# Showing that the Entries of a Matrix Arise As Inner Products

1. Nov 7, 2015

### Bashyboy

1. The problem statement, all variables and given/known data
Let $B \in M_n (\mathbb{C})$ be such that $B \ge 0$ (i.e., it is a positive semi-definite matrix) and $b_{ii} = 1$ (ones along the diagonal). Show that there exists a collection of $n$ unit vectors $\{e_1,...,e_n \} \subset \mathbb{C}^n$ such that $b_{ij} = \langle e_i, e_j \rangle$.

2. Relevant equations

3. The attempt at a solution
Note, this took a great deal of time to type, and I hope one would be so kind as to reply! :)

Now, obviously any set collection of $n$ unit vectors will satisfy the condition that $\langle e_i , e_i \rangle = 1 = b_{ii}$, for all $i \in \{1,...,n\}$. Hence, the choice of unit vectors will only depend upon the off-diagonal terms.

I tried proving the theorem in low dimensions, hoping that I might glimpse some natural choice of unit vectors that will extend to the $n \times n$ case. For instance, in the 2x2 case, we have that

$\begin{bmatrix} 1 & b_{12} \\ \overline{b_{12}} & 1 \\ \end{bmatrix} \ge 0 \implies |b_{12}| \le 1$

Letting $b_{12} = |b_{12}| e^{i \theta_{12}}$, then the choice of unit vectors would be $e_1 = \begin{bmatrix} e^{i \theta_{12}} \\ 0 \end{bmatrix}$ and $e_2 = \begin{bmatrix} |b_{12}| \\ \sqrt{1 - |b_{12}|^2} \end{bmatrix}$, both of which are obviously unit-vectors. Computing the inner-products, we arrive at $\langle e_1 , e_2 \rangle = |b_{12}| e^{i \theta_{12}} = b_{12}$ and $\langle e_2, e_1 \rangle = \overline{\langle e_1 , e_2 \rangle} = \overline{b_{12}}$, which finishes the proof in the 2x2 case.

However, things in the 3x3 become particularly unilluminating, without there being any seemingly natural choice that would help in the nxn case. So, in the 3x3 case, we need to find $e_1$, $e_2$, and $e_3$ such that $\langle e_1, e_2 \rangle = b_{12}$, $\langle e_1, e_3 \rangle = b_{13}$, etc. Because unitary matrices preserve the inner product, I know I can choose a unitary $U$ which will rotate all the unit vectors; in particular, I can rotate $e_1$ so that it becomes $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, which simplify things by reducing the number of variables. Now, let

$e_2 = \begin{bmatrix} z_1 \\ z_2 \\ z_3 \\ \end{bmatrix}$

$e_3 = \begin{bmatrix} z_4 \\ z_5 \\ z_6 \end{bmatrix}$,

demanding that $||e_2|| = 1$ and $||e_3|| = 1$ be true. So, we want to find $z_1,...,z_5$ such that

$\langle e_1 , e_2 \rangle = b_{12} \implies z_1 = b_{12}$

$\langle e_1, e_3 \rangle = b_{13} \implies z_4 = b_{13}$

$\langle e_2 , e_3 \rangle = b_{23} \implies \overline{z_4} z_1 + \overline{z_5} z_2 + \overline{z_6} z_3 = b_{23}$ or upon substitution $\overline{b_{13}} b_{12} + \overline{z_5} z_2 + \overline{z_6} z_3 = b_{23}$

As one can easily perceive, this is a hideous mess, and will only grow more hideous with larger dimensions. My question is, is there some clever trick or elegant theorem I could be employing? If so, would you mind providing some hints? The only clever trick I could come up with was the rotation with $U$.

2. Nov 7, 2015

### micromass

Staff Emeritus
Consider the inner product $\mathbf{x}B\mathbf{y}^T$.

3. Nov 7, 2015

### Bashyboy

Where x and y are just any vectors in $\mathbb{C}^n$? Why are you taking the transpose of $y$? The inner-product I am using is $\langle x,y \rangle := y^* x$.

4. Nov 8, 2015

### Bashyboy

If anyone else has any suggestions, I would appreciate them being shared.