Matrix gymnastics

Algebra Level 4

For integers $1 \le k \le n$ let $S = \left \{ M_k \right \}$ be a set of at least three $k \times k$ matrices. If $\text{det} \left( M_k \right)$ denotes the determinant of $M_k$ and $\text{det} \left( M_k \right) = D$ for all $k$ , let $\displaystyle a \text{Ж} S = \sum_k \text{det} \left( a M_k \right)$ for $a \in \mathbb{C} \text{ \ } \left \{ 0 \right \}$ .

If $a \text{Ж} S$ is invariant under $\text{Ж}$ for all $a$ , find $D$ .

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\text{Not enough information}

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1 solution

Akeel Howell
Jul 9, 2018

We know that $\displaystyle a \text{Ж} S = \sum_k \text{det} \left( a M_k \right)$ .

Since $\text{det} \left( M_k \right) = D$ for all $k$ and $M_k$ is a $k \times k$ matrix, we can observe that $\text{det} \left( a M_k \right) = D \cdot a^k.$

Thus, $\displaystyle \sum_k \text{det} \left( a M_k \right) = D \sum_{k \le n} a^k.$

Therefore, if $a \text{Ж} S$ is invariant under $\text{Ж}$ for all nonzero $a \in \mathbb{C}$ , then we see that $\displaystyle a \text{Ж} S = D \sum_{k \le n} a^k = r$ , where $r \in \mathbb{C}.$

The set $M_k$ contains at least three matrices so $D \left( a + a^2 + a^3 \right) = \cdots = D \left( a + a^2 + \cdots + a^{n-1} + a^n \right) = r.$

The only solution to the system of equations above is $r = 0$ , which would require that either $D$ or $a$ be equal to zero. Since $a$ could be any nonzero complex number, we must have $D = 0.$

Matrix gymnastics

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