Eigen vector of the adjoint operator. Error in this approach?

Question

Reisz Representation Theorem: If $V$ is finite-dimensional and $\phi$ is a linear functional on $V$. Then, there is a unique vector $v \in V~|~\phi(v) = \langle v,u\rangle~\forall~v \in V$

Definition: If $T \in L(V,W) $, then $T^* \in L(W,V)$ represents the adjoint of the linear map $T$. Then, by definition :

$\langle ~Tv, w~\rangle = ~\langle v, T^*w ~\rangle$.

By the Riesz Representation theorem: $T^*w$ is the unique vector in $V$ such that the above equation holds.

This means if for a given $v ~,~\exists ~x \in V ~|~\langle Tv,w~\rangle=~\langle v,x~\rangle$, then $T^*w=x$ as $T^*w$ must be unique by the Riesz Representation theorem.

Am I making a fundamental mistake in the above argument?

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As an example, suppose $T \in L(V,V)$ and $T$ has eigenvalue and eigen vector respectively as $\lambda,v$, then : $~\langle Tv,v~\rangle = ~\langle \lambda v,v ~\rangle = ~\langle v, \bar{\lambda}v~\rangle = ~\langle v, T^*v~\rangle$

By the above arguments, this would mean that $v$ is always an eigenvector for $T^*$ which is certainly not true. So, could someone tell me what is the error that I am making?

Another example: Suppose $v \in V$ is an eigen vector of $V$ and $w \in V$ is arbitrary. Define : $\phi(w) = \langle T v,w \rangle = \langle \lambda v, w\rangle = \langle v, \bar{\lambda} w \rangle = \langle v , T^*w \rangle$. Now, $v$ is fixed and $w$ is any arbitrary vector. Therefore, by uniqueness : $T^*w=\bar{\lambda} w ~\forall~w \in V$ which cannot be true. What could be the error here then? Thanks

Answer 1

Yes, you are making a fundamental (quite common) mistake. The uniqueness statement means that given a linear functional $\phi$, there exists a unique vector $u \in V$ such that $\phi(v) = \left< v, u \right>$ for all $v \in V$.

When you apply the uniqueness to the adjoint mapping, you fix a $w \in V$ and consider the linear functional $\phi(v) = \left< Tv, w \right>$. Then you are guaranteed a unique $x$ such that $\phi(v) = \left< Tv, w \right> = \left< v, x \right>$ and this $x$ is called $T^{*}w$. In other words, if, for a fixed $w \in W$ you have $\left< Tv, w \right> = \left< v, x \right>$ for all $v \in V$, then $x = T^{*}w$.

In your eigenvector argument, you are trying to deduce uniqueness from the value of $\left< Tv, v \right>$ for a specific $v$ and a specific $w = v$ but this is not what the uniqueness requires!

Answer 2

Suppose $v \in V$ is an eigen vector of $V$ and $w \in V$ is arbitrary. Define : $\phi(w) = \langle T v,w \rangle = \langle \lambda v, w\rangle = \langle v, \bar{\lambda} w \rangle = \langle v , T^*w \rangle$. Now, $v$ is fixed and $w$ is any arbitrary vector. Therefore, by uniqueness : $T^*w=\bar{\lambda} w ~\forall~w \in V$ which cannot be true because you are taking $v$ as eigen vector not any random vector in $V$.

Simple fact: you can prove it easily that: $$<v,w>=<v,w'>$$ for all $v$in $V$ then, $$w=w'$$ holds!!