How to show RANK of matrix $xy^* \in \Bbb C^{n\times n}$, where vectors $x, y \in\Bbb C^n$ are arbitrary vectors.

Question

I have a class in numerical mathematics, and I received several tasks I should answer. I am not a mathematician, and this is a bit out of my mind range, and I would be grateful for answers. Question is as follows:

Let $x, y \in \Bbb{C}^n$ be arbitrary vectors. Show what can be the rank of matrix $xy^* \in \Bbb{C}^{n×n}$. If rank of matrix $xy^*$ is equal to zero, describe (in most general terms) what should vectors $x$ and $y$ look like.

Edit:

From what I know, multiplying vector $x$ with conjugate transpose of vector $y$, both of which are dimension $n$, will result in square matrix of dimension $n\times n$. Maximum number of diagonal elements that can be independent are $n$, so rank of that matrix can vary from zero to n. (Please correct me if I am completely wrong in this) And only way the rank can be zero is if all diagonal elements are zero, so $x$ and $y$ should be null vectors. These are my thoughts, but I am not sufficiently familiar with linear algebra to be sure. Please help

Answer 1

It is a little unclear to me exactly what is denoted by $y^\ast$. This being said, I assume until further notice that

$0 \ne x, y \in \Bbb C^n \tag 1$

are column vectors, and that the notation $y^\ast$ denotes a transposed form of $y$, so that $y^\ast$ is a $1 \times n$ row vector. For example, we might have

$y^\ast = y^T, \tag 2$

the transpose of $y$, or perhaps

$y^\ast = \overline{y^T}, \tag 3$

the conjugate transpose or Hermitian adjoint of $y$. It really doesn't matter which of these options we interpret $y^\ast$ to be, the argument is the essentially the same in either case. But we do need to affirm that $y^\ast$ is a $1 \times n$ row vector, for we are given that

$xy^\ast \in \Bbb C^{n \times n}, \tag 4$

that is, $xy^\ast$ is a square matrix in size $n$.

We may find the range of $xy^\ast$ as follows: for any column vector

$z \in \Bbb C^n \tag 5$

we have

$(xy^\ast)z = x(y^\ast z); \tag 6$

now

$y^\ast z \in \Bbb C, \tag 7$

whence

$x(y^\ast z) = (y^\ast z)x \tag 8$

is a scalar multiple of the vector $x$; that is,

$x(y^\ast z) \in \; \text{span}\{x\} \subset \Bbb C^n, \tag 9$

the one-dimensional subspace generated by $x$. By re-scaling $z$ if necessary, we can force $y^\ast z$ to take any value in $\Bbb C$, provided $y \ne 0$; thus in fact

$\text{range}(y^\ast z) = \text{span}\{x\}, \tag{10}$

which is one-dimensional unless $x = 0$ or $y = 0$, in which case it is simply $\{0\}$.

Note Added Saturday 18 July 2020 5:20 PM PST: On the rank of $xy^\ast$: Upon reading the comments made to this answer by our OP Geux, I would like to point out that the above argument effectively solves the rank question as well; indeed, for any square matrix $A$,

$\text{rank}(A) = \dim \text{range}(A); \tag{11}$

this follows from the observation that each column of $A$ maps to a vector in $\text{range}(A)$ under the action of right multiplication by a column vector of the form

$e_i = (\delta_{ij}), \; 1 \le j \le n, \tag{12}$

that is, $e_i$ is an $n \times 1$ matrix whose $i$-th component is $1$ and which has $0$ for all other entries. If we write $A$ in columnar form,

$A = [A_1 \; A_2 \; \ldots \; A_n], \tag{13}$

then it is easy to see that

$Ae_i = A_i; \tag{14}$

as $1 \to i \to n$, every $A_i$ is thus mapped into $\text{range}(A)$, and it is clear that arbitrary column vectors map to linear combinations of the columns of $A$, which shows that $\text{range}(A)$ is in fact the span of the column space of $A$, that is, of $\{A_1, A_2, \ldots, A_n \}$; thus (11) binds.

We now apply these considerations to $xy^\ast$ and immediately deduce via (10) that

$\text{rank}(xy^\ast) = \dim \text{range}(xy^\ast) = \dim \text{span} \{x\} = 1, \tag{15}$

as per request. End of Note.

Answer 2

Let $A = x y^*$ and $v\in \mathbb{C}^n$. Try to find a different representation of $Av = (xy^*) v$ using the scalar product. Then it should be clear