L. A. G. 2015. fifth intermediate test, june 17, 2015

(1)

L. A. G. 2015. fifth intermediate test, june 17, 2015

1. Is this matrix invertible





1 1 2

2 1 0

1 0 −1



 ? If the answer is in the affirmative, find the inverse matrix .

Solution. Yes. Inverse





1 −1 2

−2 3 −4

1 −1 1





2. Compute dimension and bases of null-space and range of the following linear transformation:

T : V

⁴

→ V

³

,





 x y z t





 7→





x − y + z + t x + y + 2z + t x − 5y − z + t





(b) Compute a basis of the null-space of the linear transformation S : V

⁴

→ V

³

such that S(e

1

) =



 1 1 1



, S(e

2

) =





−1 1

−5



, S(e

3

) =



 1 2

−1



, S(e

4

) =



 1 1 1



 (where {e

1

, e

2

, e

3

, e

4

} is the standard basis of V

4

).

Solution. (a) N (T ) = L((3, 1, −2, 0), (−1, 0, 0, 1)) dim N (T ) = 2.

T (V

4

) = L((1, 1, 1), (−1, 1, −5)), rk(T ) = 2.

(b) This S is exactly the previous T .

3. Let V be the place of real polynomials of degree ≤ 2. For each of the following functions T : V → V say whether they are linear transformations or not. If not, explain why. If yes, write down the representing matrix, with respect to the basis C = {1, x, x

²

} both in the source and in the target.

(a) T (P (x)) = P (x − 1); (b) T (P (x)) = P (x) + 3x.

Solution. (b) Not linear. Indeed T (P

1

(x) + P

2

(x)) = P

1

(x) + P

2

(x) + x 6= P

1

(x) + x + P

2

(x) + x = T (P

1

(x)) + T (P

2

(x))

(a) This is linear: T (P

1

(x) + P

2

(x)) = P

1

(x − 1) + P

2

(x − 1) = T (P

1

(x)) + T (P

2

(x)).

T (λP (x)) = λP (x − 1) = λT (P (x)).

We have that T (1) = 1 = 1 · 1 + 0 · x + 0 · x

²

, T (x) = x − 1 = −1 · 1 + 1 · x + 0 · x

²

. T (x

²

) = (x − 1)

²

= 1 · 1 − 2 · x + 1 · x

²

. Therefore m

_C,C

(T ) =





1 −1 1

0 1 −2

0 0 1



.