MHB Basis for $U$ using $\operatorname{null}A$

[Dethrone]

Prove: Let $A$ be an $n$ by $n$ matrix of rank $r$. If $U={}\left\{X \in M_{nn}|AX=0\right\}$, show that $\dim U=n(n-r)$.

Proof:
Clearly, $\dim \operatorname{null}A=n-r$, and let $X=[c_1,c_2,...,c_n]$ where $c_i \in \Bbb{R}^n$ are column vectors. Then since $AX=0$, using block multiplication, $Ac_i=0, \forall i$. Thus, $c_i \in \operatorname{null}A.$

I am not sure what to do now, but I think it has something to do with $n\cdot (n-r)$. Any help is appreciated!

 

[Jameson]

Hi Rido12,

I'm a little confused by the phrasing of $\dim U=n(n-r)$. Usually we talk about the dimension of the column space (rank) or the dimension of the null space (nullity). If we just talk about the dimension of a matrix, then we we list it as rows by columns, usually $m \times n$. However usually your notation of something like $M_{nn}$ then maybe this is simply asking for the dimension of $U$.

If this is indeed what they are asking, then we know that there are $n$ rows $X$ such that $AX=0$ and there are $n-r$ solutions to this equation, so $U$ should be of size $n(n-r)$.

 

[Opalg]

Prove: Let $A$ be an $n$ by $n$ matrix of rank $r$. If $U={}\left\{X \in M_{nn}|AX=0\right\}$, show that $\dim U=n(n-r)$.

Proof:
Clearly, $\dim \operatorname{null}A=n-r$, and let $X=[c_1,c_2,...,c_n]$ where $c_i \in \Bbb{R}^n$ are column vectors. Then since $AX=0$, using block multiplication, $Ac_i=0, \forall i$. Thus, $c_i \in \operatorname{null}A.$

I am not sure what to do now, but I think it has something to do with $n\cdot (n-r)$. Any help is appreciated!

You are correct to start by saying that if $X\in U$ then each column of $X$ must be in $\operatorname{null}A$. Suppose that $\{v_1,\ldots,v_{n-r}\}$ is a basis for $\operatorname{null}A$. For $1\leqslant i\leqslant n$ and $1\leqslant j\leqslant n-r$, let $X_{ij}$ be the matrix whose $i$th column is $v_j$, and all its other columns are $0$. Show that $\{X_{ij} : 1\leqslant i\leqslant n,\,1\leqslant j\leqslant n-r\}$ is a basis for $U$.