Finding norm of a bounded linear operator defined on a Hilbert space Question: Let $H$ be a Hilbert space, and $\dim(H)>1$. Let $\{v_1,v_2\}\subset H$ be an orthogonal set of nonzero vectors. Suppose that for all $x\in H$,
$$ Tx=\langle x,v_1\rangle v_2+\langle x,v_2\rangle v_1.$$
Prove that $T$ is bounded linear operator and compute the norm of the operator $T$.
I could prove that, $T$ is bounded through proving this, for all $x\in H$,
$$\|Tx\|\leq 2\|v_1\|\|v_2\|\|x\|.$$
Thus, $\|T\|\leq 2\|v_1\|\|v_2\|$.
Next, using orthogonality of $v_1,v_2$, I could show that, $\|T(\dfrac{v_1}{\|v_1\|})\|=\|v_1\|\|v_2\|$, which gives $\|v_1\|\|v_2\|\leq \|T\|$.
But then how to find the norm of $T$? Help please.
 A: Set
$$
v_1=\|v_1\|e_1,\,\,v_2=\|v_2\|e_2.
$$
Clearly, $\{e_1,e_2\}$ is an orthonormal set and
$$
Tx=\langle x,v_1\rangle v_2+\langle x,v_2\rangle v_1
=\|v_1\|\|v_2\|\big(\langle x,e_1\rangle e_2+\langle x,e_2\rangle e_1\big),
$$
and hence
$$
\|Tx\|=\|v_1\|\|v_2\|\big(\langle x,e_1\rangle^2+\langle x,e_2\rangle^2\big)^{1/2}\le \|v_1\|\|v_2\| \|x\|,
$$
since
$$
\big(\langle x,e_1\rangle^2+\langle x,e_2\rangle^2\big)^{1/2}\le \|x\|.
$$
Hence $\|T\|\le \|v_1\|\|v_2\|$.
Also
$$
Tv_1=\langle v_1,v_1\rangle v_2=\|v_1\|^2v_2
$$
which implies
$$
\|Tv_1\|=\langle v_1,v_1\rangle v_2=\|v_1\|\|v_2\| \cdot \|v_1\|
$$
and hence $\|T\|\ge \|v_1\|\|v_2\|$.
Altogether $\|T\| = \|v_1\|\|v_2\|$.
A: Hint: $T(av_1+bv_2)=a\|v_1\|^{2}v_2+b\|v_2\|^{2}v_1$ so $\|T(av_1+bv_2)\|^{2}=\|v_1\|^{2}\|v_2\|^{2} (a^{2}\|v_1\|^{2}+b^{2}\|v_2\|^{2})$. The norm of the operator is simply the square root of  the  supremum of $\|v_1\|^{2}\|v_2\|^{2} (a^{2}\|v_1\|^{2}+b^{2}\|v_2\|^{2})$ over all $a,b$ such that $\|av_1+bv_2\|^{2}=a^{2}\|v_1\|^{2}+b^{2}\|v_2\|^{2} \leq 1$. So we get $\|T\|=||v_1||\|v_2\|$.
