I am thinking on the following problem and have some results but I need a little help. Can some one give me a little hint?
Suppose $a_{ij}= \cos (i +j )$ in the matrix $A$ find $\operatorname{rank}(A)$.
Here is my attempt:
We know that:
$${\cos(\theta)} = {{e^{i \theta} + e^ {-i \theta} }\over{2}}$$
so we can write :
$$
A={1 \over 2} \begin{pmatrix}
e^{2i} + e^ {-2i} & e^{3i} + e^ {-3i} & \cdots & e^{i(2+n-1)} + e^ {-i(2+n-1)} \\
e^{3i} + e^ {-3i} & e^{4i} + e^ {-4i} & \cdots & e^{i(3+n-1)} + e^ {-i(3+n-1)} \\
\vdots & \vdots & \ddots & \vdots \\
e^{i(2+n-1)} + e^ {-i(2+n-1)} & e^{i(3+n-1)}+ e^ {-i(3+n-1)} & \cdots & e^{i(2+2n-2)}+ e^ {-i(2+2n-2)} \\
\end{pmatrix}
$$
Now we try to show that $\det(A)=0$ , which show us $\operatorname{rank}(A) < n$. First we show that for odd $n$ , suppose $n=2k+1$, we use middle column of matrix , to change first and last column of the matrix to compute determinant, the middle column will be $(k+1)^\mathrm{th}$ column so suppose :
$$
C_{k+1}={1 \over 2} \begin{pmatrix}
e^{i(2+k)} + e^ {-i(2+k)} \\
e^{i(2+k+1)} + e^ {-i(2+k+1)} \\
\vdots \\
e^{i(2+k+n-1)} + e^ {-i(2+k+n-1)} \\
\end{pmatrix}
$$
Now if we compute $C_{k+1} \times (-e^{-k})+C_1$ we have :
$$
{1 \over 2} \begin{pmatrix}
e^ {-2i} – e^{-i(2+2k)} \\
e^ {-3i} – e^{-i(2+2k+1)} \\
\vdots \\
e^ {-i(2+n-1)} – e^{-i(2+2k+n-1)} \\
\end{pmatrix}
$$
and now if we compute $C_{k+1} \times (-e^k) +C_n$ we have :
$$
{1 \over 2} \begin{pmatrix}
e^{-i(2+2k)} – e^{-2i} \\
e^{-i(2+2k+1)} – e^{-3i} \\
\vdots \\
e^{-i(2+2k+n-1)} – e^{-i(2+n-1)} \\
\end{pmatrix}
$$
So it is clear if we add $C_{k+1} \times (-e^k) +C_n$ to the $C_1$, $A$ will be $0$ which mean $\det(A)=0$.
For the case $n$ even, I was able to cease a column with $n-1$ zeroes but I could not show that $\det(A)=0$, but I used mathematica and verified for $n \in \{ 3,4,5 \}$ determinant is $0$, also at first I thought $\operatorname{rank}(A)=n$ but now I know it is not and get a little bit confused, can some one give me a little hint?
Thanks.
Best Answer
Notice that your first matrix shows that $A$ is the sum of two matrices (split each entry at the $+$ sign) that each have rank only $1$! This strongly suggests that $A$ has rank $2$.
I'm sure there's a way to prove this from your representation. However, I suggest using the identity $$ \cos(i+j) = \cos i \cos j - \sin i \sin j $$ as a way of working out, for example, the column space of $A$.