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Linear transformation

Kaspelek

New member
Apr 21, 2013
26
Where T(p(x)) = (x+1)p'(x) - p(x) and p'(x) is derivative of p(x).

a) Find the matrix of T with respect to the standard basis B={1,x,x^2} for P2.

T(1) = (x+1) * 0 - 1 = -1 = -1 + 0x + 0x^2
T(x) = (x+1) * 1 - x = 1 = 1 + 0x + 0x^2
T(x^2) = (x+1) * 2x - x^2 = 2x + x^2 = 0 + 2x + x^2

So, the matrix for T with respect to B equals
[-1 1 0]
[0 0 2]
[0 0 1].


b) Find a basis for kerT and hence write down dim(kerT).

c) Find a basis for ImT and hence write down dim(ImT).

d) Does the transformation have an inverse?


I've done part a, so any guidance on the rest would be greatly appreciated.
 

TheBigBadBen

Active member
May 12, 2013
84
Where T(p(x)) = (x+1)p'(x) - p(x) and p'(x) is derivative of p(x).

a) Find the matrix of T with respect to the standard basis B={1,x,x^2} for P2.

T(1) = (x+1) * 0 - 1 = -1 = -1 + 0x + 0x^2
T(x) = (x+1) * 1 - x = 1 = 1 + 0x + 0x^2
T(x^2) = (x+1) * 2x - x^2 = 2x + x^2 = 0 + 2x + x^2

So, the matrix for T with respect to B equals
[-1 1 0]
[0 0 2]
[0 0 1].


b) Find a basis for kerT and hence write down dim(kerT).
As you've correctly stated, we may express the transformation in the form
\(\displaystyle
A_T=\left[ \begin{array}{ccc}
-1 & 1 & 0 \\
0 & 0 & 2 \\
0 & 0 & 1 \end{array} \right]
\)

Now as for \(\displaystyle ker(T)\), we proceed to find the kernel of this transformation in the same way as we would for any other transformation. That is, we would like to solve \(\displaystyle A_T x=\left[ \begin{array}{ccc}0 & 0 & 0\end{array} \right]^T\), which requires we put the augmented matrix
\(\displaystyle \left( A_T|0 \right) =
\left[ \begin{array}{ccc}
-1 & 1 & 0 & | & 0 \\
0 & 0 & 2 & | & 0 \\
0 & 0 & 1 & | & 0 \end{array} \right]
\)
in its reduced row echelon form (of course, this problem could be solved with a little intuition instead of row reduction, but the upshot of this method is that it works where your intuition might fail). You should end up with
\(\displaystyle \left[ \begin{array}{ccc}
1 & -1 & 0 & | & 0 \\
0 & 0 & 1 & | & 0 \\
0 & 0 & 0 & | & 0 \end{array} \right]
\)
Which tells you that the kernel of the transformation is the set of all polynomials \(\displaystyle a+bx+cx^2\) such that \(\displaystyle a-b=0\) and \(\displaystyle c=0\). Thus, we may state that the kernel of \(\displaystyle T\) is spanned by the vector
\(\displaystyle \left[ \begin{array}{ccc}1 & 1 & 0\end{array} \right]^T\), which is thus the basis of the kernel of T. Since there is one vector in the basis, the dimension of the kernel is 1.

Any questions about the process so far?
 
Last edited:

TheBigBadBen

Active member
May 12, 2013
84
c) Find a basis for ImT and hence write down dim(ImT).

d) Does the transformation have an inverse?
For c), what we need to do is find the largest possible set of linearly independent column-vectors. If you wanted to use the rref (reduced row echelon form) that we computed previously, you simply choose the column vectors corresponding to the 1's (i.e. the "pivots") of the reduced matrix. That is, choosing the first and third columns, we find that \(\displaystyle \left[ \begin{array}{ccc}-1 & 0 & 0\end{array} \right]^T\) and \(\displaystyle \left[ \begin{array}{ccc}0 & 2 & 1\end{array} \right]^T\) form a basis of the image. It follows that \(\displaystyle dim(Im(T))=2\).

For d), we simply note that the kernel of this transformation is not of dimension 0. This is enough to state that the transformation does not have an inverse.
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Something better to express the solutions as vectors of $P_2$ instead of coordinates. That is, $B_{\ker T}=\{1+x\}$ and $B_{\mbox{Im }T}=\{-1,2x+x^2\}$.
 

Kaspelek

New member
Apr 21, 2013
26
That helped a lot guys, thanks a lot.