Alg1231W8T1 - Eigenvectors and Eigenvalues

Sunday, 9 September 2018

6:46 PM

 

 

Machine generated alternative text:
The eigenvectors of a square matrix A are all 
for some scalar A, called the eigenvalue 
non-zero 
O 
vectors v such that 
For example; if A 
o 
-2 
3 
-5 
-3 
-3 
5 
2 then we can directly calculate that 
2 
1 
1 
0 
-2 
3 
-5 
-3 
-3 
5 
2 
2 
5 
1 
1 
So we have demonstrated that 
is an eigenvector with eigenvalue 
1 
Note: the Maple notation for the vector 
Similarly we can check that 
2 
3 
1,2,3 > 
1 
1 
1 
1 
is an eigenvector with eigenvalue 
is an eigenvector with eigenvalue

 

Machine generated alternative text:
-0.87 
-0.5 
Consider the matrix 
This matrix has eigenvalues 2 and 3 with respective eigenvectors 
Note: 
0.12 
• the Maple notation for the vector 
is < 
0.12, 0.34 > 
0.34 
• make sure VI corresponds to the eigenvalue 2 and v2 corresponds to the eigenvalue 3.

 

Machine generated alternative text:
0.97 
0.26 
Consider the matrix 
This matrix has eigenvalues —1 and 3 with respective eigenvectors 
112

 

Machine generated alternative text:
For a square matrix A, the statement 
is equivalent to the statement 
Say your friend claims that the matrix A 
equations 
10 
1 
-1 
2 
11 
1 
-2 
-1 
9 
"A has an eigenvalue " 
"(A — M)v = O has a non-zero solution v ' 
has eigenvalue = 9 and wants our help finding the eigenvector We proceed by attempting to solve the system of linear 
(A — 91)v = O. 
That is, we try to solve the system represented by the augmented matrix 
1 
1 
—1 
This matrix can be row-reduced to 
1 
o 
o 
Recalling our experience in Math1131 we can see this has a unique solution; namely v 
1 
Note: the Maple notation for the vector 2 1 , > 
2 
2 
1 
2 
3 
o 
—2 
—2 
1 
o 
3 
Since an eigenvector MUST be non-zero 
our friend should double check their work 
Cl the eigenvector is O 
that 9 is not an eigenvalue of the matrix A. 
O , we conclude that [select all that apply]

 

Machine generated alternative text:
Your friend from the last question claims that the matrix B 
7 
1 
-1 
2 
8 
1 
-2 
-1 has eigenvalue 
6 
5 (your friend swears they double-checked this time). We can check this claim by 
attempting to solve the system of linear equations represented by the augmented matrix 
This can be row-reduced to the matrix 
and so we can find a non-zero solution v = 
1 
Note: the Maple notation for the vector 2 is < 1 
3 
From this we can conclude that [select all that apply] 
5 really is an eigenvalue of the matrix 
our friend might just pass this course 
5 is not an eigenvalue of the matrik 
2 
1 
—1 
2 
o 
o 
2 
3 
1 
2 
2 
o

 

Machine generated alternative text:
For a square matrix A, the statement 
"(A — = O has a non-zero solution v ' 
is equivalent to the statement 
"det(A - M) = O 
The quantity det(A — M) is a polynomial in A, known as the characteristic polynomial of A, and its roots are the eigenvalues of A. This gives us a technique for finding the eigenvalues of 
t2 — 
Let's check our understanding 
1 
i) The characteristic polynomial of A 
2 
2 
1 
det(A - ti) 
2 
is the cubic 
2 
= det 
2 
VI ¯ 
and v 2 ¯ 
det(B - ti) - 
(I-t)A2 - 4 
O . Ordered the same way, the associated eigenvectors are 
Ordered tl < t2, the roots ofthis roots are tl — 
1 
Note: the Maple notation for the vector 
is < 
ii) The characteristic polynomial of B 
2 
5 
-1 
o 
-1 
5 
0 
1, 
0 
-3 
Hint: try expanding the determinant along the last row or last column. 
Ordered tl < t2 < t3 this has roots (and hence B has eigenvalues) 
-3,

 

 

MATH1131 -
The determinant of a matrix

 

 

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