Proving $(\text{ ker } A)^\perp \subset \text{ image } A^T$

Question

Let $A \in \mathbb{R}^{m \times n}$.

Let $x \in (\text{ker} A)^\perp$ so that $\langle x, y \rangle = 0$ for all $y \in \text{ ker } A$. Let $\beta = \{b_1, ..., b_k \}$ be a basis of $ \text{ image } A^T$. This basis can be extended to be a basis of $\mathbb{R}^n$ to get $\beta' = \{b_1, ..., b_k, b_{k+1}, ..., b_n \}$ so that

$$x = \sum_{i=1}^n c_i b_i = A^T v + \sum_{i=k+1}^n c_i b_i \,\,\,\,\,\,\,\,\,\text{ ( Eq 1 )}$$ for some $v \in \mathbb{R}^m$.

Then, $$\langle x , y \rangle = \langle A^T v, y \rangle + \langle \sum_{i=k+1}^n c_i b_i, y \rangle = 0 $$

$$\langle v, Ay \rangle + \sum_{i=k+1}^n c_i \langle b_i, y \rangle = 0 $$

The first term is clearly zero so we have:

$$\sum_{i=k+1}^n c_i \langle b_i, y \rangle = 0$$

I want to complete the proof by showing that the second term in $\text{(Eq 1)}$ is zero. Can anyone help?

Thanks.

Answer 1

$$x\perp \text{image}(A^T)$$$$\iff\langle x,A^Tz\rangle =0,\forall z\in\Bbb R^m$$$$\iff z^T(Ax)=x^TA^Tz=0,\forall z\in \Bbb R^m$$ $$\bigg[\text{Being a real number }\langle x,A^Tz\rangle=x^TA^Tz \text{ is equals to its transpose }z^TAx\bigg]$$$$\iff Ax=0$$$$\iff x\in \text{ker}(A).$$ So, $$\big(\text{image}(A^T)\big)^\perp=\text{ker}(A)\iff\text{image}(A^T) =\bigg(\big(\text{image}(A^T)\big)^\perp\bigg)^\perp=\big(\text{ker}(A)\big)^\perp.$$

I have used the fact that, for any subspace $V$ of a finite dimensional inner product space we have, $(V^\perp)^\perp=V$.