How to find T* if the product inner space is not explicitly

Question

I have a little question. I was studying for my Linear Algebra test and I tryed the following problem:

If $T\in\mathcal{L}(K^n)$ is such that $T(z_1,\ldots,z_n)=(0,z_1,\ldots,z_{n-1})$. Determine a formula for $T^*$.

Well, I took the matrix representation of $T$ respect to canonical basis of $K^n$, $\left\{e_1,\ldots,e_n\right\}$. Then I considered it's transpose and the problem was "done" $\blacksquare$

I'm not sure of this because, yes, I know that matrix representation of $T^*$ is the conjugate transpose of $T$, but I don't have how to verify that $T^*$ is right by considering just the definition of $T^*$ because I don't have an explicit inner product defined. But yes, when I take some standard examples of inner products it works.

Answer 1

If we take $u=(u_1,\dots,u_n), v=(v_1,\dots,v_n)$ then $$\begin{align*} (Au,v)&=((0,u_1,\dots, u_{n-1}),(v_1,\dots v_n))\\ &=(\sum_{i=1}^{n-1}u_ie_{i+1},(v_1,\dots,v_n))\\ &=\sum_{i=1}^{n-1}u_i(e_{i+1},(v_1,\dots,v_n))\\ &=\sum_{i=1}^{n-1}u_i(e_{i+1},\sum_{j=1}^nv_je_j)\\ &=\sum_{i=1}^{n-1}\sum_{j=1}^{n}u_i\bar{v_j}(e_{i+1},e_j) \end{align*}$$

Now just do the same thing (i.e. express as a linear combination of $(e_i,e_j)$) for $(u,A^*v)$ and verify that they are in fact equal.

Answer 2

The inner product on $K^n$, by default, is the "dot-product". That is, we take $$ \langle(x_1,\dots,x_n),(y_1,\dots,y_n)\rangle = \sum_{k=1}^n x_k\overline{y_k} $$ It is only with respect to this inner product that the conjugate-transpose happens to give the adjoint.