Prove unitary matrices require orthonormal basis

May 2010
13
3
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
Aug 2007
4,039
2,789
Leeds, UK
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 ... .
 
May 2010
13
3
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
Aug 2007
4,039
2,789
Leeds, UK
\(\displaystyle (AA^*)_{ij}= <x_i,x_j> \)
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)