$A$ is a linear transformation with eigenvalues Let $A:\mathbb{R}^2\rightarrow\mathbb{R}^2$ be a linear transformation with eigenvalues $\frac{2}{3}$ amd $\frac{9}{5}$. Then, there exists a non zero vector $v\in\mathbb{R}^2$ such that 


*

*$||A(v)||>2||v||$

*$||A(v)||<\frac{1}{2}||v||$

*$||A(v)||=||v||$

*$Av=0$


Clearly fourth option is wrong as this linear transformation already have two eigen values and if $Av=0$ then $0$ will also be an eigen value of $A$ a contradiction..
Consider $A=\begin{bmatrix}\frac{2}{3}&0\\0&\frac{9}{5}\end{bmatrix}$ then  for $v=(v_1,v_2)$ we have $Av=(\frac{2}{3}v_1,\frac{9}{5}v_2)$. Then, $||Av||=\sqrt{\frac{4}{9}v_1^2+\frac{81}{25}v_2^2}$.
For $||Av||=||v||$ we need $\frac{4}{9}v_1^2+\frac{81}{25}v_2^2=v_1^2+v_2^2$  i.e., $\frac{5}{9}v_1^2=\frac{56}{25}v_2^2$.. We can simply assume $v_1=\sqrt{\frac{56}{25}}$ and $v_2=\sqrt{\frac{5}{9}}$ then we have $v$ with the property that $||Av||=||v||$..
This was actually surprising to me..  I thought third option would be definitely wrong unless i solved it on paper..
For this matrix it works but i do not know how to see this for general matrix.. I know that it is diagonalizable so i thought i should give a try  for diagonal matrices first...
With this diagonal matrix, i see all three options are correct..
I some how feel first two options are not true in general..
Help me to clear this..
 A: First, look at this problem: given two values $a, b$ with $a \ge b$, what are the maximum and minimum values of $z = ax + by$, for various $x, y \ge 0$, with $x + y = 1$?
To solve it, let $y = 1 - x$, then $z = ax +b(1 -x) = b + (a - b)x$. Now $x \ge 0$ and $1 - x \ge 0$, so $x \le 1$. The maximum occurs at the maximum value of $x$, which is $1$. The minimum occurs at the minimum value $x$, which is $0$. So the maximum value of $z$ is $a$ and the minimum value is $b$. 
Further, if $z_0$ is any value $b \le z_0 \le a$, then we can solve $x = \frac{z_0 - b}{a - b}$ and get a value of $x \in [0, 1]$ such that $z = z_0$.

Now to apply this to your problem. Since you have trouble with more general proofs, I will rewrite this for only two dimensions.
When $$A = \begin{bmatrix} a_1 & 0\\ 0 & a_2\end{bmatrix}$$ and (for convenience) $|a_1| \le |a_2|$, consider the affect of $A$ on unit vectors. I.e., vectors $v$ with $\|v\| = 1$.Then
$$\|Av\|^2 = a_1^2v_1^2 + a_2^2v_2^2$$
This fits the pattern of the maximization/minimization problem above. $v_1^2, v_2^2 \ge 0$ and because $\|v\| = 1$, we have $v_1^2 + v_2^2 = 1$. Therefore by the argument given above, the largest value $\|Av\|^2$ can take on for any $v$ with $\|v\| = 1$ is $a_2^2$, and the smallest value is $a_1^2$. And for any value $a^2 \in [a_1^2, a_2^2]$, there must be some unit vector $v$ such that $\|Av\|^2 = a^2$. Taking the square roots gives:
Among all unit vectors $v$,


*

*the maximum value of $\|Av\|$ is $|a_2|$,

*the minimum value of $\|Av\|$ is $|a_1|$.

*If $|a_1| \le a \le |a_2|$, then there is some unit vector $v$ such that $\|Av\| = a$.


Now if $v$ is an arbitrary non-zero vector, then $\frac v {\|v\|}$ is a unit vector. And $$\frac{\|Av\|}{\|v\|} = \frac 1{\|v\|}\|Av\| = \left\|\frac 1{\|v\|}Av\right\| = \left\|A\left(\frac 1{\|v\|}v\right)\right\|= \left\|A\left(\frac v{\|v\|}\right)\right\| $$ Hence we can also say:
Among all non-zero vectors $v$,


*

*the maximum value of $\frac{\|Av\|}{\|v\|}$ is $|a_2|$,

*the minimum value of $\frac{\|Av\|}{\|v\|}$ is $|a_1|$.

*If $|a_1| \le a \le |a_2|$, then there is some unit vector $v$ such that $\frac{\|Av\|}{\|v\|} = a$.


While I've shown this only for diagonal matrices, it is actually true for all normal matrices (a square matrix is normal if $A^TA = AA^T$). The values of $\frac{\|Av\|}{\|v\|}$ form the interval $[a, b]$ where $a$ is the smallest absolute value of the eigenvalues of $A$, and $b$ is the largest absolute value of the eigenvalues of $A$.
In your problem, the largest eigenvalue is $\frac 9 5$, and the smallest eigenvalue is $\frac 2 3$. So the values of $\frac{\|Av\|}{\|v\|}$ that are possible are those in the interval $[2/3, 9/5]$. $1$ is in this interval. $2$, $1/2$ and $0$ are not. So


*

*There are no vectors $v$ for which $\|Av\| > 2\|v\|$.

*There are no non-zero vectors $v$ for which $\|Av\| < \frac12\|v\|$.

*There are vectors $v$ for which $\|Av\| = 1\|v\| = \|v\|$.

*There are no non-zero vectors for which $\|Av\| = 0\|v\| = 0$.

