Unformatted text preview: Find the coordinate vector of p (i) Consider p ,p , 2 . relative to B. 3 for standard basis C = 1, , p 101
10
1
10
1
101
Then 0 2 2 ~ 0 2
2~0 1
1 ~ 0 1 1 and the columns form a basis for
330
03
3
01
1
002
the IMT.
is isomorphic to
, so the corresponding polynomials form a basis for
.
(ii) Working in coordinates relative to the standard basis again, b b b
1
101
022
330 2
10
3~0 2
1
03 1
2
3 0 1 2
0
3~
5
0 1 1 1 1 100
~0 1 0
001
So p 23/12
19/12
1/12 2
101
3
011
2~
5
002
3
23/12
19/12
1/12 p by p, so
2
101
3
011
2~
1
001
6 2
3
2~
1
12 Part IV. True/false questions [2pts each]: Write “True” or “False” in the blank. No justification is
needed. TRUE 1. If and are similar matrices, then det = det . , then there is a 3 x 3 matrix A such that H = Col A. TRUE 2. If H is a subspace of FALSE 3. If A is m x n and rank A = m, then the linear transformation TRUE 4. If 0, then det is an n x n matrix and x x is onetoone. 0. FALSE 5. An n x n matrix with n linearly independent eigenvectors is invertible. FALSE 6. A plane in is a twodimensional subspace of TRUE 7. For a 5 x 7 matrix A, dim Nul A . 2. FALSE 8. Row operations on a matrix can change the null space of the matrix. TRUE 9. The matrix 5
0
0
0
0 3
2
0
0
0 4
1
6
0
0 7
0
0
3
0 8
2
1 is diagonalizable.
4
1 FALSE 10. If A is a 3 x 3 matrix, then det 3A = 3 det A....
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This note was uploaded on 01/17/2014 for the course ESE 309 taught by Professor Staff during the Fall '08 term at Washington University in St. Louis.
 Fall '08
 Staff

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