Question

Computational Example Let T:R 5

→R 4
be the linear transformation with matrix respect to the standard bases: A= ⎣
⎡
​
2
1
3
1
​
3
3
6
0
​
1
1
0
2
​
4
2
6
2
​
7
1
8
6
​
⎦
⎤
​
The image of T is clearly spanned by T(e 1
​
),T(e 2
​
),T(e 3
​
),T(e 4
​
),T(e 5
​
), which correspond to the columns of A. ( Note that, for a general linear transformation, dim(T(U)) is the dimension of the column space. ) Since A (1)
and A (2)
are not multiples of each other, they are linearly independent. On the other hand, A (3)
=2A (1)
−A (2)
, A (4)
=2A (1)
, and 6A (1)
− 3
5
​
A (2)
=A (5)
. Hence dim(T(U))=2. The ker(T) is the vector space of all solutions of the homogeneous system of linear equations Ax=0. Via Gaussian elimination, one finds that a solution has the form: x 1
​
x 2
​
x 3
​
x 4
​
x 5
​
​
=−2r−2s−6t
=r+ 3
5
​
t
=r
=s
=t
​
That is, ker(T) is spanned by ⎝
⎛
​
−2
1
1
0
0
​
⎠
⎞
​
, ⎝
⎛
​
−2
0
0
1
0
​
⎠
⎞
​
, ⎝
⎛
​
−6
3
5
​
0
0
1
​
⎠
⎞
​
39 These three vectors are easily seen to be linearly independent, and hence dim(ker(T))=3. The dimension of the domain space is 5 , and 3+2=5, consistent with the Rank plus Nullity Theorem. Exercise 37. Let U=F[x], the F vector space of polynomials in the variable x having coefficients in F. Let T∈L(U,U) be defined by T(f)=xf for all f∈F[x]. What is ker (T) ? What is T(U) ? Is T injective? Is T surjective?

Answer 1

Let's consider T: R5 → R4 be the linear transformation with matrix A. And let's follow the steps to answer all the questions.Exercise 37: Let U = F[x], the F vector space of polynomials in the variable x having coefficients in F. Let T ∈ L(U, U) be defined by T(f) = xf for all f ∈ F[x].What is ker(T)?The kernel of T (ker(T)) is the set of all polynomials f ∈ F[x] such that xf = 0. It means that f must be a polynomial that has x as a factor, that is, f = xg for some polynomial g ∈ F[x]. So, ker(T) = g ∈ F[x].What is T(U)?For a polynomial f ∈ F[x], T(f) is given by T(f) = xf. Therefore, T(U) is the set of all polynomials that are multiples of x, that is, T(U) = f ∈ F[x].Is T injective?T is not injective because T(x) = x² = T(x²) while x ≠ x².Is T surjective?T is not surjective because x is not in the range of T. Therefore, the range of T is not equal to the codomain of T.