Some of the proofs need properties of [[Complex Inner Product]] --- ### **Theorem Statement** > $ > (\exists x: Ax = b) \iff \neg (\exists\; y^Tb\neq \mathbf 0 : A^Ty = \mathbf 0) > $ Observe that: $A^Ty = \mathbf 0$ implies that $y\perp \text{ran}(A)$ or equivalently $\ y\in \text{null}(A^T)$, and if that is the case, they $b\perp y$ will have to be true so that $y\in \text{ran}(A)$, asserting the existence of the a solution for $Ax = b$ and vice versa. Take this as a proof for the theorem. --- #### **Lin Alg | Only the Real** * Let's say that: $A \in \mathbb{R}^{m\times n}$, then we will know that: $y^TAx = xA^Ty$ replace the transpose with $*$ so it's compatible with complex matrix. * Consider the expression: $A^Ty = 0$, we can say that: * $y$ is a zero eigen vector of $A^T$, or $y\in \text{null}(A^T)$. * $y\in \text{null}(A^T) \implies xA^Ty = 0$ by def. * Choose $x\notin \text{null}(A)$ making $Ax\in \text{range}(A)$ * then $y\perp Ax$ because $y^TAx = 0$ * which is $\text{null}(A^T) \perp \text{range}(A)$ and amazingly it's also true that $\text{null}(A) \perp \text{range}(A^T)$ **(1)** * And you know what this means for solving $Ax = b$??? It means that $b\perp \text{null}(A^*)$ implies that the system is solvable because $b\in range(A)$ by (1). * Then the space is not spanned by the $A^T$, the adjoin operator is * the null space of the original operator called: $A$ * The complex equivalent for the Fredholm Alternative is simply changing transpose to Conjugate Transpose. --- #### **Another proof but more concise**. Given any $A\in \mathbb{C}^{n\times n}$, let's start with the statement: > $y^HAx = (x^HA^Hy)^*$ Choose $y$ such that $A^Hy = \mathbb{0}$, then: $x^HA^Hy = 0 \quad \forall x$ $y\in \text{null}(A^H)$ But at the same time: $y^HAx = 0 \quad \forall x$ Which is saying: $ y \perp \text{range}(A) $ $ \text{null}(A^H) \perp \text{range}(A) $ So, let's frame this results here, this results is obtained by exploiting the properties of the Hermitian Adjoin: > $ > \text{null}(A^H) \perp \text{range}(A) \hspace{2em} \text{null}(A) \perp \text{range}(A^H) > $