Prove unitary matrices require orthonormal basis

• May 26th 2010, 09:44 PM
kaelbu
Prove unitary matrices require orthonormal basis
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.
• May 27th 2010, 12:15 AM
Opalg
Quote:

Originally Posted by 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!

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 27th 2010, 12:53 AM
kaelbu
Quote:

Originally Posted by Opalg
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!
• May 27th 2010, 03:15 AM
Opalg
Quote:

Originally Posted by kaelbu
$\displaystyle (AA^*)_{ij}= <x_i,x_j>$
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)