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I was thinking in this way,

Let V be a real inner product space of dimension 1 and T is a linear map on this space i.e. $T\in\mathcal L(V)$.

If $\lambda$ is an (real) eigenvalue and $v$ is corresponding non-zero eigenvector, then, $$ T(v) = \lambda v\\$$

Let $(v_1)$ is a basis for $V$ vector space.

Then $$v=av_1$$ for some $a\in R$

$a\neq 0$ because $v$ is non-zero.

Therefore,

\begin{aligned} T(av_1)&=\lambda av_1\\ aT(v_1)&=a\lambda v_1\\ T(v_1)&=\lambda v_1\\ ||T(v_1)||&=\lambda||v_1||\\ \lambda &=\frac{||T(v_1)||}{||v_1||} \end{aligned}

That means any non-zero vector is an eigenvector for 1 dimensional inner product space with eigenvalue equal to $\frac{||T(v_1)||}{||v_1||}$.

Is there anything wrong in this approach?

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    You don't use the inner product! You just need a normed linear space. – 2017-02-11
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    @sujoy Your reasoning is correct. – 2017-02-11
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    It is not correct ! Look at $T(v)=-v$ – 2017-02-11
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    Your formula would induce that you can only have **positive** eigenvalues... – 2017-02-11

1 Answers 1

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If $V$ is a normed space, then you only can conclude: $|\lambda |=\frac{||T(v_1)||}{||v_1||}$

With your notation: there is a scalar $t_0$ such that $T(v_1)=t_0v_1$. If $v \in V$, then $v=sv_1$ with some scalar $s$, hence

$T(v)=T(sv_1)=sT(v_1)=st_0v_1=t_0v$. Thus $T$ has the form

$T(v)=t_0v$.

Consequence: the only eigenvalue of $T$ is $t_0$ and each nonzero $v \in V$ is eigenvector.