3
$\begingroup$

My book states that a 3$\times$3 unitary matrix has 9 degrees of freedom, but I have trouble seeing why that is true. So I want to try and show that an $n\times n$ matrix has $n^2$ degrees of freedom.

Base case $n=1$: This is just a matrix with one entry which has length 1. It is clear that any entry with length 1 will do. So I have $1 = 1^2$ degree of freedom

$n \implies n+1$: Suppose we have an $n \times n$ unitary matrix and it has $n^2$ degrees of freedom. I want to now show that a $(n+1) \times (n+1)$ unitary matrix has $(n+1)^2$ degrees of freedom. If I add a $(n+1)$th row and column to the $n \times n$ unitary matrix, I have added $(n+1) + n = 2n + 1$ entries. For every entry $a_{ij}$ I have added, I need to ensure that each $a_{ij}^2 = 0$ in order for the matrix to be unitary. Since there are numerous ways to choose $a_{ij}$ such that $a_{ij}^2$ = 0, I have added $2n+1$ degrees of freedom. So the total degrees of freedom is $n^2 + 2n + 1 = (n+1)^2$.

Can I get some feedback on this proof? Thanks in advance!

Note It seems like my approach above is at a dead end.

  • 0
    I realize that I have an error when picking the entries $a_{ij}$ for there to be a unitary matrix. Is there an easy fix?2012-11-19
  • 0
    I don't see a way to fix this. The plan seems flawed: you cannot start with a given $n\times n$ unitary matrix and add something to get an $(n+1)\times(n+1)$ matrix. Most of the $(n+1)\times(n+1)$ unitary matrices don't have an $n\times n$ unitary submatrix.2012-11-19
  • 0
    Induction is not really a natural approach to this problem in my opinion. Your $n-1 \times n-1$ submatrix is generally not unitary (because you cannot add a non-zero entry to a unit length vector and have it remain unit length). A more fundamental problem is: What exactly do you mean by "degrees of freedom"? It is _not_ dimension because the unitary matrices do not form a subspace. So what exactly is it are you trying to prove?2012-11-19
  • 0
    It would be good to know what exactly you mean by degrees of freedom. How strictly is this notion defined in the book? If I had to formulate the question rigorously, I'd have to use the notion of a [manifold](http://en.wikipedia.org/wiki/Differentiable_manifold). Does your book do that or does it formulate things less rigorously?2012-11-19
  • 0
    @DanShved: You're right. So is there an easy way to see that an $n \times n$ unitary matrix has only $n^2$ degrees of freedom? My book says that an $n \times n$ matrix in general has $2n^2$ degrees of freedom since there are $n^2$ entries and each entry has 2 real numbers: one for the complex and one for the real part.2012-11-19
  • 0
    @Student As I (and EuYu) said, the real question is how the book defines degrees of freedom in the first place. Or does it start using these words right away, without a definition? This is important to us, because otherwise we are at risk of giving you the answer that will be much more technical and convoluted than you need.2012-11-19
  • 0
    @DanShved: My book uses the words right away without a formal definition.2012-11-19
  • 2
    Here is a hand waving viewpoint: The unitary matrix is characterized by the constraints $u_i^* u_j = \delta_{ij}$ for $i\geq j$, where $u_i$ are the columns of $U$. The constraints $u_i^* u_i = 1$ form 3 'real' constraints and the others form 3 'complex' constraints. Hence in terms of a real parameterization, there are 18 real parameters for a 3x3 matrix, with 3 real constraints and three complex constraints. Since you can view a complex constraint as two real ones, this corresponds to a remaining $18-3-2\cdot 3 = 9$ real parameters (loud sounds of hand waving in background).2012-11-19

1 Answers 1