Given $\mathbb C^2$ with the standard inner product, an operator $T(x,y) = (3x+4y, -4x+3y)$. Find $T^{*}$ and prove that $T$ is normal.
So, I took the standard basis $B = \{(1,0),(0,1)\}$ and we know it's orthogonal in respect to the standard inner product. I displayed $T$ in the $B$ basis, then transposed and conjugated it and got $[T^*]_B = \begin{bmatrix} 3 & -4 \\ 4 & 3 \end{bmatrix}$. But how exactly do we know what $T$ does on a general vector $(x,y)$ from that?
And I am also interested to know if there is another method.
Note that $T^*$ is uniquely determined by the equation $<Tz,w>=<z,T^*w>$ for $z,w \in \mathbb{C^2}.$
Now let $z=(x,y)$ and $w=(a,b)$. Then $\begin{equation} <Tz,w>=<(3x+4y,-4x+3y),(a,b)> = 3x\bar{a}+4y\bar{a}-4x\bar{b}+3y\bar{b} \end{equation}$ $= <(x,y),(3a-4b,4a+3b)> = <z,T^*w>$
From the above equation it can be seen $T^*(a,b)= (3a-4b,4a+3b)$. Now just compute to check that $TT^*=T^*T$ so that $T$ becomes normal.