Solutions To Linear Algebra Practice Problems Page 12
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1
2
3
4
5
6
7
8
9
10
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182
2
Thus, an eigenvector corresponding to λ = 1 is
. Any multiple
1
of this vector is also an eigenvector.
Now, we find the eigenvalue corresponding to λ = 2:
2
4
2
1
2
0
A
2I =
2
4
2
0
0
1
0
0
1
0
0
0
2
Thus, an eigenvector corresponding to λ = 2 is
1
. Any multiple
0
of this vector is also an eigenvector.
1
2
Thus, the eigenvalues and eigenvectors are λ = 0,
1
, λ = 1,
2
,
0
1
2
and λ = 2,
1
0
11. Consider the following matrix:
6
4
A =
6
4
Find a general formula for the entries of A . (
Diagonalize A.)
Answer: The trace of A is 2 and the determinant is 0. This means that
the sum of the eigenvalues is 2, and the product is 0, so the eigenvalues
must be λ = 0, 2.
We find the eigenvector corresponding to λ = 0:
6
4
3 2
A
0I =
6
4
0 0
2
Thus, an eigenvector corresponding to λ = 0 is
. Any multiple
3
of this vector is also an eigenvector.
12
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