Let $v$ and $w$ be non-zero column vectors in $\Bbb R^n$ and let $A = vw^t$ so that $A$ is an $n\times n$ matrix. Use the rank theorem to show dim Nul $A = n − 1$
Here Nul(A) represents the Null space of A and w^t represent the transpose of the column vector w
I know to prove that the dimNul$(A) = n-1$, I can show that the column space of A has dimension 1, by Rank Theorem. I got up to showing that $Ax=(vw^t)x$ where $x$ belongs to the Column space of A and now I am unsure of where to proceed from here. Have I gone in the right direction? Feel free to help edit my question for clarity, Thanks!
We have a theorem of rank as: $$ \operatorname{Rank}(AB)\leqslant\min{(\operatorname{Rank}(A),\operatorname{Rank}(B))} $$ Hence $$ \operatorname{Rank}(A)\leqslant \min{(\operatorname{Rank}(v),\operatorname{Rank}(w))}=1 $$ for $\operatorname{Rank}(v)=\operatorname{Rank}(w)=1$. And $$ \dim{\operatorname{Nul}(A)}=n-\operatorname{Rank}(A)=n-1 $$ If $\operatorname{Rank}(A)=1$, then there is only $1$ independent column in $A$. Other columns are linear combination of it. Suppose column one is $x$. So $$ A=(x, a_1x,\cdots,a_{n-1}x)=x(1,a_1,,\cdots,a_{n-1})=xv^T=vx^T $$ where $a_i$ are numbers.