# Prove unitary matrices require orthonormal basis

#### kaelbu

Problem: Let A be an nXn matrix with complex entries. Prove that AA*=I if and only if the rows of A form an orthonormal basis for $$\displaystyle C^n$$

Thoughts: If AA*=I then A*=A^-1, but I don't understand how (or if) that implies that the rows of A are an orthonormal basis, or how the reverse is true, or, for that matter, much of anything involving linear algebra, gah!

Thanks for any help.

#### Opalg

MHF Hall of Honor
Problem: Let A be an nXn matrix with complex entries. Prove that AA*=I if and only if the rows of A form an orthonormal basis for $$\displaystyle C^n$$

Thoughts: If AA*=I then A*=A^-1, but I don't understand how (or if) that implies that the rows of A are an orthonormal basis, or how the reverse is true, or, for that matter, much of anything involving linear algebra, gah!
If you have a matrix product AB, then the definition of matrix multiplication says that the (i,j)-element of AB is the inner product of row i of A with the complex conjugate of column j of B. (The reason for the complex conjugation is that the formula for matrix multiplication gives $$\displaystyle (AB)_{ij} = \textstyle\sum a_{ik}b_{kj}$$, but the formula for an inner product uses the complex conjugate of the second vector: $$\displaystyle \langle x,y\rangle = \textstyle\sum x_k\overline{y_k}$$.)

In the adjoint matrix A*, column j is the complex conjugate of row j of A.

Putting those two facts together, you see that the (i,j)-element of AA* is the inner product of row i of A with row j of A. If that is equal to the (i,j)-element of the identity matrix then ... .

• HallsofIvy and kaelbu

#### kaelbu

Putting those two facts together, you see that the (i,j)-element of AA* is the inner product of row i of A with row j of A. If that is equal to the (i,j)-element of the identity matrix then ... .
$$\displaystyle (AA^*)_{ij}= <x_i,x_j>$$
if $$\displaystyle i=j <x_i, x_i> must = 1$$ therefore $$\displaystyle ||x_i||=1$$ -> normal
if $$\displaystyle i \not=j <x_i, x_j> must = 0$$ therefore $$\displaystyle x_j$$ is perpendicular to $$\displaystyle x_i$$ -> orthogonal
=> orthonormal

Going the opposite direction:
if the rows of A are orthonormal
$$\displaystyle <x_i,x_j> = 0$$
except when i=j when $$\displaystyle <x_i, x_j>=1$$
therefore $$\displaystyle (AA^*)_{ij} = a_i \overline{a_j} = 1$$ if $$\displaystyle i=j$$
or = 0 if $$\displaystyle i \not=j$$which is the definition of the identity.

One more question: is this enough exposition? That is, did I prove it well enough?
Thank you!

#### Opalg

MHF Hall of Honor
$$\displaystyle (AA^*)_{ij}= <x_i,x_j>$$
if $$\displaystyle i=j <x_i, x_i> must = 1$$ therefore $$\displaystyle ||x_i||=1$$ -> normal
if $$\displaystyle i \not=j <x_i, x_j> must = 0$$ therefore $$\displaystyle x_j$$ is perpendicular to $$\displaystyle x_i$$ -> orthogonal
=> orthonormal

Going the opposite direction:
if the rows of A are orthonormal
$$\displaystyle <x_i,x_j> = 0$$
except when i=j when $$\displaystyle <x_i, x_j>=1$$
therefore $$\displaystyle (AA^*)_{ij} = a_i \overline{a_j} = 1$$ if $$\displaystyle i=j$$
or = 0 if $$\displaystyle i \not=j$$which is the definition of the identity.

One more question: is this enough exposition? That is, did I prove it well enough?
Yes, that looks good to me. (Rock)