Alg1231W8T4 - Eigenvalues of some special matrices

Thursday, 13 September 2018

7:05 PM

Machine generated alternative text:
Consider the matrix 
1 
1 
1 
1 
3 
2 
1 
3 
This is a special type of matrix called a transition matrix, which are use to model simple random processes know as Markov chains. The two properties that define a 
transition matrix P are: 
1. Every entry in P is non-negative (i.e. greater than or equal to zero) 
2. Every column of P sums up to 1. 
Being a rather special type of matrix, transition matrices have rather special eigenvalues. 
Consider what happens when we multiply the transpose p T by the vector of all I's 
1 
1 
This proves that p T has an eigenvector with eigenvalue 1 
eigenvalue. 
1 
1 
o 
1 
2 
9 
1 
1 
o 
o 
1 
1 
1 
O . Recalling what we learnt in the video, this means P also has the same corresponding 
Let's find the corresponding eigenvector of the matrix P above. By considering the augmented matrix 
(P -110) 
we row-reduce and back-substitute to get an eigenvector <4,6,1> 
-1/2 
1/4 
1/4 
1/3 0 0 
-1/3

 

Machine generated alternative text:
We can see this is quite different from 
will be. 
Note: the Maple notation for the vector 
1 
I 
1 
1 
2 
3 
. So while P and PT have matching eigenvalues, it is hard to guess what (if any) the relationship between their eigenvectors

 

Machine generated alternative text:
Consider a matrix A with the special property that Am = I for some positive integer m. Such a matrix is special, as such we might expect it to have special eigenvalues. 
Say v is an eigenvector of A with eigenvalue t. By left-multiplying m times by the matrix A we get 
Am v = Iv = v. 
Since v is an eigenvector with eigenvalue t, this implies that 
Since eigenvectors are non-zero O the only way this can be satisfied is if 
t 
implying that the eigenvalues of A must be mth roots of unity 
An example of such a matrix is 
o 
1 
—1 
Here m 
Recall: 
O (please enter the smallest possible m) and the eigenvalues are the set 
{1,-1} 
• a set in Maple notation is something of the form {1 ,1+2*1 , 3-2* I} 
• remember to use capital i (I) to represent i — 
—1 
can be entered using the Maple syntax sqrt (2)

 

Machine generated alternative text:
Consider a matrix A with the special property that Am = Z for some integer m. Here Z is the zero-matrix (i.e. the matrix filled with zeroes). Such a matrix A is called 
nilpotent, and as it is clearly special, it will have special eigenvalues. 
Say v is an eigenvector of A with eigenvalue t. By left-multiplying m times by the matrix A we get 
Am v = Zv = O. 
Since v is an eigenvector with eigenvalue t, this implies that 
tm v = O. 
Since eigenvectors are non-zero O the only way this can be satisfied is if 
implying that the eigenvalues of A must be zero 
An example of a nilpotent matrix is 
B 
t 
o 
o 
o 
1 
o 
o 
—1 
1 
o 
since Bm 
o 
o 
o 
o 
o 
o 
o 
0 
o 
where m 
3

 

Machine generated alternative text:
Consider the matrix 
5 
o 
o 
o 
1 
—5 
o 
o 
30 
4 
9 
o 
—5 
120 
1300 
300 
This is a special type of matrix known as an upper triangular matrix. All the non-zero entries appear either on or above the main diagonal. 
The eigenvectors of this matrix may not be particularly obvious, but the eigenvalues are. To see why, let's calculate the characteristic polynomial of A. 
p(t) — det(A — tI) 
= det 
o 
o 
1 
o 
o 
30 
4 
o 
By expanding down the first column we get 
p(t) 
Repeating this procedure we get 
—5 
120 
1300 
300 - t 
o 
4 
o 
120 
1300 
300 - t 
1300 
o 
300 - t 
And so we can see the roots of p(t), and hence the eigenvalues of A , are the numbers in the set 
Note: enter your answer as a set (e.g. {1,2,3,4}). 
The same argument shows that the eigenvalues of an upper triangular matrix are 
the diagonal entries

 

Machine generated alternative text:
Consider a matrix A with the special property that Am = A for some positive integer m. Such a matrix is special, and as such we might expect it to have special 
eigenvalues. 
Say v is an eigenvector of A with eigenvalue t. By left-multiplying m times by the matrix A we get 
Am v — Av 
Since v is an eigenvector with eigenvalue t this translates into 
Since eigenvectors are non-zero O the only way this can be satisfied is if 
implying that the eigenvalues of A must be either zero; or (m-l)th roots of unity 
An example of such a matrix is 
tmv 
tm 
-2 
1 
-1 
tv. 
-5 
3 
o 
3 
-2 
—1 
Here m 
Recall: 
5 and the eigenvalues are the set 
• a set in Maple notation is something of the form {1 ,1+2*1 , 3-2* I} 
• remember to use capital i (I) to represent i — 
—1 
x./ä can be entered using the Maple syntax sqrt (2)

 

 

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