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

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  • $\begingroup$ I apologize on the lack of format, I am not sure how to get it right $\endgroup$ Oct 2, 2013 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, 2013 at 19:47
  • $\begingroup$ fixed the original format $\endgroup$ Oct 2, 2013 at 19:51
  • $\begingroup$ Nice. Have you already studied about matrix representations of linear transformations? $\endgroup$
    – DonAntonio
    Oct 2, 2013 at 19:52
  • $\begingroup$ to be honest, I'm in a theoretical physics class and have no idea what's going on... $\endgroup$ Oct 2, 2013 at 19:55

3 Answers 3

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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...

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  • $\begingroup$ my orignial parameters were wrong, and I have input in rref $\endgroup$ Oct 2, 2013 at 20:09
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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 ${\cal B}$

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  • $\begingroup$ how do I determine the kernel and Image though? $\endgroup$ Oct 2, 2013 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, 2013 at 20:22
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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 $T$.

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