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Problem 1873

National Olympiad, first round

Algebra Difficulty 6.4 Prove it Romanian Mathematical Olympiad - District Round · Romania

Suppose $A$ is a singular matrix of order $n$ with complex entries, all of which having absolute value equal to $1$ .

a. Let $n = 3$ . Show that two lines or two columns of the matrix $A$ are proportional.

b. Find, with proof, if the above claim holds for $n = 4$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

a. By suitable multiplication on each row and column, the matrix $A$ can be written as $k \begin{pmatrix} 1 & 1 & 1 \\ 1 & a & b \\ 1 & c & d \end{pmatrix}$ , where $a, b, c, d, k$ are complex numbers of absolute value $1$ .

The relation $\det(A) = 0$ gives $(a-1)(d-1) = (b-1)(c-1)$ . Take the complex conjugates to get $\overline{ad}(a-1)(d-1) = \overline{bc}(b-1)(c-1)$ .

If $(a-1)(d-1) = 0$ , then $(b-1)(c-1) = 0$ and two rows — or columns — are equal to $(1 \ 1 \ 1)$ and the claim is reached.

Suppose $(a-1)(d-1) = (b-1)(c-1) \neq 0$ . Then $\overline{ad} = \overline{bc}$ and $ad = bc$ . From $(a-1)(d-1) = (b-1)(c-1)$ we get $a+d = b+c$ , hence $\{a,d\} = \{b,c\}$ or $a=b=c=d$ . It follows that the bottom two rows or the rightmost two columns are equal, hence the claim.

b. Notice that $A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & i & -i \\ 1 & -i & 1 & -i \\ 1 & i & i & -1 \end{pmatrix}$ is a singular matrix and any two rows or columns are not proportional. The above claim fails for $n = 4$ .

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