0
$\begingroup$

Let V be a 4-d vector space. $T:V \rightarrow V $is a linear operator whose effect on basis {$e_1, e_2,e_3,e_4$} is

$Te_1= 2e_1- e_4$

$Te_2= -2e_1 + e_4$

$Te_3= -2e_1 + e_4$

$Te_4= e_1$

Find a basis for Ker T and Image T. Calculate the rank and nullity of T.

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

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

$\endgroup$
  • $\begingroup$ I apologize on the lack of format, I am not sure how to get it right $\endgroup$ – James Hendren Oct 2 '13 at 19:46
  • $\begingroup$ Read a little about LaTeX, or get into any other question and do "edit" to see how the formating is done. $\endgroup$ – DonAntonio Oct 2 '13 at 19:47
  • $\begingroup$ fixed the original format $\endgroup$ – James Hendren Oct 2 '13 at 19:51
  • $\begingroup$ Nice. Have you already studied about matrix representations of linear transformations? $\endgroup$ – DonAntonio Oct 2 '13 at 19:52
  • $\begingroup$ to be honest, I'm in a theoretical physics class and have no idea what's going on... $\endgroup$ – James Hendren Oct 2 '13 at 19:55
1
$\begingroup$

The matrix corresponding to $\;T\;$ and the given basis is in fact

$$\begin{pmatrix}2&-2&-2&-2\\0&0&0&0\\0&0&0&0\\-1&1&1&1\end{pmatrix}$$

From this we can see the matrix rank = the transformation's image's dimension, is one, and thus its kernel (both of the matrix and the transformation) has dimension three.

If you don't understand this I can't see a way to explain you any further without solving completely the question and you don't understanding a thing...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ my orignial parameters were wrong, and I have input in rref $\endgroup$ – James Hendren Oct 2 '13 at 20:09
1
$\begingroup$

First, notice that if you write the matrix $A$ defined as $Ax = T(x)$

Then if we consider the basis ${\cal B} = \{e_1,e_2,e_3,e_4\}$ we have that

$$A = [T(e_1)\quad T(e_2)\quad T(e_3)\quad T(e_4)]$$

That's because every $x\in V$ can be writed as $x = \alpha_1e_1+\alpha_2e_2+\alpha_3e_3+\alpha_4e_4$, with $\alpha_i\in\mathbb{K}$ (with $\mathbb{K}$ )

Then, as $T$ is linear

\begin{eqnarray*} T(x) &= \alpha_1T(e_1)+\alpha_2T(e_2)+\alpha_3T(e_3)+\alpha_4T(e_4)\\ &= [T(e_1)\quad T(e_2)\quad T(e_3)\quad T(e_4)]\,\begin{pmatrix}\alpha_1\\\alpha_2\\\alpha_3\\\alpha_4\end{pmatrix} \end{eqnarray*}

Then you must determine the Kernel and Image of $A$ and then write those vectors on basis ${\cal B}$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ how do I determine the kernel and Image though? $\endgroup$ – James Hendren Oct 2 '13 at 20:12
  • $\begingroup$ For the image, you must determinate which vectors can be writted as Ax for some x, so the image is equal to the generated space of the l.i. vectors of the columns of A and the kernel are those who solve the equation Ax = 0 $\endgroup$ – MathGuest Oct 2 '13 at 20:22
0
$\begingroup$

Remember that $Rank(T) + Kern(T) = Dim(V)$. So if $Dim(V) = 4$ and $Rank(T) = 1$, finding the Kernel should be pretty easy.

You can think of Rank as the dimension of the subspace of $V$ given by $T$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.