenvectors, or the n
eigenvectors you get may not be LI.) If this happens, the
matrix is not diagonalizable.
Theorem 2
Let 1, 2,…, k be distinct eigenvalues of a matrix A.
If X1, X2,…, Xk are eigenvectors corresponding to 1, 2,…, k ,
then {X1, X2,…, Xk} is a LI set.
Proof (by induction):
for k=1: X1 is clearly LI
Assume true for k-1: r1X1+r2X2+…+rk-1Xk-1=0 implies r’s =0
Show true for k:
r1X1+r2X2+…+rkXk=0
(1)
Left mult by A:
r1AX1+r2AX2+…+rkAXk=0
Since AXi=iXi,
r1 1X1+r2 2X2+…+rk kXk=0
(2)
Mult. (2)- k(1) :
r1 (1 - k)X1+…+rk (k-1 - k)Xk-1=0
Theorem 2
Have:
r1 (1 - k)X1+…+rk (k-1 - k)Xk-1=0
We know X1,…,Xk-1 are LI, so
r1 (1 - k)=…= rk-1 (k-1 - k)=0
Since ’s are distinct, r1=…=rk-1=0
Which leaves us in (1) with rkXk=0.
Since Xk≠0 (by def of eigenvalues), rk=0. •
Theorem 3
An (n x n) matrix with n distinct eigenvalues is
diagonalizable.
This is a natural conclusion from Thm’s 1 and 2
Example
Show that A is diagonalizable:
1

A  0


0
0
1
0
1 

0

2 

Just need to show that there are n distinct eigenvalues.
To note
Recall that in our first example, we had a diagonalizable matrix
which did not have n distinct eigenvalues.
It is not true that if a matrix does not have n distinct eigenvalues
it is not diagonalizable.
So if you are asked to show whether a matrix is diagonalizable,
it is not enough to show that it does not have n distinct
eigenvalues.
Multiplicity
An eigenvalue  has multiplicity m if it is repeated as a root m
times in the characteristic equation. For example:
cA(x) = (x+2)2(x-1)
 = -2 has multiplicity 2
 = 1 has multiplicity 1
Dimension of the eigenspace
Each eigenvalue will correspond to an eigenspace which has a
certain set of vectors as its basis. The number of vectors in
the basis is the dimension of the eigenspace.
Recall example 1:
1

A  1


0
0
2
0
0 

1

1 

Here, cA(x) = (x-1)2(x-2) so 1=1, 2=2.
E1=span{X1,X2}, E2=span{X3}
 1
 1
0 
 
 
 
X1  0 , X2  1 , X3  1
 
 
 



 1 

 0 

0 

E1 has dimension 2.
E2 has dimesion 1.
Theorem 4
A (n x n), c A ( x )  ( x   1 ) m ( x   2 ) m    ( x   k ) m
1
2
Where 1, 2, …, k are distinct real eigenvalues of A with
multiplicity m1, m2, …, mk respectively.
d i  dim( E  )
i
Then the following are equivalent:
1. A is diagonalizable
2. d1 + d2 +…+dk = n
3. di = mi for each i.
(We will prove after thm 5)
k
Theorem 5
If  is an eigenvalue of A of multiplicity m, then dim (E)≤m
Proof of Theorem 4
By thm 5, di ≤ mi for each I.
Also m1 + m2 +… + mk = deg(cA(x)) = n
(1)(2) Since A is diagonalizable, by thm 1, it has n LI
eigenvectors. Let B be the set of these n LI eigenvectors, and
let ti of these lie in E
. Then ti ≤ di = dim E


n = t1 + t2 +…+ tk ≤ d1 + d2 +…+ dk ≤ m1 + m2 +…+ mk = n
So d1 + d2 +…+ dk = n.
i
i
Proof of Theorem 4
(2)(3) n = d1 + … + dk ≤ m1 + … + mk = n
So d1 + … + dk = m1 + … + mk .
Also di ≤ mi for each i (by thm 5). So di = mi for each i.
Proof of Theorem 4
(3)(1) Let Bi be a basis for E for each i, and let B be the set
of all vectors in all of these bases. This set will consist of
d1 + … + dk = m1 + … + mk = n vectors (by 3)
We just have to show that this set B is LI as we will then have
that A has n LI eigenvectors which means it is diagonalizable.
Let Bi={vi1,vi2,…,vidi}
Let Yi= ai1vi1+ai2vi2+…+aidividi be a linear combination of
vectors in Bi.
Let Y1+Y2+…+Yk=0 -- we will prove that this means that the
coeff’s (a’s) must be 0 proving the vectors LI.
i
Proof of Theorem 4
Each Yi lies in its corresponding eigenspace since it is just a
linear combination of the vectors.
Since the i are distinct, the Yi are LI by thm 2.
But the Yi sum to zero, so each Yi must be zero (otherwise we
would have linear dependence).
Therefore, Yi = ai1vi1+ai2vi2+…+aidividi = 0
Since this is a linear combo of vectors from the basis, and
these vectors are LI, the coefficients must be 0.
This tells us that all of the vectors in B are LI and therefore A
is diagonalizable. •
Use of thm 4
Now to show that a matrix is diagonalizable, all we need to
show is that di = mi for each i. In other words, we need to
show that the dimension of each eigenspace is the same as the
eigenvalue’s multiplicity.
Example
Show that A is not diagonalizable.
1

A  0


0
1
1
0
0 

0

2 

Just need to find the eigenvalues and the corresponding
eigenspaces.
We compare the multiplicity of the eigenvalues to the
dimension of their corresponding eigenspaces. If they are not
equal in all cases, then the matrix is not diagonalizable.
Notation
diag (1, 2,…, n) denotes the (n x n) diagonal matrix with
1, 2,…, n in order down the main diagonal.
Example
Suppose 2 = 2 for every  of a diagonalizable A. Show
that A2 = 2A.
Proof: P-1AP = D = diag (1, 2,…, n)
Since i2 = 2i for each i, we get:
D2 = diag (12, 22,…, n2) = 2 diag (1, 2,…, n) =2D
So, A2 = (PDP-1)2 = PD2P-1 = P(2D)P-1 = 2(PDP-1) = 2A •
Example
Suppose a sequence of numbers x0,x1,x2,… is determined by the
condition that x0= x1 = 1 and then each successive xn is:
xn+2 = 6xn + xn+1 n ≥ 0 so x2 = 7, x3 = 13, etc.
We will find
 x n 
V n  

x n  1 
V1 = AV0,
V2 = AV1 = A(AV0) = A2V0
Vn = AnV0
Vn 1
x n  1 
 


x n  2 
x 0 
V 0    
x 1 
0

6
1 x n 
x   AV n
1 n  1 
1
 
1
Then An calc is easier by
diagonalizing.