# Let $v$ and $w$ be eigen vectors of $T$ corresponding to two distinct eigen values $\lambda _1$ and $\lambda_ 2$ respectively

Problem:Let $v$ and $w$ be eigen vectors of $T$ corresponding to two distinct eigen values $\lambda _1$ and $\lambda_ 2$ respectively

Then which of the following is true ?

$1)$ For non zero scalars $\alpha_ 1$ , $\alpha_ 2$, the vector $\alpha_ 1v + \alpha_ 2 w$ is not eigen vector of $T$

$2)$ For all scalars $\alpha_ 1$ , $\alpha_ 2$, the vector $\alpha_ 1v + \alpha_ 2 w$ is not eigen vector of $T$

$3)$ $\alpha_ 1v + \alpha_ 2 w$ is eigen vector of $T$, if $\alpha_ 1 = \alpha_ 2$

$4)$ $\alpha_ 1v + \alpha_ 2 w$ is eigen vector of $T$, if $\alpha_ 1 = -\alpha_ 2$

Solution:

I know about eigen values and eigen functions

$Tv=\lambda_{1}v$ and $Tw=\lambda_{2}w$ with $\lambda_1 \neq \lambda_2$. Since $T$ is (presumably) linear, $$T\left(\alpha_{1}v+\alpha_{2}w\right)=\alpha_{1}Tv+\alpha_{2}Tw=\alpha_{1}\lambda_{1}v+\alpha_{2}\lambda_{2}w.$$ Since $\lambda_{1}$ and $\lambda_{2}$ are distinct, we cannot write this as $\lambda\left(\alpha_{1}v+\alpha_{2}w\right)$. (1) is true.

Take $\alpha_{1}=0$. Then clearly $\alpha_{1}v+\alpha_{2}w$ is an eigenvector of $T$. (2) is false.

(3) and (4) are false by (1).

Let $T$ act on what you're asking about. For example, what is equal to $T(\alpha_1 v + \alpha_2 w)$? You can simplify this expression.

$Tv = \lambda_1 v$ and $Tw = \lambda_2 w$.

It follows: $T(\alpha_1 v + \alpha_2 w) = \alpha_1 \lambda_1 v + \alpha_2 \lambda_2 w$ . Since $\lambda_1$ and $\lambda_2$ are distinct, $\alpha_1 v + \alpha_2 w$ cannot be an eigenvector for all non-zero scalars $\alpha_1$ and $\alpha_2$.

Hence, the first option is true.