Proving that $\operatorname{rank}(\mathbf{A}) = 1$ if and only if $\mathbf{A}$ is the product of a non-zero column vector and a non-zero row vector

Question

$\newcommand{\mn}[1]{\mathbf{#1}}$ $\newcommand{\vn}[1]{\mathbf{#1}}$ $\newcommand{\rk}[1]{\operatorname{rank}(#1)}$ $\newcommand{\t}{\mathrm{T}}$

As a beginner in linear algebra, I encountered an interesting problem in my textbook. I would like to have my solution verified.

Problem

For a matrix $\mathbf{A}$, prove that $$\operatorname{rank}(\mathbf{A}) = 1$$ if and only if $$\mathbf{A} = \mathbf{a}\mathbf{b}^{\mathrm{T}}$$ for some non-zero column vector $\mathbf{a}$ and non-zero row vector $\mathbf{b}^{\mathrm{T}}$.

My proof

Let:

• $m \times n$ be the dimensions of $\mn A$;

• $\mn F$ be the $m \times n$ canonical matrix with rank $1$;

• $\vn i_k$ be the $k$-dimensional unit vector $(1, 0, \dotsc, 0)^\t$.

Hence, $\mn F = \vn i_m \vn i_n^\t$.

The only if direction

It is known that $\rk{\mn A} = 1$. Therefore, there exists invertible matrices $\mn P_{(m \times m)}$ and $\mn Q_{(n \times n)}$, such that $$ \mn P \mn A \mn Q = \mn F. $$ Thus, $$ \mn A = \mn P^{-1} \mn F \mn Q^{-1} = \mn P^{-1} \vn i_m \vn i_n^\t \mn Q^{-1}.$$ Let $\vn a = \mn P^{-1} \vn i_m$ and $\vn b^\t = \vn i_n^\t \mn Q^{-1}$, so that $\mn A = \vn a \vn b^\t$.

Because $\det \mn P^{-1} \neq 0$, no column of $\mn P^{-1}$ consists exclusively of zeros. $\vn a$ is just the first column of $\mn P^{-1}$, so $\vn a \neq \vn 0$. Similarly, $\vn b^\t \neq \vn 0$. Therefore, $\mn A = \vn a \vn b^\t$ for non-zero vectors $\vn a$ and $\vn b^\t$. $\blacksquare$

The if direction

It is known that $\mn A = \mn a \mn b^\t$ for non-zero vectors $\mn a$ and $\mn b$. Let $a_i$ be one non-zero element of $\vn a$. By swapping $a_1$ and $a_i$ (possibly $i = 1$), we transform $\vn a$ into $\vn a'$. Then, build the block matrix $$ \mn C = \begin{bmatrix} \vn a' & \begin{matrix} \vn 0 \\ \mn I_{m-1} \end{matrix} \end{bmatrix} $$ and swap its first row with its $i$th row to obtain $\mn D$, so that the first column of $\mn D$ is equal to $\vn a$. Hence, $$ \det \mn D = \pm \det \mn C = \pm a_i \neq 0. $$ Let $\mn P = \mn D^{-1}$, so $$ \vn a = \mn D \vn i_m = \mn P^{-1} \vn i_m. $$ By the same process, we find an invertible matrix $\mn Q$ such that $$ \vn b^\t = \vn i_n^\t \mn Q^{-1}. $$ Therefore, $$ \mn A = \vn a \vn b^\t = \mn P^{-1} \vn i_m \vn i_n^\t \mn Q^{-1} = \mn P^{-1} \mn F \mn Q^{-1}, $$ so $$ \mn P \mn A \mn Q = \mn F, $$ which shows that $\rk{\mn A} = 1$. $\blacksquare$

Concerns

• Is my proof correct? Did I misuse any theorem?

• Is my proof clear and concise? Does any step warrant further explanation? Can the notation be improved?

• Are there simpler beginner-accessible solutions?

• What else can be improved?

Answer 1

Alternative proof :

Denote by $L_1, ..., L_n$ the rows of $A$.

Then $\mathrm{rank}(A)=1$ means that all the rows are proportional to a a nonzero one, i.e. that there exists $i \in \lbrace 1, ..., n \rbrace$ such that for every $k \in \lbrace 1, ..., n \rbrace$, there exists $\alpha_k$ such that $L_k=\alpha_k L_i$.

Now let $a=L_i$ and $b=(\alpha_1, ..., \alpha_n)$ and you are done.

Answer 2

I would suggest that you focus more on the definitions than trying to compute and work with factorizations.

For example, in the if direction, suppose that $A=ab^T$ and let $v$ be any vector. Then $Av=a(b^Tv)$. Since $b^Tv$ is just a scalar, we find that $Av$ is a scalar multiple of $a$. Hence, $Av\in\operatorname{Span}(a)$. From the definition, this implies that the rank of $A$ is at most one.

Now, if you let $v=b$, then $b^Tb=\|b\|$, which is nonzero if and only if $b$ is nonzero. Therefore, the image of $A$ is not only the zero vector, i.e., has dimension at least one.