findind the image and kernel of a transformation..

**transgalactic** · June 24th 2011, 05:58 AM

$T:R_{3}[x]->R_{3}[x]$ is lenear transformation given by $A=(\begin{array}{ccc}<br /> 4 & 2 & 0\\<br /> 2 & 3 & 2\\<br /> 0 & 2 & 2\end{array})$ with respect to $B=(1+x,1+x^{2},x+x^{2})$ basis
find ker T and ImT ?
How i tried to solve it:
$T(v_{1})=4(1+x)+2(1+x^{2})+0$
$T(v_{2})=5x^{2}+4x+5$
$T(v_{3})=2+2x+4x^{2}$
so i got the image of three vectors.
how to find the Im T and KerT

**HallsofIvy** · June 24th 2011, 10:58 AM

You are given the matrix form of T and can determin image and kernel from that.

A vector, $\begin{bmatrix} u & v & w\end{bmatrix}$ , is in the image of T if and only if there exist a vector, $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ such that
$\begin{bmatrix}4 & 2 & 0 \\ 2 & 2 & 2 \\ 0 & 2 & 2 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}u \\ v \\ w\end{bmatrix}$

Of course, you can solve that equation by row-reducing the augmented matrix
$\begin{bmatrix}4 & 2 & 0 & u \\ 2 & 2 & 2 & v \\ 0 & 2 & 2 & w\end{bmatrix}$

The vector $\begin{bmatrix} u \\ v \\ w\end{bmatrix}$ is in the image of T if and only if row reduction does not result in an "impossible row"- a row where the first three numbers are 0 and the fourth is not.

A vector $\begin{bmatrix} x \\ y \\ z\end{bmatrix}$ is in the kernel of T if and only if
$\begin{bmatrix}4 & 2 & 0 \\ 2 & 2 & 2 \\ 0 & 2 & 2 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

which you can determine by row-reducing the augmented matrix
$\begin{bmatrix}4 & 2 & 0 & 0 \\ 2 & 2 & 2 & 0 \\ 0 & 2 & 2 & 0\end{bmatrix}$

For this particular problem the easy way to do it is to look at the determinant! If the determinant is non-zero, then the matrix is invertible and so maps $R^4$ one to one onto $R^4$ . Its image is all of $R^4$ and its kernel is {0}.

**topspin1617** · June 25th 2011, 03:02 AM

Originally Posted by HallsofIvy

For this particular problem the easy way to do it is to look at the determinant! If the determinant is non-zero, then the matrix is invertible and so maps $R^4$ one to one onto $R^4$ . Its image is all of $R^4$ and its kernel is {0}.

What do you mean by this? The original matrix certainly has zero determinant... unless you meant something else by this comment.

**HallsofIvy** · June 25th 2011, 05:48 AM

You are right! I had accidently changed a number.

**Deveno** · June 25th 2011, 08:03 AM

transgalactic, i'm not sure what you mean by "find im(T)" (not that i don't know what it is, but rather what counts as "finding it").

trivially, one can say $im(T) = \{T(a+bx+cx^2):a,b,c \in \mathbb{R}\}$ .

but even more to the point, there may be several equivalent ways of describing im(T) more explicitly.

what is usually more helpful, is finding dim(im(T)) = rank(A), and exhibiting a basis for im(T).

similarly, one finds dim(ker(T)) = nullity(A), and exhibits a basis for ker(T).

bases are in general not unique, but the dimensions of im(T) and ker(T) are.

now, if we row-reduce the matrix A, we get:

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

this tells us the first two columns of A form a basis for im(T), so im(T) is their span.

more explicitly, $im(T) = \{(6a+5b) + (4a+4b)x^2 + (2a+5b)x^2: a,b \in \mathbb{R}\}$ .

how did i get this? it is just $a(4,2,0)_B + b(2,3,2)_B = (4a+2b,2a+3b,2b)_B$

$= (4a+2b)(1+x) + (2a+3b)(1+x^2) + (2b)(x+x^2)$ .

the rank-nullity theorem tells us that 3 = dim(im(T)) + dim(ker(T)) = 2 + dim(ker(T)), so the kernel of T has dimension 1.

so all we need to do is find one non-zero vector in ker(T), to have a basis.

applying the rref(A) matrix to $(a,b,c)_B$ and setting the result = 0

gives us a - c/2 = 0, and b + c = 0. that is: a = c/2, b= -c, so ker(T) is all elements of the form

$(c/2,-c,c)_B$ . we can, if we wish, for convenience choose c = 2, so that a basis for ker(T) is $\{(1,-2,2)_B\}$ ,

and since $(1,-2,2)_B = (1+x) - 2(1+x^2) + 2(x+x^2) = 3x-1$ ,

{3x-1} is a basis for ker(T).

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