Question #219962

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


1
Expert's answer
2021-07-27T08:40:04-0400

Solution;

Idetify the linear transformation R3 to R3 by the mapping;

a+bx+cx2 in which a,b,cϵ\epsilonR

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=[201210011]\begin{bmatrix} 2&0& 1 \\ -2 &1& 0\\ 0&1&1 \end{bmatrix}

The formula for the transformation is;

L(ax2,bx,c)=x[201210011]x\begin{bmatrix} 2&0 & 1\\ -2 & 1&0\\ 0&1&1 \end{bmatrix} [1xx2]\begin{bmatrix} 1\\ x\\ x^2 \end{bmatrix}

b)

Find Kernel L

L[abxcx2]\begin{bmatrix} a \\ bx\\ cx^2 \end{bmatrix} =0R30_{R^3}

Transform matrix A into a reduced row achelon;

[1012011000]\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;

(12-\frac12 ,-1,1)

L is not a monomorphism because

1211-\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)][(2,-2,0),(0,1,1),(1,0,1)]

d)

Find the basis of

L-1[A]

A=x2

Matrix of L=[201210011]\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)






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