Prove that there exists a vector $v$ such that $\|v\|=\sqrt{2}$ and $\|Tv\|=5$ From Axler's "Algebra Done Right":

Suppose that $T$ is a normal operator on $V$ and that $3$ and $4$ are eigenvalues
  of $T$. 
Prove that there exists a vector $v$ in vector space $V$ such that $\|v\|=\sqrt{2}$ and $\|Tv\|=5$.

Not sure how to approach this one. I know that $TT^{*} = T^{*}T$ because normal operators.
Hence 
$\sqrt{\left< 3v, 3v \right>} = 5$ or $\sqrt{\left<4v, 4v\right>} = 5$. 
This is what I can clearly derive from the question. Naturally I have to use the fact that the operator is normal - likely something very simple but I'm having a block.
Any help is appreciated.
 A: Let $v_1 \in V$ such that $T(v_1) = 3v_1$ and $v_2 \in V$ such that $T(v_2) = 4v_2$. Since $T$ is normal and the eigenvectors $v_1,v_2$ are associated to different eigenvalues, we must have $v_1 \perp v_2$. By replacing $v_i$ with $\frac{v_i}{\| v_i \|}$, we can assume that $\| v_i \| = 1$.
Let us try to find a $v \in V$ which satisfies the required conditions and is also a linear combination of $v_1,v_2$. This is a natural thing to do because we don't know anything else about $T$ and $V$ and actually it might be the case that $V = \operatorname{span} \{ v_1, v_2 \}$ so there aren't any other options. So, let $v = av_1 + b_2$. Then we want:
$$ \| v \| = \sqrt{|a|^2 + |b|^2} = \sqrt{2}, \\
\| Tv \| = \| aTv_1 + bTv_2 \| = \|3a v_1 + 4b v_2 \| = \sqrt{(3|a|)^2 + (4|b|)^2} = 5.
$$
This is satisfied if $|a| = |b| = 1$.
A: Let $u_1, u_2$ be the eigenvectors of $T.$ asociated with the eigenvalues $3,4$ respectively
$Tu_1 = 3 u_1\\Tu_2 = 4u_2$
Then if $v = \frac {u_1}{\|u_1\|}+\frac {u_2}{\|u_2\|}$ and $u_1$ is normal to $u_2$
$\|v\| = \sqrt 2$ and $\|Tv\| = \|3\frac {u_1}{\|u_1\|}+4\frac {u_2}{\|u_2\|}\| = 5$
