Suppose square matrix B has nxn dimensions, has $Null(B) = Col(B)^\perp$ and rank(B)=1, how can we prove the matrix B is equal to its transpose?
I'm not sure how to get started I think I have to start with the eigen vectors and then somehow prove its symmetry but I'm not sure
On
Since $rank(B)=1$, all the columns of $B$ are scalar multiples of the same $n\times 1$ vector $b$, that is, $$B=bx^\intercal$$ for some $n\times 1$ vector $x$. By the properties of the four fundamental subspaces (https://www.statlect.com/matrix-algebra/four-fundamental-subspaces) and the stated assumption, we have $$Col\left(B^\intercal\right)=Null\left(B\right)^\perp=\left(Col\left(B\right)^\perp\right)^\perp=Col\left(B\right)$$ The transpose of $B$ is a matrix whose columns are all scalar multiples of $x$: $$B^\intercal=xb^\intercal$$ But since the column spaces of $B$ and $B^\intercal$ coincide (as just demonstrated), it must be that $x=kb$ for some scalar $k$. Therefore, $$B=kbb^\intercal$$which is clearly symmetric.
Hint: If the rank of $B$ is $1$, then there are non-zero column-vectors $u,v$ for which $$ B = uv^T. $$ What is the relationship between the vectors $u,v$ and the column/null spaces of $B$? Note that $B$ will be a symmetric matrix and only if $u$ is a multiple of $v$.