Alg1231W8T3 - Diagonalisation

Wednesday, 12 September 2018

10:12 PM

Machine generated alternative text:
If we have a matrix A and we construct a matrix M , whose columns are just eigenvectors of A , then we always have 
AM = MD 
for some diagonal matrix D However, if we wish to use this matrix to write down a nice factored form for A, we must take care when constructing M to ensure that M is an invertible 
matrix. 
-11 
Consider the matrix A 
7 
If we define 
-14 
This has eigenvectors VI — 
10 
and v 2 
AVI = 3 VI 
and 
where 
AV2 — 
—4". 
10 
—10 
(i.e define L to be the matrix whose columns are VI and 10V1)v then we can calculate 
1104) ( 
—11 
Note: the Maple notation for the matrix 
112 >,< 314 
3 
o 
10 
-10 
3 
Since the columns of L are linearly dependent; the matrix L 
If we instead define 
10 
-10 
whose columns are the linearly independent eigenvectors above, then we get 
—11 
AM 
is not invertible! So we cannot use this to diagonalise A. 
1104) ( 
3

 

Machine generated alternative text:
Since the columns of M are linearly independent, the matrix M is invertible. Namely M has inverse 
By multiplying on the right by M we get A = MDM¯I where 
In this case A is said to be diagonalisable

 

Machine generated alternative text:
Two eigenvectors with different eigenvalues are always linearly independent To see why, take a matrix A with eigenvectors VI and v2 such that 
AVI = AV2 = tv,2 where s # t. 
Recall that VI and v2 must be non-zero vectors, since being non-zero is part of the definition of an eigenvector 
Now consider the equation 
PVI + qu = O. 
(1) 
lution to (1) isp= q = O , then we will have shown VI and v2 are linearly independent 
If we can show that the only 
Since PVI = —qv,2 we have 
spvl = A(pvl) = A(—qv2) —tqv2 = tPV1 • 
Then, by rearrangement 
p(s — t)V1 = O. 
There are three cases to consider: 
VI = O, which is impossible since eigenvectors are non-zero by definition; 
O @twhich is impossible since the eigenvalues were assumed to be different, 
• orp=0. 
Only one of these three cases is possible, so we conclude p = 0_ Then equation (1) becomes 
qv2 = O. 
Since v2 must be a non-zero vector the only way to solve this is to have q = 0. Thus we have show that 
v2 must be linearly independent. 
the only solution 
O 
IS p 
0, and hence the vectors VI and

 

Machine generated alternative text:
The number of linearly independent eigenvectors for the eigenvalue is equal to nullity(A — M). 
• if A 
then there is one eigenvalue = 2 and nullity(A — 21) = 2 
• ifB 
then there is only one eigenvalue = 2 and nullity(B — 21) = 1 
Hence; 
O and the number of linearly independent eigenvectors for = 2 is 2 
O 
and the number of linearly independent eigenvectors for = 2 is 1 
Because the matrix B does not have enough linearly independent eigenvectors, it is called a defective matrix, and cannot be diagonalised

 

Machine generated alternative text:
Which of the following matrices are diagonalisable? 
4 10 
any n x n matrix with n linearly independent eigenvectors 
any 2 x 2 matrix with two distinct eigenvalues

 

Machine generated alternative text:
1 
1 
We are told that A 
has eigenvalues 5 and 3 with respective eigenvectors 
-2 
and 
. So we can diagonalise A by finding an invertible matrix 
—1 
and diagonal matrix D such that 
5 
i) If we wanted D to be the matrix D 
o 
M-IAM 
then we should take 
3 
ii) If instead we wanted D to be the diagonal matrix D 
then we should take 
iii) One of the advantages of having a diagonal matrix is that matrix powers are easy to take, for example 
6 
The relation for ATI is not nearly as simple. 
iv) Once we write A = MDM¯I we observe that 142 
MD2M¯1 .43 = MD3M¯1, and so on. Thus, using parts (i) and (iii), we can easily calculate that the top 
left entry of 146 is 30521 
o. 
1 
Recall: the Maple notation for the matrix 
3 
2 
is 
4

 

 

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