It says: Show that v is an eigenvector of A and find the corresponding eigenvalue.
A =
[3 0 0
0 1 -2
1 0 1]
v =
[2
-1
1]
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It says: Show that v is an eigenvector of A and find the corresponding eigenvalue.
A =
[3 0 0
0 1 -2
1 0 1]
v =
[2
-1
1]
v is an eigen vector iffQuote:
Originally Posted by nicolem1051
where
is the eigenvalue for v.
thank you so much, that's the answer I got but I don't understand why it is 3. For some reason I am not getting it(Doh)
to find the eigenvalues all u have to do is find the characteristic equation det(A-λI)=0
for example..the matrix A[ 3 0 0
0 1 -2
1 0 1]
u write it like this A[ 3-λ 0 0
0 1-λ -2
1 0 1-λ]
and then you expand by Column 2
and therefore u get (1-λ)(3-λ)(1-λ)=0
hence the eigenvalues are λ=1, λ=3, λ=1.
You can now find the corresponding eigenvectors u, v and w.
Let u=[a,b,c]. If Au=λu where λ=1 and 3 in this case. u solve the system Au=λu and u find the eigenvectors.