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

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

• And whathave you tried ? Jul 17, 2020 at 17:47
• You need to do some work, inspiration doesn't pop out of nowhere. Try a few examples in $\mathbb{R}^2$ first before going nuclear on MSE. Jul 17, 2020 at 17:48
• 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 x 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.
– Geux
Jul 17, 2020 at 18:25

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.

• Vector $y^*$ denotes conjugate transpose of vector $y$. So, if I am reading this correctly, if I expand $xy^*$ with a vector $z$ that has the same dimension as $x$ and $y$, we will obtain a scalar value that is $∈C$. If $y≠0$ then that scalar can be any value of $C$
– Geux
Jul 18, 2020 at 12:54
• Thus, we can say that range of matrix is equal to span of $x$.
– Geux
Jul 18, 2020 at 13:08
• Thank you, I was trying to find the range of matrix, while the task was to find rank of matrix, which I realized when trying to replicate the results. I made a mistake while translating from my language to english, and I am sorry for confusion.
– Geux
Jul 18, 2020 at 13:10
• @Geux: I added some material to my answer to show how the rank may also be deduced. Hope it helps. Cheers! Jul 19, 2020 at 2:04

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

• This was supposed to be a comment... Jul 17, 2020 at 19:30