Conjecture: If matrix $M$ has entries (left to right, then top to botom) $\sin 1,\sin 2,\sin 3,\dots,\sin (n^2)$, where $n\ge 3$, then $\det M = 0$.

Question

According to my calculator,

$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 \\ \sin 4 & \sin 5 & \sin 6 \\ \sin 7 & \sin 8 & \sin 9\end{bmatrix}=0$

$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 & \sin 4 \\ \sin 5 & \sin 6 & \sin 7 & \sin 8 \\ \sin 9 & \sin 10 & \sin 11 & \sin 12 \\ \sin 13 & \sin 14 & \sin 15 & \sin 16 \end{bmatrix}=0$

$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 & \sin 4 & \sin 5 \\ \sin 6 & \sin 7 & \sin 8 & \sin 9 & \sin 10 \\ \sin 11 & \sin 12 & \sin 13 & \sin 14 & \sin 15 \\ \sin 16 & \sin 17 & \sin 18 & \sin 19 & \sin 20 \\ \sin 21 & \sin 22 & \sin 23 & \sin 24 & \sin 25\end{bmatrix}=0$

I conjecture that, for $n\ge 3$,

$\det \begin{bmatrix} \sin 1 & \sin 2 & \sin 3 & \dots & \sin n \\ \sin (n+1) & \sin (n+2) & \sin (n+3) & \dots & \sin (2n) \\ \sin (2n+1) & \sin (2n+2) & \sin (2n+3) & \dots & \sin(3n) \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \sin ((n-1)n+1) & \sin ((n-1)n+2) & \sin ((n-1)n+3) & \dots & \sin (n^2) \end{bmatrix}=0$

Is my conjecture true?

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I have only been able to prove the case with $n=3$.

$\sin 5 + \sin 7 + [\sin 1 + \sin (-1)] + [\sin 3 + \sin (-3)]$
$=\sin 5 + \sin 7 + [\sin 1 + \sin (-1)] + [\sin 3 + \sin (-3)]$

Rearrange each side:

$[\sin (-3) + \sin 5] + [\sin 1 + \sin 3] + [\sin (-1) + \sin 7]$
$=[\sin (-1) + \sin 3] + [\sin (-3) + \sin 7] + [\sin 1 + \sin 5]$

Use the product-to-sum formulas:

$(\sin 1)(\cos 4) + (\sin 2)(\cos 1) + (\sin 3)(\cos 4)$
$=(\sin 1)(\cos 2) + (\sin 2)(\cos 5) + (\sin 3)(\cos 2)$

Subtract $(\sin 1)(\cos 14)+(\sin 2)(\cos 13)+(\sin 3)(\cos 12)$ from both sides:

$(\sin 1)(\cos 4 - \cos 14) + (\sin 2)(\cos 1 - \cos 13) + (\sin 3)(\cos 4 - \cos 12)$
$=(\sin 1)(\cos 2 - \cos 14) + (\sin 2)(\cos 5 - \cos 13) + (\sin 3)(\cos 2 - \cos 12)$

Use the product-to-sum formulas again:

$(\sin 1)(\sin 5)(\sin 9)+(\sin 2)(\sin 6)(\sin 7)+(\sin 3)(\sin 4)(\sin 8)$ $=(\sin 1)(\sin 6)(\sin 8)+(\sin 2)(\sin 4)(\sin 9)+(\sin 3)(\sin 5)(\sin 7)$

which is equivalent to

$\det\begin{bmatrix}\sin 1 & \sin 2 & \sin 3 \\ \sin 4 & \sin 5 & \sin 6 \\ \sin 7 & \sin 8 & \sin 9\end{bmatrix}=0$

Answer 1

Every one of the functions $\sin(x), \ldots, \sin(x+n)$ solves the ODE $y'' + y = 0$. But the solution space of a linear (homogeneous) second order ODE is only two-dimensional, which tells you that these sine functions must be linearly dependent if you take at least $3$ of them. That is, there exist $\lambda_1,\ldots,\lambda_n \in \mathbb{C}$ not all $0$ such that $$\sum\limits_{j = 1}^n \lambda_j \sin(x+j) = 0$$ for all $x$. Now plug in $x = 0$, $x = n$ and so on. This shows that your columns are linearly dependent for $n \geq 3$ and hence the determinant must be $0$.

Answer 2

Hint

Use the formula $\sin(a+2)= \sin a \cos 2 + \cos a \sin 2$ in the last column, the formula $\sin(a+1) = \sin a \cos 1 + \cos a \sin 1$ in the penultimate column and develop the determinant. You’ll get a linear combination of 4 determinants all having two identical columns if $n \ge 3$. Hence the given determinant vanishes.

Answer 3

By applying the Sum-to-product formulas we get that

$\sin\!\big(2n+i\big)+\sin i=2\sin\!\big(n+i\big)\cos n$

for any $\,i\in\big\{1,2,\cdots,n\big\}\,.$

Consequently ,

$\sin\!\big(2n+i\big)=\color{red}{-1}\!\cdot\!\sin i+\color{red}{2\cos n}\!\cdot\!\sin\!\big(n+i\big)$

for any $\,i\in\big\{1,2,\cdots,n\big\}\,.$

It means that the third row of the matrix $M$ is a linear combination of the first two rows.

So the rows of the matrix $M$ form a linearly dependent set, hence the determinant of $M$ is equal to $\,0\,.$

Answer 4

No computation is necessary. The entries of $M$ are given by \begin{eqnarray*} M(k,l) = \sin(1+nk+\ell) = \sin(1+nk)\cos\ell + \cos(1+nk)\sin\ell. \end{eqnarray*} This formula shows that $M$ is the sum of two matrices with rank $1$, so rank$(M) \le 2$. Hence $M$ is not invertible if $n \ge 3$.

Answer 5

Using complex matrices. Indeed, let $N\in \mathbb C^{n\times n}$ such that $N_{k,\ell} = e^{i(n(k-1)+\ell)}$ it is clear that the $\ell$ column of $N$ is $e^{i\ell}$ times the first column. So $N$ is rank $1$ matrix. Since $M={\mathrm{ Im}}(N)=(N-\overline{N})/2i$. You have $\mathrm{rank}(M) \le \mathrm{rank}(N) + \mathrm{rank}(\overline N)= 2$ so $\det(M)=0$ if $n\ge 3$