Answer to Question #219962 in Linear Algebra for harry

Question

Let L : R

3 −→ R3[x] be a linear transformation such that

L(1, 0, 0) = 2x + x3, L(0, 1, 0) = −2x + x2, and L(0, 0, 1) = x2 + x3

.

(a) Find a formula for L(a, b, c), where a, b, c ∈ R.

(b) Find a basis for ker L. Is L a monomorphism?

(c) Find a basis for Im L. Is L an epimorphism?

(d) Find a basis for L−1[A], where A = x^2

Answer 1

Solution;

Idetify the linear transformation R³ to R³ by the mapping;

a+bx+cx² in which a,b,c$\epsilon$R

By the standard bases given;

Since L:R³$\to$ R³[x] ,we can factor out x to get,

L(1,0,0)=(2x+x³)=(2+x²)x

L(0,1,0)=(-2x+x²)=(-2+x)x

L(0,0,1)=(x²+x³)=(x+x²)x

The standard matrix for L is,

A=$\begin{bmatrix} 2&0& 1 \\ -2 &1& 0\\ 0&1&1 \end{bmatrix}$

The formula for the transformation is;

L(ax²,bx,c)=$x\begin{bmatrix} 2&0 & 1\\ -2 & 1&0\\ 0&1&1 \end{bmatrix}$ $\begin{bmatrix} 1\\ x\\ x^2 \end{bmatrix}$

b)

Find Kernel L

L$\begin{bmatrix} a \\ bx\\ cx^2 \end{bmatrix}$ =$0_{R^3}$

Transform matrix A into a reduced row achelon;

$\begin{bmatrix} 1 &0& \frac12\\ 0 &1&1\\ 0&0&0 \end{bmatrix}$

The corresponding system is;

1+0.5x²=0

x+x²=0

0=0

Basis of Kernel(L) is;

($-\frac12$ ,-1,1)

L is not a monomorphism because

$-\frac12\neq -1\neq 1$

c)

Find image of L;

By definition;

L(x)=Ax=b

b is the image of the transformation.

The basis of image L is

$[(2,-2,0),(0,1,1),(1,0,1)]$

d)

Find the basis of

L^-1[A]

A=x²

Matrix of L=$\begin{bmatrix} 2&0&1 \\ -2 & 1&0\\ 0&1&1 \end{bmatrix}$

det =0

The matrix is not invertible.

The basis (0,0,0)