How to prove $U^{\perp }=${$x|x $ in $ ℝ^n , Ax^T=0$}?

Question

Let $U =$ span{$v_1,...,v_k$}, $v_i $ in $ℝ^n$ and A is $k\times n $ matrix with $v_i$ as rows.

How to prove $U^{\perp }=${$x|x $ in $ ℝ^n , Ax^T=0$} ?

By theorem, $U^{\perp }=${$x|x $ in $ ℝ^n , x\cdot y=0$, $\forall y\in U$}

In this case how do you distinguish $x$ and $y$?

Where does the $x^T$ come from?

Answer 1

I think when you write a vector $x$, you mean a row vector.

Because $$A=\left(\begin{array}{c}v_1\\v_2\\\vdots\\v_k\end{array}\right),$$ for any $x\in \mathbb{R}^n$, $Ax=0$ if and only if $v_j\cdot x=0$ for all $1\leq j\leq k$, but this implies $y\cdot x=0$ for all $y\in U$.

Answer 2

$Ax=0\iff v_j=0$ for all $j$. In this problem, the vector is transposed so that it lines up properly for vector multiplication. You are probably defining $\mathbb{R}^n$ as row vectors, but you need a column vector for the multiplication.