Shortest and most elementary proof that the product of an $n$-column and an $n$-row has determinant $0$

Question

Let $\bf u$ be any column vector and $\bf v$ be any row vector, each with $n \geq 2$ arbitrary entries from a field. Then it is well known that

${\bf u} {\bf v}$ is an $n \times n$ matrix such that $\det({\bf uv})=0$.

I am curious to know the $\bf shortest$ and $\bf most~elementary$ proof of this result, say understandable by a (good) high school student. I have one in mind, but to make this interesting, perhaps I should let you present your version first?

UPDATE: Someone already presented (below) the same proof I had in mind. But let's see if there is a simpler proof; finding one is the main motivation here.

Answer 1

If $$u=(a,b)^T\text{ and }v=(c,d),$$

Taking the product $u.v$ we have

$$ \begin{pmatrix} a\cdot c & a\cdot d \\ b\cdot c & b\cdot d \end{pmatrix}.$$

But this shows the first and the second lines are proportional:

$$a\cdot \begin{pmatrix} c & d \end{pmatrix}$$ $$b\cdot \begin{pmatrix} c & d \end{pmatrix}$$

Thus $\det(u\cdot v)=0$.

In the general case, $v$ represents the rows of the product matrix and $u$ the scaling factors, so all the lines are proportional and $\det(u\cdot v)=0$.

Answer 2

Here is an elementary proof which can be taught to high school students. It turns $u$ and $v$ into $n\times n$ matrices by adding zero rows and columns: $$uv= \begin{bmatrix} u\mid 0 \end{bmatrix}_{n\times n} \begin{bmatrix} v\\ \hline 0 \end{bmatrix}_{n\times n} $$ Thus $\det(uv)=\det(AB)=\det(A)\det(B)=0$, where $A$ and $B$ are the above $n\times n$ matrices obtained from $u$ and $v$.

PS The multiplicative property of the determinant function for square matrices is definitely the easiest property of determinant that students can remember. Year ago, I used this trick in a presentation for students with limited knowledge of linear algebra to avoid the notion of rank of matrices and I think it served this purpose very well. However, the only downside is that it is based on a "fact" which is not usually proved for high school students, which is precisely the multiplicative property of the determinant.

Answer 3

I am not sure what the average "good" high school student understands, but from what I learned about matrices in high school this is the proof that would have been simplest to me:

Let u=$[u_1, u_2, ... u_n]^T$ and v=$[v_1, v_2, ..., v_n]$. Then we can write $$\mathrm{uv}= \begin{bmatrix} u_1v_1 & u_1v_2 & \cdots & u_1v_n\\ u_2v_1 & u_2v_2 & \cdots & u_2v_n\\ \vdots & \vdots & \ddots & \vdots\\ u_nv_1 & u_nv_2 & \cdots & u_nv_n \end{bmatrix}.$$ Now add $\frac{u_1}{u_2}$ times the second row to the first row. This doesn't change the value of the determinant (this is where a high school student may get lost depending on how well he or she knows the definition of determinant), and we get

$$\det(\mathrm{uv})= \det\left(\begin{bmatrix} u_1v_1 & u_1v_2 & \cdots & u_1v_n\\ u_2v_1 & u_2v_2 & \cdots & u_2v_n\\ \vdots & \vdots & \ddots & \vdots\\ u_nv_1 & u_nv_2 & \cdots & u_nv_n \end{bmatrix}\right)=\det\left(\begin{bmatrix} 0 & 0 & \cdots & 0\\ u_2v_1 & u_2v_2 & \cdots & u_2v_n\\ \vdots & \vdots & \ddots & \vdots\\ u_nv_1 & u_nv_2 & \cdots & u_nv_n \end{bmatrix}\right)$$ which it is easy to see is equal to $0$.

Answer 4

Take any non-zero vector $w$ orthogonal to $v$. Then we have

$$ (u\cdot v)\cdot w = u\cdot (v\cdot w) = u\cdot 0 = 0,$$

so $u\cdot v$ must be singular.

Answer 5

Each row of $uv$ is a multiple of $v$. If the first row is all zeros, the determinant is zero. If not, then the second row is a multiple of the first. The determinant is unchanged if we subtract that multiple of the first row from the second row. The second row is now all zeros and so the determinant is zero.

Answer 6

By associativity of the matrix product one has $$\bigl({\bf u}{\bf v}\bigr){\bf x}={\bf u}\bigl({\bf v}\>{\bf x}\bigr)=\phi_{\bf v}({\bf x})\>{\bf u}$$ for all column vectors ${\bf x}$. It follows that the rank of ${\bf u}\>{\bf v}$ is $<2$, whence ${\rm det}({\bf u}\>{\bf v})=0$.

Answer 7

We will use two elementary thing.

$ \quad (1)$ Let $A$ an $m \times n$ matrix. The rank of $A$ is the smallest integer $k$ such that $A$ can be factored as $A=CR$, where $C$ is an $m \times k$ matrix and $R$ is an $k \times n$ matrix. Note that $A=CR$ is the rank decomposition of $A$.

$ \quad (2)$ Let $A$ an $n \times n$ matrix. Then $\operatorname{rank}(A)=n$ if and only if $\det(A) \neq 0$. This is because of the fundamental theorem on inverses (Theorem 2.3.4 here).

Because of $(1)$ we know that $\mathbf{uv}$ has rank $1$, since $\mathbf u$ is an $n \times 1$ and $\mathbf v$ is an $1 \times n$ matrix. Since $n \geq 2$, we know that $1 = \operatorname{rank}(\mathbf{uv}) \neq n$, so by using $(2)$ we get that $\det(\mathbf{uv})=0$.

In one sentence: Because $\mathbf{uv}$ is not a full rank matrix $\det(\mathbf{uv})=0$.

Answer 8

This question can be easily answered by recalling determinant properties. Specifically: $det(A_1,A_2,\dots,cA_j,\dots,A_n)$ = $c$$(det(A_1,A_2,\dots,A_j,\dots,A_n))$

Therefore if $u = \begin{bmatrix}u_1\\ u_2\\ \vdots \\ u_n \end{bmatrix}$ and $v = \begin{bmatrix} v_1,& v_2,& \dots,& v_n\end{bmatrix}$

Then:

$uv$ = $\begin{pmatrix} u_1v_1 & u_1v_2 & \cdots & u_1v_2 \\ u_2v_1 & u_2v_2 & \cdots & u_2v_n \\ \vdots & \vdots & \ddots & \vdots \\ u_nv_1 & u_nv_2 & \cdots & u_nv_n \end{pmatrix} $

Then, by the property mentioned, $det(uv)$ = $v_1v_2$$\cdots v_n$ $det$ $\begin{pmatrix} u_1 & u_1 & \cdots & u_1 \\ u_2 & u_2 & \cdots & u_2 \\ \vdots & \vdots & \ddots & \vdots \\ u_n & u_n & \cdots & u_n \end{pmatrix} $

Clearly this determinant is zero, therefore $det(uv)$ = $0$.

Answer 9

Let $c_{ij}= a_i \cdot b_j$

The terms in the expansion of the determinant are

$$ c_{i \sigma(1)} c_{2 \sigma(2)}\ldots c_{n \sigma (n)} = a_1 b_{\sigma(1)} \ldots a_n b_{\sigma(n)} = a_1\ldots a_n b_1 \ldots b_n$$

so all the terms are equal and the signs $\sum \text{sign}(\sigma)$ sum up to $0$.