Math 125,
Fall 2005
N. Kerzman
DIAGONALIZATION AND FUNCTIONS OF MATRICES.
A ROAD MAP INVOLVING MATLAB.
We use MatLab notation without warning.
Sometimes we use A*B for matrix multiplication (in MatLab style),
and sometimes we just write AB (in math style).
Matrices A are n by n. Their entries may be real or complex. The
eigenvalues and eigenvectors of A are often complex even when A
is real.
We only consider n by n matrices A that have n distinct
eigenvalues. This is the “most frequent” case, as explained in
class.
A need not be symmetric.
References:
Linear Algebra by Otto Bretscher, Prentice Hall. A nice
undergraduate textbook.
Matrix
Theory , Vols. I, II, by F. Gantmacher, Chelsea. A
thorough graduate level classic and reference. Very useful.
-------1.
Diagonalization of A.
What is it?
Starting with any n by n matrix A, the command
(1.1)
[V,R]=eig(A)
produces two n by n matrices V and R. The 1st. column of V is an
eigenvector of A, called v1. It corresponds to an eigenvalue r1
of A. Similarly for the second column, etc.
R is a diagonal
matrix. Its main diagonal has entries, r1, r2, …, rn. They are
the eigenvalues of A.
What we have just said can be written down more compactly
as
(1.2)
AV  VR
1
This is an immediate consequence of the definitions of an
eigenvalue, an eigenvector, and of matrix multiplication.
(Please verify it!)
Notice:
a) R is called R as a reminder that it goes on the right of V in
(1.2).
b) Even if A has all real entries, V and R often have complex
entries.
When A has n distinct eigenvalues r1,…,rn,
the n
eigenvectors v1,…,vn are linearly independent and hence V has an
inverse V^(-1)=inv(V). This is basic linear algebra. See,
e.g., Reference 1. Hence (1.2) can be rewritten
A  VRV 1
(1.3)
Conclusion: An arbitrary matrix A need not be diagonal, of
course. But if it has n distinct eigenvalues, then A can be
expressed via (1.3) in terms of an associated diagonal matrix R.
In the main diagonal of R sit precisely the eigenvalues of A. In
mathematics language: “A has been diagonalized to R via the
matrix V and (1.3)”
Example 1.1
A =
7
2
0
2
6
2
-1
2
8
>> [V,R]=eig(A)
V =
0.5524
-0.7580
0.3467
0.7071
-0.0000
-0.7071
0.1283
0.5611
0.8177
0
8.0000
0
0
0
9.3723
R =
3.6277
0
0
2
Checking now, we see that (1.3) holds:
>> V*R*inv(V)
ans =
7.0000
2.0000
0.0000
2.0000
6.0000
2.0000
-1.0000
2.0000
8.0000
The answer is A.
Example 1.2.
A =
0
-1
1
0
>> [V,R]=eig(A)
V =
0.7071
0 + 0.7071i
0.7071
0 - 0.7071i
R =
0 + 1.0000i
0
0
0 - 1.0000i
Checking (1.3):
>> V*R*inv(V)
ans =
0
-1
1
0
This example shows how starting with a very simple real matrix,
you may end up having to deal with complex eigenvalues and
eigenvectors. In this case the eigenvalues are r1=i, r2=-i.
--------
3
2.
What’s the point of diagonalization.
What do you gain
with it?
When a matrix D is diagonal , it is extremely easy to write
D^2, D^3,…, inv(D), and more . Namely, if D  diag ([ d1 , d 2 ,..., d n ]) ,
then
D 2  diag ([d1 , d 2 ,..., d n ]) ,
2
2
2
D 3  diag ([d1 , d 2 ,..., d n ])
3
3
3
D 1  diag ([1 / d1 ,1 / d 2 ,...,1 / d n ]) --the last one assuming that all the
d j  0 , which is the only situation in which
D 1 exists.
Also
expm( D)  e D  diag ([ e d 1 , e d 2 ,..., e d 3 ]) .
All this is obvious when you focus your attention on what matrix
multiplication means. Hence, for diagonal matrices D, the
function f(D) of the matrix D, e.g., f ( D)  D 2 , is obtained by
simply replacing each d j by f ( d j ) .
When the n by n matrix A is not diagonal, but it has n
distinct eigenvalues, then (1.3), allows you to obtain f(A)
working in the easiest possible way with R, exactly as we did
with D above.
SOMETHING IMPORTANT TO NOTICE :
(2.1)
(1.3) implies
A2  VRV 1VRV 1  VR 2V 1
because VV 1  I
Similarly,
(2.2)
A3  VR 3V 1 ,
A4  VR 4V 1 ,
etc.
And
A1  VR 1V 1 ,…,
(2.3)
that R 1 exists
A4  VR 4V 1 ,
Please verify these formulas by yourself.
understand each step.
etc. -- assuming
Make sure you
Also for the exponential matrix e A , that by definition is
A A2
 ... , using (2.1) and (2.2) you get
the series e  I  
1! 2!
A
e A  Ve RV 1
4
Example 2.1
Find as many square roots as you can of the
matrix A in Example 1.1.
Here the function f ( A)
is
A.
This looks difficult because A is not diagonal. But R is
diagonal. R has many (eight) obvious square roots. You get them
by taking the square root of each R(1,1), R(2,2), R(3,3). There
are two choices for each , e.g., 
For each choice of 
(2.4)
R(1,1) .
(of the eight possible ones), the matrix
 R(1,1)

M V *
0

0

0
 R( 2,2)
0

0

1
0
 *V
 R(3,3) 
is a square root of A.
Reason: when you do M 2  M * M , you get
M * M  V * R * V 1 * V * R * V 1  VRV 1  A .
--------
3.
An
m-file for Example 2.1
%square roots of a 3 by 3 matrix
A=[7 2 -1;2 6 2;0 2 8];
[V,R]=eig(A);
%A=V*R*inv(V)
%Compute the eight sqrt(R), call them S1, S2, etc.
%S1=diag(w1); w1=[sqrt(R(1,1),sqrt(R(2,2),sqrt(R(3,3)], etc.
w1=[sqrt(R(1,1)),sqrt(R(2,2)),sqrt(R(3,3))];
S1=diag(w1);M1=V*S1*inv(V);
w2=[sqrt(R(1,1)),sqrt(R(2,2)),sqrt(R(3,3))];S2=diag(w2);M2=V*S2*inv(V);
w3=[sqrt(R(1,1)),sqrt(R(2,2)),sqrt(R(3,3))];S3=diag(w3);M3=V*S3*inv(V);
w4=[sqrt(R(1,1)),-sqrt(R(2,2)),sqrt(R(3,3))];S4=diag(w4);M4=V*S4*inv(V);
w5=[sqrt(R(1,1)),sqrt(R(2,2)),sqrt(R(3,3))];
S5=diag(w5);M5=V*S5*inv(V);
w6=[-sqrt(R(1,1)),sqrt(R(2,2)),sqrt(R(3,3))];S6=diag(w6);M6=V*S6*inv(V);
5
w7=[-sqrt(R(1,1)),sqrt(R(2,2)),sqrt(R(3,3))];S7=diag(w7);M7=V*S7*inv(V);
w8=[-sqrt(R(1,1)),-sqrt(R(2,2)),sqrt(R(3,3))];S8=diag(w8);M8=V*S8*inv(V);
--------
4. Questions and answers
a) Are the eight roots of A we got, namely
different?
M1, M2,…, M8 all
Yes. Reason: say,e.g, M1=M2. Then S1=S2 necessarily because
S1=inv(V)*M1*V and S2=inv(V)*M2*V.
But the eight matrices S1,
S2,…,S8 are different because they arise from different vectors
w1,w2,…,w8. (the signs are different in some slot). Hence the
Mk are also different.
b) Are there any other square roots of A thay we did not think
of?
No: there are only these eight, but this is more difficult to
show and we won’t do it here.
5.
Complex numbers: roots and logarithms
The matrix A in Example 1.2 has complex eigenvalues i and –i. If
you want to get square roots of A, you will need to know what are
 i . See the Notes on Complex Numbers on this web page.
i and
After this hurdle, you proceed exactly as before and get the four
square roots of A as
 i
A V *
 0
with i  e
i / 4

 * inv (V )
  i
0
 i  e  i / 4
,
Similarly, if you want to obtain log(A)
0 
log( i )
log( A)  V * 
* inv (V )
log( i )
 0
with log( i )  i

2

i 2k ,
log( i )  i
6

2
 i 2q ,
k  0,  1,  2,.... and q  0,  1,  2,.... can be chosen independently of
where
each other.
We have written only four square roots of A but infinitely many
logarithms of A!!
7