eigenvalues and eigenvectors Watch

dfsdf3333
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#1
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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
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mqb2766
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#2
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#2
The eigenvalues have the wrong sign. Can you check them?
The eigenvector calculation won't work with your values.
Please upload your working as it helps to give a bit of advice.
(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; 5 days ago
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dfsdf3333
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#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))]

which leads to
[-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.
Please upload your working as it helps to give a bit of advice.
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mqb2766
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#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.

Put the working down and try not to jump to conclusions.

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))]

which leads to
[-1, 1;
-5, 5]
so the eigenvector for an eigenvalue of -3 is
[-1;
1]?
Last edited by mqb2766; 5 days ago
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dfsdf3333
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#5
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#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.

Put the working down and try not to jump to conclusions.

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

Do the others?
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mqb2766
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#6
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#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
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