# Projector matrices

1. May 4, 2010

### math8

Consider the matrix $$A = u v^{\ast }$$ where $$u, v \in \textbf{C}^{n}$$. Under what condition on u and v is A a projector?

A is a projector if $$A^{2}=A$$, so we have $$u v^{*} u v^{*}= u v^{\ast }$$.

Does this imply $$u v^{\ast } = I$$ ? And what exactly are the conditions on u and v that they are asking?

do we have that $$u_{i} v^{\ast }_{i}=1$$ and $$u_{i} v^{\ast }_{j}=0$$ for $$i\neq j$$ ?

2. May 4, 2010

### Staff: Mentor

Can you clarify what uv* means and how this could be a matrix?

3. May 4, 2010

### jbunniii

Note that $v^{*}u$ is a scalar, so you rearrange the terms of the product:

$$u v^* u v^* = u (v^* u) v^* = (v^* u) (u v^*)$$

4. May 4, 2010

### math8

To Mark44, u is an nx1 column vector, v* is the conjugate transpose of v, where v is an nx1 column vector. So uv* is an nxn square matrix.

5. May 4, 2010

### Staff: Mentor

OK, that makes more sense.

A2 = A <==> A(A - I) = 0. What are the conditions for the last equation? Clearly, one possibility is that A = I (which is to say that uv* = I). But there are other possibilities (plural).

6. May 4, 2010

### math8

The only other possibility that I see is that uv* is the 0 matrix. In this case, $$u_{i}v^{*}_{j} = 0$$ for all i and all j.

7. May 4, 2010

### Staff: Mentor

There's another possibility. AB = 0 does not necessarily imply that either A = 0 or B = 0. For example,
$$\left[\begin{array}{c c} 0 & 1 \\ 0 & 0 \end{array}\right]\left[\begin{array}{c c} 0 & 1 \\ 0 & 0 \end{array}\right] = \left[\begin{array}{c c} 0 & 0 \\ 0 & 0 \end{array}\right]$$

8. May 4, 2010

### math8

Oh true! But I am lost here, I am not sure what would be the conditions on u and v then.

9. May 4, 2010

### Staff: Mentor

If AB = 0 (meaning the 0 matrix), then |AB| = 0 (the number).