I'm trying to solve the following exercise but can't get anywhere:
The problem:
$E$ an euclidian space
$f \in \mathscr{L}(E), (. | .) $ a scalar product on $E$
$\forall x \in E, (f(x) \:|\: x) \: = 0$
Compare $\operatorname {ker} f$ and $\operatorname {Im} f$
What I've tried :
I'm trying to find a relationship between the two sets. To do that, I've declared two variables, $x, y \in \operatorname {ker}(f) * \operatorname {Im}(f)$ and expanded $( f(x+y)\:|\: x+y) \:= 0$ using multi-linearity of the scalar product. I end up getting $(f(x)\:|\:y) + (f(y)\:|\:x) \:= \:(f(x+y)\:|\:x+y) \:= 0$ but I have no idea how to progress further.
Any help?
Thanks
Best Answer
Following your work, we deduce that for any $x,y \in E$, we have $$ (f(x) \mid y) + (f(y) \mid x) = 0 \implies (f(y) \mid x) = -(f(x)\mid y). $$ Now, suppose that $x \in \ker f$. Every element of $\operatorname{im}(f)$ can be expressed as $f(y)$ for some $y \in E$. We note that for any $x \in \ker f$ and $y \in E$, we have $$ (f(y)\mid x) = -(f(x) \mid y) = -(0 \mid y) = 0. $$ That is, the elements $x$ and $f(y)$ are orthogonal. In other words, we have deduced that $\ker f$ and $\operatorname{im} f $ are orthogonal subspaces.
By the rank nullity theorem, we know that $\dim \ker f + \dim \operatorname{im} f = \dim E$. Thus, we may deduce that $\ker f$ and $\operatorname{im} f$ are actually orthogonal complements.
Here's a solution using the adjoint transformation.
Let $f^*$ denote the adjoint of $f$. Using the fact that $(f(x)|x) = 0$ for all $x \in E$, we deduce that $(f^*(x)|x) = 0$ for all $x \in E$, and it follows that the map $g = \frac 12(f + f^*)$ satisfies $$ (g(x)\mid x) = 0 \quad \text{for all } x \in E. $$ Deduce from the fact that $g$ is self-adjoint and the above condition that $g=0$. That is, we have $f + f^* = 0 \implies f^* = -f$ (which is to say that $f$ is "skew-adjoint"). For any transformation $f$, we have $$ \ker(f) = \operatorname{im}(f^*)^\perp, $$ where $U^\perp$ denotes the orthogonal complement of a subspace $U \subseteq E$. However, because $f^* = -f$, we deduce that $\operatorname{im}(f^*) = \operatorname{im}(f)$ so that $$ \ker(f) = \operatorname{im}(f)^\perp. $$