# Describe $\Lambda$ with linear equations in the standard basis of $L(\mathbb{R}^2,\mathbb{R}^3)$

Let $V \subset \mathbb{R}^3$ be the vector space defined by $x_1+x_2+x_3=0$. In the space of linear transformations $L(\mathbb{R}^2,\mathbb{R}^3)$ there's a subspace $$\Lambda = \{ \phi \in L(\mathbb{R}^2,\mathbb{R}^3): \text{im}(\phi) \subseteq V \}$$

We want to describe $\Lambda$ with linear equations in the standard basis of $L(\mathbb{R}^2,\mathbb{R}^3)$.

First of all, $V$ is spanned by $(-1,1,0)$ and $(-1,0,1)$. $\Lambda$ can be described as a subspace of 3 by 2 matrices. Since $\text{im}(\phi) \subseteq V$ these matrices will have linear combinations of $(-1,1,0)$ and $(-1,0,1)$ in their columns. The matrices in the standard basis of $L(\mathbb{R}^2,\mathbb{R}^3)$ would be $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{bmatrix}$$ Now, my suspicion is that elements of $\Lambda$ can be written down as $$(-c_1-c_2) \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix} +(-c_3-c_4) \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix} +c_1 \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ \end{bmatrix} +c_4 \begin{bmatrix} 0 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{bmatrix} +c_2 \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ \end{bmatrix} +c_3 \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{bmatrix}$$

How can I get from this to a system of linear equations?

• What exactly is meant with "describe $\Lambda$ in linear equations"...? Dec 16 '16 at 10:04
• This what I've been wondering from the get go. The problem itself doesn't seem complicated but, just as you, I find it somewhat ambiguous. Dec 16 '16 at 10:09

I'm not sure what exactly is meant with "describe $\Lambda$ in linear equations" (see comments to the question), here's an idea...
A linear transformation $\phi$ in $L$ has a matrix of the form: $$\begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix}$$ An element $\mathbf{w} \in \text{im}(\phi)$ is of the form: $$\mathbf{w}=\begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} ax+by \\ cx+dy \\ ex+fy \end{bmatrix}$$ Now $\phi \in \Lambda$ if $\text{im}(\phi) \subseteq V$ so you need that $\mathbf{w} \in \text{im}(\phi) \implies \mathbf{w} \in V$. But $\mathbf{w} \in V$ means that for all $x$ and $y$; you need: \begin{align} \underbrace{(ax+by)}_{x_1}+\underbrace{(cx+dy)}_{x_2}+\underbrace{(ex+fy)}_{x_3} = 0 & \iff (a+c+e)x+(b+d+f)y = 0 \\ {} & \iff \left\{\begin{array}{l} \color{blue}{a+c+e=0} \\ \color{blue}{b+d+f=0} \end{array}\right. \end{align}