# Show that $(\text{im } M)^{\bot}=\text{ker } M^{*}$.

Hi I'm reading Logemanns book on ODEs and I wanted to prove his theorem (Theorem A.1) on page 262 which says

Let $$M \in \mathbb F^{NxP}$$. Then $$(\text{im } M)^{\bot}=\text{ker } M^{*}.$$ im $$M$$ is the image of the matrix $$M$$. $$(\cdot)^{\bot}$$ is the orthogonal complement of a subspace (in this case im $$M$$). $$M^*$$ is the hermitian transpose of $$M$$. Ker $$M^*$$ is the set described below: $$\text{ker } M^* =\{x\in \mathbb F^N : M^*x=0\}$$ The set im $$M$$ is described below $$\text{im } M = \{Mx: x\in \mathbb F^P \}$$

What I have done:

Let $$S=:\text{im } M$$. If $$S^\bot$$ is the orthogonal complement of $$S$$ then the inner product $$\langle x,y \rangle=0 \quad \forall x \in S, \forall y \in S^\bot$$. Hence all I have to show is the following:

$$(a)\quad \langle Mx,y\rangle = 0\, \, \text{for every } x \text{ iff } y\in \text{Ker }M^*$$

Proof:

Assume $$\langle Mx,y\rangle=\langle x,M^*y\rangle=0$$ Since $$x$$ is chosen arbitrary then $$M^*y=0$$ hence $$y\in \text{Ker }M^*$$.

Now assume $$y\in \text{Ker }M^*$$ which means $$M^*y=0$$ then $$\langle Mx,y\rangle=\langle x,M^*y\rangle=\langle x,0\rangle=0$$.

This feels wrong because if $$x=(1,-1)$$ and $$M^*y=(1,1)$$ the the inner product would be zero while $$y\notin \text{Ker }M^*$$. Help would be appreciated.

Let $$x\in \mathrm{Ker}\ M^*$$, then for any $$y$$ we have $$0=\langle M^*x,y \rangle = \langle x,My\rangle$$ which means that $$x \perp \mathrm{Im}\ M$$, hence $$\mathrm{Ker}\ M^* \subset\mathrm{Im}\ M^\perp$$.
Now let $$y\in \mathrm{Im}\ M^\perp$$. Then for any $$x$$ we have $$0=\langle Mx, y\rangle = \langle x,M^*y\rangle.$$ Since this has to be true for any vector $$x$$, then $$M^*y$$ has to be the zero vector, thus $$y\in \mathrm{Ker}\ M^*$$, hence $$\mathrm{Im}\ M^\perp \subset \mathrm{Ker}\ M^*.$$
$$\langle Mx , y \rangle =0$$ for every $$x$$ iff $$y \in Ker M^{*}$$. In your example you are choosing a particular $$x$$.