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

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.

• The matrix has rank $1<n$ so is singular, and therefore has determinant $0$. This can be generalized for a sum of $n-1$ rank one matrices. – Yuval Filmus Sep 17 '14 at 17:43
• A good answer to this question highly depends on the used definition of determinant, on the set of "known properties" , on allowed operations leaving the determinant unchanged, etc. – Christian Blatter Sep 19 '14 at 18:02
• It also depends on the definition of a good high school student. – Zavosh Sep 19 '14 at 18:37
• I really would like to know how determinant is defined in your high school curriculum. In my city, determinant is defined only for matrices up to 3x3 (and by the rule of Sarrus) in high schools. Having to deal with determinants of general square matrices is news to me. By the way, while high school students in my city are expected to know how to perform an outer product from the definition of matrix multiplication, most of them don't really know how because they usually only multiply wide matrices to tall ones, but not vice versa. – user1551 Sep 20 '14 at 8:54

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.

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

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 $0$.

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.

• This proof is very short, but in my opinion, the concepts of orthogonality and singularity are not the most elementary. – E W H Lee Sep 20 '14 at 3:59
• You haven't yet mentioned what properties of determinants the students have learned yet. Do they know for example, linearity in each row/column, multiplicativity, invariant under transpose, or Cramer's rule? Do they know any use for the determinant at all? – Zavosh Sep 20 '14 at 8:51
• As I said, I have one proof in mind (that someone has already presented), and I am interested in seeing if there is a simpler and more elementary proof. So for a more elementary proof, the less required (such as all the properties/rules you mentioned), the better. – E W H Lee Sep 20 '14 at 17:54

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 $det(uv)$ = $0$.

• What I like about this answer is that it is short and that it uses a defining property of determinants (linearity) rather than a derived property. Best one so far imo. – DanielV Sep 20 '14 at 14:59
• $$\det(uv) = \det\begin{bmatrix}u_1 v_1 & \dots & u_1v_n \\ \vdots & & \vdots \\ u_n v_1 & \dots & u_n v_n \end{bmatrix} = v_1\dots v_n \det\begin{bmatrix}u_1 & \dots & u_1 \\ \vdots & & \vdots \\ u_n & \dots & u_n \end{bmatrix} = 0$$ Looks like a one liner to me. Of course, it depends on how much explanation you want to add. – DanielV Sep 20 '14 at 18:42
• @DanielV: Upon looking again, I agree with you. – E W H Lee Sep 22 '14 at 18:04

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.

• Now that I read it, this may be too close to J. Gardiner's answer, but perhaps this statement is simpler. – robjohn Sep 20 '14 at 7:05

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 $0$.

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 ${\rm det}({\bf u}\>{\bf v})=0$.

• This is short, but in my opinion, the concepts of rank is not the most elementary. Do you agree? – E W H Lee Sep 20 '14 at 14:19

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 $\det(\mathbf{uv})=0$.