I'm struggling with answering this question:
"Consider a subspace $V$ in $\mathbb{R^n}$ that is defined by $n$ homogeneous linear equations:
$\begin{vmatrix} a_{11}x_1 + a_{12}x_2 + \cdots +a_{1m}x_m=0 \\ a_{21}x_1 + a_{22}x_2 + \cdots +a_{2m}x_m=0 \\ \vdots\\ a_{n1}x_1 + a_{n2}x_2 + \cdots +a_{nm}x_m=0 \\ \end{vmatrix}$
What is the relationship between the dimension of $V$ and the quantity $m - n$? State your answer as an inequality. Explain carefully."
I know this has to do with rank-nullity, but I'm having a hard time incorporating $dimV$ into the equation. Any suggestions? Thanks!
If you agree to consider each element of $\Bbb R^m$ as an column vector then you can define a map $T:\Bbb R^m\to\Bbb R^n$ by $T(x)=Ax$, where $A=(a_{ij})_{n\times m}$. Show that $T$ is a linear map and $\ker(T)=V$. Using Rank-nullity theorem, $$\text{Rank}(T)+\text{Nullity}(T)=m$$ and conclude that $$\dim (V)=m-\text{Rank}(T)\ge m-n$$ (also observe that $\text{Rank}(T)=\text{Rank(A)}$, which is easy to prove).