3
$\begingroup$

Let $u \in \mathbb{R}^m$ and $v \in \mathbb{R}^n$ and $u,v \neq 0$. Since $u$ is an $m\times1$ matrix and $v^T$ is a $1\times n$ matrix, the product $uv^{T}$ is a $m\times n$ matrix. Then:

$$\text{rank} (uv^{T}) =1$$

Just to illustrate: pick a random vector $u$ in $\mathbb{R}^4$, and $v$ in $\mathbb{R}^5$, for example: $u= (1, 2, 3, 4),$ and $v= (1, 2, 3, 4, 5)$. The rank of the matrix $m\times n$ is $1$ which can be checked by row-reducing. Chosing arbitrary vectors $u,v \neq 0$, the rank of the product matrix mxn of $uv^{T}$ is always $1$.

Why is this?

Any help would be greatly appreciated.

$\endgroup$
4
  • $\begingroup$ Rank of a matrix: number of linear independent rows or columns. So? $\endgroup$
    – Alex Silva
    Oct 24, 2016 at 9:45
  • 2
    $\begingroup$ Don't you think that $1v,2v,3v,4v$ are related ? $\endgroup$
    – user65203
    Oct 24, 2016 at 9:47
  • $\begingroup$ I think an essential question here is what definition of rank you're working with. If the rank of an $m \times n$ real matrix $A$ is the dimension of the range of $A$, i.e., $\operatorname{dim}\{Ax \mid x \in \mathbb{R}^n\}$, then K. Miller's one-line argument gets straight to the heart of the matter. If, instead, the rank of $A$ is dimension of its column space, i.e., $\operatorname{dim}\{c_1 A_1 + \cdots c_n A_n \mid c_1,\dotsc,c_n \in \mathbb{R}\}$, where $A_k \in \mathbb{R}^m$ is the $k$th column of $A$, which, therefore, is the same as the number of linearly independent columns of $A$… $\endgroup$ Oct 24, 2016 at 15:15
  • $\begingroup$ …then you can check from the definition of matrix multiplication that for any $x = (x_1,\dotsc,x_n)^T \in \mathbb{R}^n$, we have $Ax = x_1 A_1 + \dotsc + x_n A_n$, which shows that the range and column space of $A$ define the same subspace of $\mathbb{R}^m$, and hence that the two definitions of rank are equal. $\endgroup$ Oct 24, 2016 at 15:17

4 Answers 4

6
$\begingroup$

Since $v\neq 0$, the range of $uv^T$ is simply $\operatorname{span}\{u\}$, which necessarily has dimension $1$ since $u \neq 0$.

$\endgroup$
5
$\begingroup$

Write $$\vec v=\begin{pmatrix}v_1\\v_2\\\dots\\v_n\end{pmatrix}$$ where $v_i, i=1,\dots,n$ are scalars (numbers) and similarly for $\vec u$. Then, for $u\in \mathbb R^m$, $uv^T$ is indeed an $m\times n$ matrix which may be written as $$\vec{u}v^T=\begin{pmatrix}u_1\\u_2\\\dots\\u_m\end{pmatrix}(v_1,v_2,\dots, v_n)=\begin{pmatrix}v_1\begin{pmatrix}u_1\\u_2\\\dots\\u_m\end{pmatrix},\,v_2\begin{pmatrix}u_1\\u_2\\\dots\\u_m\end{pmatrix},\dots ,\, v_n\begin{pmatrix}u_1\\u_2\\\dots\\u_m\end{pmatrix}\end{pmatrix}$$ Hence any column of $uv^T$ is a multiple of the first non-zero column (the first column for which the corresponfing $v_i\neq 0$) and hence linearly dependent with it. Hence $$\text{rank }(uv^T)=\max {\{\text{rank }(u), \text{rank }(v^T)\}}=1$$ since you assumed that $u,v \neq 0$.

$\endgroup$
1
  • 1
    $\begingroup$ ... or a multiple of the first non-zero column. No guarantees that $v_1\neq0$ here. Also, note that even $\vec0$ is a multiple of said column, so you don't really need the "either-or" in there at all. Apart from some slight nit-picking, though, it's a good answer. $\endgroup$
    – Arthur
    Oct 24, 2016 at 11:08
2
$\begingroup$

The matrix product $[u]\,[v^\top]$ encodes a composition of linear maps ${\mathbb R}^n\to{\mathbb R}\to{\mathbb R}^m$. Then single ${\mathbb R}$ in the middle enforces ${\rm rank}\bigl([u]\,[v^\top]\bigr)\leq1$. On the other hand the elements of $[u]\,[v^\top]$ are the $m\cdot n$ products $u_iv_j$. By assumption on $u$ and $v$ at least one of these products is $\ne0$, and this excludes $[u]\,[v^\top]=0$.

$\endgroup$
0
$\begingroup$

Remember that if $\;AB\;$ is a well defined product of matrices, it is always true that $\;\text{rank}\,AB\le\min(\text{rank}\,A,\,\text{rank}\,B)\;$.

In your case, as matrices, $\;\text{rank}\,u=\text{rank}\,v^t=1\;$ and thus we almost get the claim ("almost" unless, as vectors, these two are orthogonal)

For (counter) example

$$(1\;0)\binom 01=(0)$$

If for you the usual vector is a column one, then $\;uv^t\;$ is an $\;n\times n\;$ matrix instead of a $\;1\times1\;$ as above...but the same applies but this time there is no possibility of "orthogonality" in the way shown above.

$\endgroup$
2
  • 4
    $\begingroup$ In the question, $u$ and $v$ are tacitly assumed to be column vectors. It follows that $uv^\top$ is an $(m\times n)$-matrix. Your counterexample does not belong here. $\endgroup$ Oct 24, 2016 at 9:49
  • $\begingroup$ @ChristianBlatter Perhaps so, but it is there for anyone who takes the other equivalente definition of "vector". In fact, I didn't even noticed what the assumption of the OP is in this case (I have the healthy though sometimes self-confusing habit not to read all the complete questions in this site since, I've been told, I've a life to live...) until I had already written my answer, so I added the rather obvious generalization with the same facts as shown in the first part about rank of product. $\endgroup$
    – DonAntonio
    Oct 24, 2016 at 9:52

Not the answer you're looking for? Browse other questions tagged .