linear-guest

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we have
⎛ ⎞
2
Lv = ⎝−2⎠ = 2v
4
so v ∈ E (2) . In general, we note the linearly independent vectors v (λ)
i with the
same eigenvalue λ span an eigenspace since for any v = ∑ i ci v (λ)
i , we have
Lv = ∑ i
c i Lv (λ)
i
= ∑ i
c i λv (λ)
i
= λ ∑ i
c i v (λ)
i
= λv.
Hint for Review Problem 9
We are( looking ) at the matrix M, and a sequence of vectors starting with
x(0)
v(0) = and defined recursively so that
y(0)
v(1) =
( ) x(1)
= M
y(1)
( ) x(0)
.
y(0)
We first examine the eigenvectors and eigenvalues of
M =
( ) 3 2
.
2 3
We can find the eigenvalues and vectors by solving
det(M − λI) = 0
for λ.
( )
3 − λ 2
det
= 0
2 3 − λ
By computing the determinant and solving for λ we can find the eigenvalues λ =
1 and 5, and the corresponding eigenvectors. You should do the computations
to find these for yourself.
When we think about the question in part (b) which asks to find a vector
v(0) such that v(0) = v(1) = v(2) . . ., we must look for a vector that satisfies
v = Mv. What eigenvalue does this correspond to? If you found a v(0) with
this property would cv(0) for a scalar c also work? Remember that eigenvectors
have to be nonzero, so what if c = 0?
For part (c) if we tried an eigenvector would we have restrictions on what
the eigenvalue should be? Think about what it means to be pointed in the same
direction.
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