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Is there a Linear Transformation from $T : \Bbb R^5 \rightarrow \Bbb R^4 $ so $$\operatorname{Ker}T = \{( x,y,z,t,w) \in \Bbb R^5 \; | \; x = 2y, \text{ and, } z = 2t = 3w\}$$ if so find an example of this transformation. It's enough to define it on base $\Bbb R^5 $. If not, explain why.

Well I tried to solve it and I'm not sure that what I've done was correct... I said that $x=2y$ , $z=3w$ and $t=\frac{3w}{2}$ so it should look like this

$$\operatorname{Ker}T = \left\{\left( 2y,y,3w,\frac{3w}{2},w \right) \right\}$$ so I got $x$ and $w$ out in order to get 2 vectors... $$\operatorname{Ker}T = \left\{y( 2,1,0,0,0)+w \left( 0,0,3,\frac{3}{2},1 \right) \right\}$$ I changed the second vector so it will be nicer and get $y$ and $w$ out so..

$$\operatorname{Ker}T = \operatorname{Sp}\{( 2,1,0,0,0),(0,0,6,3,2)\}$$

Hopefully so far I'm correct? since they are kernal i can say that:

$T(2,1,0,0,0)=(0,0,0,0)$ and $T(0,0,6,3,2)=(0,0,0,0)$

from here I defined 3 more vectors from (which I made)

$$T(0,1,0,0,0)=(0,1,0,0), T(0,0,0,1,0)=(0,0,1,0) \text{ and } T(0,0,0,0,1)=(0,0,0,1)$$

I wrote those matrix and used row reduction in order to see that it's not linearly dependent and because of that it is a base (english is not my main language so sorry if I didn't write it correctly)

$$\begin{bmatrix} 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 6 & 3 & 2 \\0 & 1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 &0 \\0 & 0 & 0 & 0 &1\end{bmatrix};$$

and we get

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\0 & 0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 &0 \\0 & 0 & 0 & 0 &1\end{bmatrix};$$

so it must be a base therefore since $T(v_i) =w_i$ for each $i=1,2...n$. and it is linear transformation... Is it correct or am I wrong? Thank you very much!

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  • $\begingroup$ Everything looks fine. $\endgroup$ – Ilan Aizelman WS Jun 6 '14 at 8:03
  • $\begingroup$ You've written a $5\times5$ matrix, which corresponds to a transformation from ${\bf R}^5$ to ${\bf R}^5$, not to ${\bf R}^4$. $\endgroup$ – Gerry Myerson Jun 6 '14 at 9:15
  • $\begingroup$ @GerryMyerson well then how can i fix that? i thought that if i proove that B(which contain those vectors) is a base for $R^5$ then each of his vectors has atleast of tranformation to $R^4$ $\endgroup$ – user2323232 Jun 6 '14 at 9:21
  • $\begingroup$ You mean I need to substract one of the KerT's vector and make it 4x5 matrix? And thus, it will be from $R^5$ to $R^4$? $\endgroup$ – Ilan Aizelman WS Jun 6 '14 at 9:21
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    $\begingroup$ Never mind. I thought you intended your matrix to be the matrix representing the transformation, but now I see you were using the matrix to prove you have a basis for ${\bf R}^5$. Since you have a basis, and since the range of the transformation clearly has dimension 3, you're done. $\endgroup$ – Gerry Myerson Jun 6 '14 at 9:28

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