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

Question

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.

Answer 1

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^*. $

Answer 2

You have sated (a) wrongly.

$\langle Mx , y \rangle =0$ for every $x$ iff $y \in Ker M^{*}$. In your example you are choosing a particular $x$.