How to explain that null $A$=(row$A$)$^\perp$? How would you explain that null$A$=(row$A$)$^\perp$?
Formally, null $A$=($x$ in $ℝ^n|Ax=0$). So suppose $A$ is $m\times n$, null$A$ spans row space. I think it is an alternative way of saying null$A$ is the orthogonal complement of rows of $A$ so that $x(r_i)=0$.
But why isn't it col A and $ℝ^m$? And it is true that null $A^T$=(col$A$)$^\perp$?
 A: Here's the quickest explanation I know:
Let $A$ be a matrix with rows $A_1,\dots,A_m$.  Let $x$ be a (column) vector with $n$ entries.  Note that the definition of matrix multiplication tells us that
$$
Ax = \pmatrix{A_1 \cdot x\\ \vdots \\ A_n \cdot x}
$$
where $\cdot$ denotes the dot product.  It follows that if $Ax = 0$, then $A_i \cdot x = 0$ for every row $i$.  That is, $x$ is perpendicular to every row-vector of $A$.  This means in turn that $x$ is in the orthogonal complement to the row space of $A$.
This logic works in reverse as well: if $x$ is in the orthogonal complement of the row-space, then clearly we'll have $A_i \cdot x = 0$ for every $i$, which is to say that $Ax = 0$.
So, null$(A) = $row$(A)^\perp$.
And yes: null$(A^T) = $col$(A)^\perp$.
A: $(Ax)_i$ is the dot product of the $i$th row of $A$ with $x$. So $Ax$ is the zero vector if and only if $x$ is orthogonal to all the rows of $A$.
You get the other statement by applying this one to $A^T$ and recalling that $(A^T)^T=A$.
A: Hint :-
For $A$, row rank $A$ = col rank $A$ = rank $A$ and null $A$ + rank $A$ = $n$, where $n$ is the dimension of the matrix.
