Projector Matrices: Conditions for A=uv*

[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 ?

 

[Mark44]

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?

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

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 ?

 

[jbunniii]

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?

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^*)

 

[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.

 

[Mark44]

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).

 

[math8]

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).

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.

 

[Mark44]

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 &amp; 1 \\ 0 &amp; 0 \end{array}\right]\left[\begin{array}{c c} 0 &amp; 1 \\ 0 &amp; 0 \end{array}\right] = \left[\begin{array}{c c} 0 &amp; 0 \\ 0 &amp; 0 \end{array}\right]

 

[math8]

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 &amp; 1 \\ 0 &amp; 0 \end{array}\right]\left[\begin{array}{c c} 0 &amp; 1 \\ 0 &amp; 0 \end{array}\right] = \left[\begin{array}{c c} 0 &amp; 0 \\ 0 &amp; 0 \end{array}\right]

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

 

[Mark44]

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