Image and kernel related to inner product Let $(V,\left\langle\cdot,\cdot\right\rangle)$ a finite dimensional euclidean vector space and $v,w\in V\setminus\{0\}$ are arbitary but confirmed. And $f\in \mathcal{L}(V,V)$ with $f(x)=\left\langle x,v\right\rangle w$.
1. Confirm the image and kernel of $f$.
2. Prove that $f$ is self adjoint iff when $v,w$ linearly dependent.
For 1. I only have an idea about the kernel: If $x=0$ then $f(x)=\left\langle 0,v\right\rangle w=0$, so $0\in \text{Kernel}(f)$ and if $x$ is a vector of orthogonal basis and v is the other vecotr of the orthogonal bais, then $f(x)=0$, besides if $v,w=0$ $f(x)$ also equals 0, is these vectors also part of the kernel? And how should I write the formly?
For 2 I have only $$\left\langle f(x),v\right\rangle=\left\langle x,f^{\text{ad}}(v)\right\rangle= \left\langle x,f(v)\right\rangle$$
$$\left\langle  \left\langle  x,v\right\rangle w,v\right\rangle= \left\langle x,f(v)\right\rangle$$ $$ \left\langle x,v\right\rangle\left\langle w,v\right\rangle =\left\langle x,w\right\rangle \left\langle v,v\right\rangle$$ and I get have any further.
Realy appreciate it if you could give some hints and suggestions!
 A: The kernel is the orthogonal complement of $v$, since it is enough for the inner product $\langle x, v \rangle$ to vanish to get $f(x) = 0$. So $\ker f = \{v\}^\perp := \{x \in V\ |\ \langle x,v \rangle = 0\}$.
The image is simply the span of $w$, i.e., $\text{Im} f = \text{span}\{w\}$.
For the adjoint (I assume you have a sesquilinear inner product which is conjugate linear in the second variable, not the first, becuase otherwise $f =   \langle \cdot,v\rangle w$ would not be linear but conjugate linear, if your inner product is a real then ignore this qualification)
$$
\langle f(x),y\rangle =
\langle  \langle  x,v\rangle w,y\rangle = 
\langle  x,v\rangle\langle w,y\rangle = 
\langle  x,\langle y,w\rangle v\rangle =
\langle  x,f^\dagger(y)\rangle,
$$
where $f^\dagger = \langle \cdot, w\rangle v$.
Assume $f = f^\dagger \iff \langle \cdot, v\rangle w= \langle \cdot, w\rangle v$, then the images span the same 1-dimensional subspace, $\text{span} \{v\} = \text{span} \{w\} \implies w = \lambda v$, for some constant $\lambda$. Since $f$ and $f^\dagger$ should coincide on each $x \in V$, i.e $\lambda \langle x, v\rangle v = \bar \lambda \langle x, v\rangle v$, we should have $\lambda \in \mathbb R$.
The converse is obvious, $\lambda \in \mathbb R,\ w = \lambda v \implies f = f^\dagger $.
A: Here's how you can prove $(b)$. It's not hard to show that $$\forall x,y\in V:\big<x,f(y)\big>=\big<f(x),y\big>\iff \forall x,y\in V:\det\begin{pmatrix}\big<x,w\big>&\big<x,v\big>\\ \big<y,w\big>&\big<y,v\big>\end{pmatrix}=0$$ If we assume $f$ is self$-$adjoint, take $x=w,y=v$ in the previous biconditional to get $$\det\begin{pmatrix}\|w\|^2&\big<v,w\big>\\ \big<v,w\big>&\big\|v\|^2\end{pmatrix}=0$$ This means $$\big<v,w\big>=\pm\|v\|\|w\|$$ which is only true if $w=cv$ for some scalar $c$. On the other hand, if $w=cv$ for some scalar $c$ then for any $x,y\in V$ we have $$ \det\begin{pmatrix}\big<x,w\big>&\big<x,v\big>\\ \big<y,w\big>&\big<x,v\big>\end{pmatrix}=\det\begin{pmatrix}\big<x,cv\big>&\big<x,v\big>\\ \big<y,cv\big>&\big<x,v\big>\end{pmatrix}=\det\begin{pmatrix}c\big<x,v\big>&\big<x,v\big>\\ c\big<y,v\big>&\big<y,v\big>\end{pmatrix}=0$$ So $f$ is self$-$adjoint.
