Question
Asked Nov 26, 2019
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I am stuck on number 14. The directions say Find the basis for the eigenspace corresponding to each listed eigenvalue.

L
A = 1,5
-1
12. A =
-3
42
4C
0
I
I
1, 2, 3
e.
13. A
-2
-2
0
I
23. E
e
0 -1
I
I
-3
4 -13
14. A =
0
,A = -2
24.
2 D
I
25.
2
3
15. A =
-1
-3
I
4
2
6
3
0
0
T
16. A=
0
A = 4
T
0
U
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L A = 1,5 -1 12. A = -3 42 4C 0 I I 1, 2, 3 e. 13. A -2 -2 0 I 23. E e 0 -1 I I -3 4 -13 14. A = 0 ,A = -2 24. 2 D I 25. 2 3 15. A = -1 -3 I 4 2 6 3 0 0 T 16. A= 0 A = 4 T 0 U

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Expert Answer

Step 1

By definition, the Eigen space corresponding to the Eigen value -2 is the null space of the matrix A+2I. Thus we have:

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EN(A 21) -11 1 0 A =1 -3 0 413 1 -1 3 A+211 -1 0 4 -13 3 1 1 0 3 1 rref (A 210 1 3 0 0 0

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Step 2

Thus, the solution x of (A+2I) x = 0 satisfy x1 + x2 + 3x3 = 0 or x1 = - x2 - 3x3

 

Thus, the null space N(A+2I) consists of vectors.

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-1 -3 r-3r 1x 0 = x x 0 1 eri

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Step 3

For any scaler x2 and x3.

 

Hence we have,

...
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-3 E N(A+21) Span 1 0

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