A First Course in Linear Algebra, 2017a

566 Selected Exercise Answers
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1
1
5.7.2 A basis for ker(T ) is
and a basis for im(T ) is .
−1
1
There are many other possibilities for the specific bases, but in this case dim(ker(T )) = 1 and dim(im(T )) =
1.
5.7.3 In this case ker(T )={0} and im(T )=R 2 (pick any basis of R 2 ).
5.7.4 There are many possible such extensions, one is (how do we know?):
⎧⎡
⎤ ⎡ ⎤ ⎡ ⎤⎫
⎨ 1 −1 0 ⎬
⎣ 1 ⎦, ⎣ 2 ⎦, ⎣ 0 ⎦
⎩
⎭
1 −1 1
5.7.5 We can easily see that dim(im(T )) = 1, and thus dim(ker(T )) = 3 − dim(im(T )) = 3 − 1 = 2.
⎡
5.8.2 C B (⃗x)= ⎣
2
1
−1
⎤
⎦.
[ ]
1 0
5.8.3 M B2 B 1
=
−1 1
⎡
5.9.1 Solution is: ⎣
−3ˆt
−ˆt
ˆt
⎤
⎦, ˆt 3 ∈ R . A basis for the solution space is ⎣
⎡
−3
−1
1
5.9.2 Note that this has the same matrix as the above problem. Solution is: ⎣
⎡
5.9.3 Solution is: ⎣
⎡
5.9.4 Solution is: ⎣
⎡
5.9.5 Solution is: ⎣
⎡
5.9.6 Solution is: ⎣
3ˆt
2ˆt
ˆt
3ˆt
2ˆt
ˆt
⎤
⎡
⎦, Abasisis⎣
⎤
−4ˆt
−2ˆt
ˆt
−4ˆt
−2ˆt
ˆt
⎦ + ⎣
⎤
⎡
−3
−1
0
⎤
3
2
1
⎤
⎦
⎦, ˆt ∈ R
⎦. Abasisis⎣
⎤
⎡
⎦ + ⎣
0
−1
0
⎤
⎡
−4
−2
1
⎦, ˆt ∈ R.
⎤
⎦
⎡
⎤
⎦
⎤
−3ˆt 3
−ˆt 3
⎦ +
ˆt 3
⎡
⎣
0
−1
0
⎤
⎦, ˆt 3 ∈ R