Find the dimension of the vector space of solutions of $3 \times 4$ matrices $N$ where $N^{T}M=0$

Question

Let $M$ be the $3 \times 4$ matrix displayed below $$ \begin{bmatrix} 1 &3&2&4\\ 2&4&3&5\\ 3&5&4&6\\ \end{bmatrix} $$ I want to find the dimension of the vector space of solutions of $3 \times 4$ matrices $N$ where $N^{T}M=0$ but I am having some trouble.

Thoughts on the matter:

I am in general, wondering if there is some sort of conceptional insight towards doing this problem, but here is what I've thought.

To say that we have a matrix $N$ where $N^{T}M=0$ is the same as saying that $M^{T}N=0$. Displayed below as $$ \begin{bmatrix} 1 &2&3\\ 3&4&5\\ 2&3&4 \\ 4&5&6 \\ \end{bmatrix} \cdot \begin{bmatrix} n_{11} &n_{12}&n_{13}&n_{14}\\ n_{21}&n_{22}&n_{23}&n_{24}\\ n_{31}&n_{32}&n_{33}&n_{34}\\ \end{bmatrix}=0$$ I can think of this as four seperate systems of equations of the form $$ \begin{bmatrix} 1 &2&3\\ 3&4&5\\ 2&3&4 \\ 4&5&6 \\ \end{bmatrix} \cdot \begin{bmatrix} n_{1j} \\ n_{2j}\\ n_{3j}\\ \end{bmatrix}=\begin{bmatrix} 0 \\ 0\\ 0\\ \end{bmatrix}$$ for $j=1,2,3,4$. I found the rank of the matrix $M^{T}$ to be $2$ using column reduction which should imply that the null space of linear map $\mathbb{C}^{3} \rightarrow \mathbb{C}^{4}$ to have dimesion $3-2=1$. I'm not sure where to go from here.

Answer 1

row reducing $ M^\top =\pmatrix{ 1 &2&3\\ 3&4&5\\ 2&3&4 \\ 4&5&6 },\, $ i get $\pmatrix{1&0&-1\\0&1&2\\0&0&0\\0&0&0}.$

therefore the null space of $M^\top$ is spanned $(1, -2, 1)^\top.$ we have $N^\top M = 0 \iff M^\top N = 0.$ that is every column of $N$ is a multiple of $(1, -2, 1)^\top.$

Answer 2

I would argue that each column of your solution matrix is independent of each other, so you may just find the direction of the Null space for M and just copy that general solution in each of the columns. You could combine then independently of each other, so I'm guessing you'd get 1*4 degrees of freedom for your space of solution matrices

Answer 3

Continuing on from your working: $M^T$ has rank $2$ and nullity $1$. So for each $j$, the quantities $n_{1j},n_{2j},n_{3j}$ will be given in terms of one parameter. So the solutions for the whole of $N$ will involve $4$ parameters, and the solution space has dimension $4$. Specifically, a basis for the space will consist of $$\pmatrix{1&0&0&0\cr-2&0&0&0\cr1&0&0&0\cr}\,,\ \pmatrix{0&1&0&0\cr0&-2&0&0\cr0&1&0&0\cr}\,,\ \pmatrix{0&0&1&0\cr0&0&-2&0\cr0&0&1&0\cr}\,,\ \pmatrix{0&0&0&1\cr0&0&0&-2\cr0&0&0&1\cr}\,.$$