# Plane Equation in R^3 from Column Space

Dear fellow mathematics enthusiasts,

I require assistance on an easy excercise. I will try to be as brief and concise as possible.

Problem description

Let matrix $\mathbf{A}$ = $\begin{bmatrix} 1 & 1 & 1 & 1 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ \end{bmatrix}$. Its column space is C(A) = span$\left(\begin{bmatrix}1 \\ 2\\ 3\\ \end{bmatrix}, \begin{bmatrix}1 \\ 1\\ 4\\ \end{bmatrix}\right)$. These vectors span a plane in $\mathbb{R}^3$. I wish to find the equation of the plane.

Solution

I have already shown in three different ways that the correct equation is $5x-y-z=0$. While two of the solutions make sense, I cannot understand why the following method works. Here goes.

From the definition of a column space C($\mathbf{A}$) = $\{\mathbf{A}\vec{x}\mid \vec{x}\in\mathbb{R}^4\}$ = $\{\vec{b}\mid \mathbf{A}\vec{x}=\vec{b},\vec{x}\in\mathbb{R}^4\}$.

Let $\vec{b}$ = $\begin{bmatrix}x_1 \\ y_1 \\z_1\\\end{bmatrix}$, then $\left[ \begin{array}{cccc|c} 1&1&1&1&x_1\\ 2&1&4&3&y_1\\3&4&1&2&z_1 \end{array} \right]$ $\implies\left[ \begin{array}{cccc|c} 1&1&1&1&x_1\\ 0&-1&2&1&y_1-2x_1\\0&1&-2&-1&z_1-3x_1 \end{array} \right]\implies$

$\implies$$\left[ \begin{array}{cccc|c} 1&0&3&2&y_1-x_1\\ 0&1&-2&-1&2x_1-y_1\\0&0&0&0&y_1+z_1-5x_1 \end{array} \right]$. The bottom row yields $0 = y_1+z_1-5x_1$.

Question

Why can one get the plane equation from the bottom row alone? Essentially, my question is why can one ignore the top two rows in the augmented matrix. Is is it not the case that they also add extra contraints on the solution set?

Extra

This is my third day studying linear algebra. Also, first time on the site. Simple, yet mathematically rigorous answers are greatly appreciated.

Thank you in advance and happy holidays, everybody!

Yours faithfully,

Linear Christmas

## 2 Answers

One of the basic ways to describe a subspace of a vector space $V$ is by specifying a set of vectors $\mathbf\alpha_i$ in the dual space $V^*$ that annihilate the subspace, i.e., by giving a homogeneous system of linear equations $\mathbf\alpha_i[\mathbf v]=0$, the solutions of which are the subspace.

In this case, you're looking for the annihilator of $C(\mathbf A)$, which will be the kernel of $\mathbf A^T$. You can find this by row-reducing $\mathbf A^T$, but the method you used is equivalent. After row-reducing the augmented matrix, the expressions in the extra column are all of the form $\mathbf\alpha[\mathbf v]$. Since we're only interested in the ones that result in $0$, we ignore the non-zero rows. This is similar to how the basis for the column space of the matrix is found by considering only the columns that have pivots and a basis for the nullspace can be read from the columns without pivots.

Let $\vec{x}\in\mathbb{R}^4, \vec{x}=(a_1,a_2,a_3,a_4)$ (I would have let $b=(b_1,b_2,b_3)$)
The first row of the row-reduced matrix means: $1a_1+0a_2+3a_3+2a_4=y_1-x_1$. This has the solution $a_1=y_1-x_1-3a_3-2a_4$. You always get that simple solution when there is a pivot, because $a_1$ is not involved in any other row.