# Let $V$ be a subspace of $\mathbb{R}^4$, spanned by $v$ and $u$. Find a linear transformation whose kernel is $V$.

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.

• Your interpretation is correct, and more specifically, every vector whose image is the zero vector must be an element of $V$. – Anthony Ter Dec 16 '18 at 2:17
• 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. – NL1992 Dec 16 '18 at 2:18
• There is an infinite number of solutions. Hint: the row space of a matrix is the orthogonal complement of its kernel. – amd Dec 16 '18 at 6:02

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 $$v$$

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 $$Ker(A)=span\{u,v\}$$

• I think you could compare dimensions... – David Apr 15 at 1:21