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And the vectors given are $v = (1,0,3,-2)$ and $u = (0,1,4,1)$.

It asks me to find the linear transformation from $\mathbb{R}^4$ to $\mathbb{R}^2$, where the kernel of that transformation is $V$.

So what I know is that: the transformation I'm trying to find, applied to every vector in the span of $(1,0,3,-2)$ and $(0,1,4,1)$, will give the zero vector.

Please let me know if that interpretation is incorrect.

I've really no idea how to get started on this question. I have the equation $Av = 0$ where $A$ is the matrix of the transformation in question, and v is any vector of the subspace V, but...I don't think that gets me anywhere. Any help is greatly appreciated.

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  • $\begingroup$ Your interpretation is correct, and more specifically, every vector whose image is the zero vector must be an element of $V$. $\endgroup$ – Anthony Ter Dec 16 '18 at 2:17
  • $\begingroup$ You can get some relation on the coefficients in a $4\times 2$ matrix from the fact that it is multiplied by 2 independent vectors to be 0. $\endgroup$ – NL1992 Dec 16 '18 at 2:18
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    $\begingroup$ There is an infinite number of solutions. Hint: the row space of a matrix is the orthogonal complement of its kernel. $\endgroup$ – amd Dec 16 '18 at 6:02
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You have to find a $2$ by $4$ matrix whose rows are linearly independent and orthogonal to the given vectors $u$ and $v$.

One such matrix is $$ A = \left[\begin{matrix} 2 \ &-1 \ &0 \ &1 \\ -3 \ & -4 \ &1 \ &0 \end{matrix} \right] $$

The desired linear transformation is defined by $ T(v)=Av$ for $v\in R^4$

Note that according to the rank theorem $$ n= nullity + rank$$ which in this case we have $$4=2+2$$ therefore the kernel of your transformation is a two dimensional subspace generated by the given vectors $u$ and $v$

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We are searching for a matrix $A$ such that $$ A = \left[\begin{matrix} a \ &b \ &c \ &d \\ x \ & y \ &z \ &t \end{matrix} \right] \left[\begin{matrix} 1\\ 0\\ 3 \\ -2 \end{matrix} \right] =\left[\begin{matrix} 0 \\ 0 \end{matrix} \right] $$

And we do the same with the other vector $u$.

So we get the equation

$$\left\{ \begin{aligned} a + 3c -2d &= 0\\ b +4c +d &= 0 \end{aligned} \right.$$

We choose arbitrary $a=1$ and $b=5$ and we get $c=-1$ and $d=-1$.

And from the other 2 equations,

$$\left\{ \begin{aligned} x + 3y -2t &= 0\\ y +4z +t &= 0 \end{aligned} \right.$$

We also choose arbitrary $x=-7$ and $y=-2$ we get that $z=1$ and $t=-2$

So, $$ A = \left[\begin{matrix} 1 \ &5 \ &-1 \ &-1 \\ -7 \ & -2 \ &1 \ &-2 \end{matrix} \right]$$

and $Av=0$ and $Au=0$

And by the rank theorem we have that $$ n= nullity + rank$$ So, in this case we get that $nullity = 2$ and so $$Ker(A)=span\{u,v\}$$

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    $\begingroup$ I think you could compare dimensions... $\endgroup$ – David Apr 15 at 1:21

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