a) Let $(\varphi_{\alpha})_{\alpha=1}^{N}$ and $(\psi_{\beta})_{\beta=1}^{M}$ be two orthonormal bases in a Hilbert space, $H$. Define an $N\times M$ matrix , $T_{\alpha \beta}=<\varphi_{\alpha},\psi_{\beta}>$. Prove that $T$ is a matrix of a bounded invertible operator from $\mathbb{C}^{M}$ to $\mathbb{C}^{N}$ (where $\mathbb{C}^{\alpha}$ means $l^2$)
b) Prove $N=M$
My candidate to be $T$ is:
$T((x_1,\ldots, x_M))=(\sum_{j=1}^{M}x_j<\varphi_1,\psi_j>,\ldots, \sum_{j=1}^{M} x_j <\varphi_N,\psi_j>).$
how prove $T$ is invertible?
$T$ is a unitary matrix, and thus invertible. To see this, $$ [T^*T]_{ab} = \sum_{b=0}^N \overline{T}_{ca}T_{cb} = \sum_{c=0}^N\overline{\left<\varphi_c,\psi_a\right>}\left<\varphi_c,\psi_b\right> = \sum_{c=0}^N\left<\psi_a,\varphi_c\right>\left<\varphi_c,\psi_b\right> = \left<\psi_a,\psi_b\right> = \delta_{ab}, $$ where the orthonormality of the bases is used in the last two equalities. A similar argument shows $TT^* = I$, and $N=M$ shouldn't be too hard from there.