Why is the 2nd norm of a complex vector equal to the norm of the same vector multiplied from left by its in dimensions corresponding fourier matrix?

Question

The question is:

with the fourier matrix defined as

and the supplied solution is

I get the first step, because the fourier matrix multiplied by its inverse results in the identity matrix. But I am lost by the second. Can someone show me why the dot product of two vectors is equal to the dot product of a multiplication of an invertable matrix to both vectors?

Update after answered with relevant definitions

This matlab script demonstrates the definition used in the second step

N = 4;
X = complex(rand(N, N), rand(N,N));
z = [3;5;7;8]
dot(X*z,z)
dot(z,ctranspose(X)*z) % the result of these last two lines is always equal

1. The conjugate transpose $A^*$ of a matrix $A$ is $$(A^*)_{i,j} = (A^H)_{i,j} = \overline {A_{j,i}}$$

2. A complex square matrix $U$ is unitary if its conjugate transpose $U^*$ is also its inverse $$UU^* = UU^{-1} = I$$

Answer 1

In the step

$$ \langle z, z\rangle = \langle F^*F z,z\rangle, $$

the fact that the DFT matrix $F$ is unitary is used: $F^*F = I$ so that $z = F^*F z$.

In the step

$$ \langle F^* F z,z\rangle = \langle Fz, Fz\rangle, $$

the definition of the adjoint is used: $\langle A^* x,y\rangle = \langle x, Ay\rangle$ so that $\langle F^* Fz, z\rangle = \langle Fz, Fz\rangle$.