eigenvalues and eigenvectorsWatch

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#1
the question: http://prntscr.com/np3xrq

im not sure where to start with this because i dont much understand eigenvectors or eigenvalues.
from habit i i did this matrix
[(-4 - lambda), 1:
-5, (2- lambda)]
then did the determinant = 0 to find lambda =3 and -1

then i try to find the eigenvectors by replacing lambda first with 3
i get equations
-7x + 1
and
-5 - 1
i set both of these to 0 and this is where im stuck, i find x and y to be 0. though i reckon its because i have no idea what im doing for this question thats the real problem lol
0
1 month ago
#2
The eigenvalues have the wrong sign. Can you check them?
The eigenvector calculation won't work with your values.
(Original post by dfsdf3333)
the question: http://prntscr.com/np3xrq

im not sure where to start with this because i dont much understand eigenvectors or eigenvalues.
from habit i i did this matrix
[(-4 - lambda), 1:
-5, (2- lambda)]
then did the determinant = 0 to find lambda =3 and -1

then i try to find the eigenvectors by replacing lambda first with 3
i get equations
-7x + 1
and
-5 - 1
i set both of these to 0 and this is where im stuck, i find x and y to be 0. though i reckon its because i have no idea what im doing for this question thats the real problem lol
Last edited by mqb2766; 1 month ago
0
#3
i did put the wrong signs u were right.

to get my eigenvalues i did ad-bc = 0,
(-4 - lambda)(2 - lambda) - (1 * -5) = 0,
lambda^2 + (2*lambda) - 3 = 0,
(lambda - 1)(lambda + 3) = 0,
so eigenvalues = -3, +1.
after that im still not sure what to do, i plug in -3 for lambda in the matrix to get
[(-4 - (-3)), 1;
-5, (2 - (-3))]

[-1, 1;
-5, 5]
so the eigenvector for an eigenvalue of -3 is
[-1;
1]?

(Original post by mqb2766)
The eigenvalues have the wrong sign. Can you check them?
The eigenvector calculation won't work with your values.
0
1 month ago
#4
No (eigenvalues are ok). For the eigenvectors you have
(A - lam*I)*v = 0

Row one gives
-v1 + v2 = 0
so v1 = v2
so the eigenvector is
[1, 1]'
subject to a scaling constant.

As a quick check
A*[1,1]' = [-3, -3]'
which is -3 * [1,1]', so correct.

Do the other?

(Original post by dfsdf3333)
i did put the wrong signs u were right.

to get my eigenvalues i did ad-bc = 0,
(-4 - lambda)(2 - lambda) - (1 * -5) = 0,
lambda^2 + (2*lambda) - 3 = 0,
(lambda - 1)(lambda + 3) = 0,
so eigenvalues = -3, +1.
after that im still not sure what to do, i plug in -3 for lambda in the matrix to get
[(-4 - (-3)), 1;
-5, (2 - (-3))]

[-1, 1;
-5, 5]
so the eigenvector for an eigenvalue of -3 is
[-1;
1]?
Last edited by mqb2766; 1 month ago
0
#5
plug in +1 for lambda into the matrix to get
[(-4 - 1), 1;
-5, (2 - 1)]

[-5, 1;
-5, 1]
so

-5v1 + v2 = 0
v2 = 5v1

so eigenvector for eigenvalue of 1 is
[5;
1]?
the reason i ask is because no solution was given on my example

(Original post by mqb2766)
No (eigenvalues are ok). For the eigenvectors you have
(A - lam*I)*v = 0

Row one gives
-v1 + v2 = 0
so v1 = v2
so the eigenvector is
[1, 1]'
subject to a scaling constant.

As a quick check
A*[1,1]' = [-3, -3]'
which is -3 * [1,1]', so correct.

Do the others?
0
1 month ago
#6
To check, calculate
A*v
and see if it equals
lam*v
which is what I did?
(Original post by dfsdf3333)
plug in +1 for lambda into the matrix to get
[(-4 - 1), 1;
-5, (2 - 1)]

[-5, 1;
-5, 1]
so

-5v1 + v2 = 0
v2 = 5v1

so eigenvector for eigenvalue of 1 is
[5;
1]?
the reason i ask is because no solution was given on my example
0
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