Find a T from $R^3$ to $R^4$ given an equation for a subspace in $R^4$ A subspace V in $R^4$ is defined by the equation $x_{1}$-$x_{2}$+$2x_{3}$+$4x_{4}$=0. I need to find T such that Ker(T)=zero vector, and Im(T)=V. How do I approach this problem? As I understand, the equation given to me is a set of points that are solutions to Im(T)=V, so in a sense they vectors of that plane are elements of Ker(V), but I can't go any further. Thanks.
 A: Hint: Find three linearly independent vectors in $\mathbb{R}^4$ that span $V$.
A: I am going to treat "matrix" and "linear transformation" as synonyms in this answer, because we are working in $\mathbb{R}^n$ and have the standard basis at our disposal.  
Rephrasing your problem: you are given a subspace $V$ of $\mathbb{R}^4$, defined by a homogeneous linear equation.  The problem is to find a matrix $M$ whose nullspace consists only of the zero vector and whose column space is $V$.
Here are some useful tools to have in your toolbox:
(1) Given a subspace $W$ of $\mathbb{R}^n$ defined by a system of $m$ homogeneous linear equations, write down a matrix $A$ whose nullspace is precisely $W$.
[Solution, making use only of the definitions of the words and matrix multiplication: rewrite the system of homogeneous linear equations $\sum_{j=1}^n a_{ij} x_j = 0$, $1 \leq i \leq m$, as a single matrix equation $A x = 0$, where $A$ is the matrix whose row $i$, column $j$ entry is $a_{ij}$, and where $x$ is the column vector whose $i$th entry is $x_i$.  The matrix $A$ will do.]
(2) Given an $m \times n$ matrix $A$, find a basis for the nullspace of $A$.
[Solution: somewhat hard to express in words if you haven't seen it--- look in a textbook.  The standard way this is done is to solve $Ax = 0$ using Gaussian elimination, to introduce a "free variable" for every non-pivot variable column, and thus to write the general solution to $Ax = 0$ in terms of "free variables".  If you rewrite this general solution in vector form--- as a linear combination of fixed numerical vectors, with the coefficients the free variables--- a basis falls right out.]
(3) Given a basis $v_1, \dots, v_k$ for a subspace $W$ of $\mathbb{R}^m$, find a matrix $T$ whose nullspace is $\{0\}$ and whose range is $W$.
[Solution, based only on the definitions of the words and matrix multiplication: write the $v$s as column vectors and make them columns of a matrix.  In general, if $M$ is any matrix, the nullspace of $M$ is $\{0\}$ precisely when its columns are linearly independent, and the range of $M$ is precisely the span of its columns--- so as the $v_1, \dots, v_k$ are a basis for $W$, the matrix $T$ formed with these vectors as columns has the two properties you want.]
In this example, doing (1) I see that your $V$ is the nullspace of the $1 \times 4$ matrix $A = (1,-1,2,4)$.
Doing (2) I want to find the general solution to $Ax = 0$, with $A = (1,-1,2,4)$ and $x$ the $4 \times 1$ column vector of variables.  The augmented matrix for this system is $(1,-1,2,4,0)$ and it is already in row echelon form: the variables $x_2 = s$, $x_3 = t$ and $x_4 = u$ are "free", and when they have been assigned values $x_1$ must then satisfy $x_1 - s + 2t + 4u = 0$, so that $x_1 = s - 2t - 4u$.  The general solution to $Ax = 0$ is thus the set of all vectors of the form $(s - 2t - 4u, s, t ,u) = s (1,1,0,0) + t (-2, 0, 1, 0) + u(-4, 0, 0, 1)$; in other words, it is the span of the vectors $v_1 = (1,1,0,0)$, $v_2 = (-2,0,1,0)$, and $v_3 = (-4,0,0,1)$.  By the general magic of this approach, these vectors are linearly independent, so they are a basis for the nullspace of $A$ (and hence a basis of $V$).
(3) I form the matrix $T = \begin{pmatrix} 1 & -2 & -4 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ with $v_1$, $v_2$, and $v_3$ as columns, and I see that it has all of the desired properties.  Going back into the language of linear transformations, the linear transformation from $\mathbb{R}^3$ to $ \mathbb{R}^4$ whose matrix with respect to the standard basis is this matrix $T$, will do the job.
It is worth pointing out that in step (2) of this procedure (finding a basis for the nullspace of $A$) we are making an arbitrary choice in how we do that (any vector space has many bases; somebody with a different method for calculating the basis of a nullspace of a matrix may well get a different set of vectors).  So there are some degrees of freedom here and this problem does not have a unique solution.  If you think about it a while you may be able to describe the set of all possible solutions to this problem.  
