Inverse of a diagonal matrix with nonzero complex entries.

Question

If $z_1,\dots,z_n$ are nonzero complex numbers, prove that $$ \begin{bmatrix} z_1 & \\ & \ddots \\ & & z_n \end{bmatrix} \cdot \begin{bmatrix} \bar{z}_1/|z_1|^2 & \\ & \ddots \\ & & \bar{z}_n//|z_n|^2 \end{bmatrix} =I_n. $$

Answer 1

This question has more to do with complex numbers than matrices.

If $z = a + ib$, then $z \bar{z} = a^2+b^2 = |z|^2$. Thus, if $z$ is nonzero (i.e., $|z| \ne 0$), then $$ z\frac{\bar{z}}{|z|^2} = 1. $$
By virtue of the mechanics of matrix multiplication, in particular that of diagonal matrices, we have $$ \begin{bmatrix} z_1 & \\ & \ddots \\ & & z_n \end{bmatrix} \cdot \begin{bmatrix} \bar{z}_1/|z_1|^2 & \\ & \ddots \\ & & \bar{z}_n//|z_n|^2 \end{bmatrix} =I_n. $$