# When every term of the determinant is zero

The determinant of $A$ is defined as $$\det(A) = \sum \pm a_{1,i_1} a_{2,i_2} \cdots a_{n,i_n}$$ Suppose $A$ is a real matrix such that every term in the above sum is zero. Is it true that $A$ has a zero row or zero column?

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Counterexamples have been given, yet since the determinant of the matrix is zero we get a singular matrix, so we can find a similar matrix (say, by row reduction) with a row of zeros. – DonAntonio Oct 23 '12 at 10:42

Counterexamples: $A=\begin{pmatrix}1&0&1\\0&1&0\\0&1&0\end{pmatrix}$ or $B=\begin{pmatrix}1&1&1\\0&1&0\\0&1&0\end{pmatrix}$.

If you look at matrix $B$:

• every product of the above form contains an element from the first column
• if it contains $0$, we are fine
• if it contains $1$ from the position $(1,1)$ then the rest of the product must be some of the summands in the subdeterminant $\begin{vmatrix}1&0\\1&0\end{vmatrix}$.
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More generally, an $n \times n$ matrix with $k$ columns, each of which has $n - k + 1$ zeroes at prescribed positions (e.g. the first and third columns in $A$ and $B$ above for $k = 2$), will have every term zero. – Lord_Farin Oct 23 '12 at 8:21
If I understand this correctly, the proof would probably use the characterization of rank using non-vanishing minors. Perhaps you could post your comment as an answer (with adding some more details)? – Martin Sleziak Oct 23 '12 at 8:30

Building upon the answer of @MartinSleziak, I saw a more general pattern emerge. I'm not sure yet if my observed method will yield a necessary and sufficient condition, but at least it is sufficient.

So let $A$ be an $n \times n$ matrix, such that there are $k$ columns $a_i, i \in I$, with the property that:

$\quad \forall j \in J: a_{i,j} = 0$

(with $I,J \subseteq \{1, \ldots, n\}, \#(I) = k, \#(J) = n-k+1$).

Now let $\sigma \in S_n$, so that $\displaystyle \prod_{l = 1}^n a_{l, \sigma(l)}$ is a term of $\det A$ (disregarding sign as it will become zero anyway).

Consider the set $\{\sigma(i): i \in I\} \subseteq \{1, \ldots, n\}$. It has $k$ elements, so since $J$ has $n-k+1$ elements, there is necessarily an $i \in I$ such that $\sigma(i) \in J$.

By the above assumption on $J$, it follows that $a_{i, \sigma(i)} = 0$, and we conclude:

$\quad \displaystyle \prod_{l = 1}^n a_{l, \sigma(l)} = 0$

Since $\sigma$ was arbitrary, all terms of $\det A$ are zero.

Swapping $k$ for $n-k+1$ proves the same argument for rows. Probably, there is a connection with non-vanishing minors, as was already pointed out.

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I've checked manually that it is impossible to have all zero terms without satisfying the condition provided for $n < 5$. – Lord_Farin Oct 23 '12 at 9:16