Find the rank of the following matrix

Question

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.

Answer 1

Let $$ A=\begin{pmatrix} \cos(2) & \cos(3) & \cdots & \cos(2+n-1) \\ \cos(3) & \cos(4) & \cdots & \cos(3+n-1) \\ \vdots & \vdots & \ddots & \vdots \\ \cos(2+n-1) & \cos(2+n) & \cdots & \cos(2+n-1+n-1) \\ \end{pmatrix} $$

Suppose $V_i$ is the $i$th row of the $A$ , it is clear: $$ V_i=\begin{pmatrix} \cos(2+i-1) & \cos(2+i-1+1) & \cdots & \cos(2+i-1+n-1) \\ \end{pmatrix} $$

Using $\cos(i+j)=\cos(i)\cos(j)-\sin(i)\sin(j)$ and $\cos(i)-\cos(j)=-2\sin({(i-j) \over 2})\sin({(i+j) \over 2})$ we show that :

$$ V_i-{{V_i-V_{i-2}} \over 2}-V_{i-1}\cos(1)=0 $$

With computin ${{V_i-V_{i-2}} \over 2}$ we have : $$ {{V_i-V_{i-2}} \over 2}=\begin{pmatrix} -\sin(1)\sin(2+i-2) & -\sin(1)\sin(2+i-1) & \cdots & -\sin(1)\sin(2+i-2+n-1) \\ \end{pmatrix} $$ No using the identities we can write : $$ V_i=\begin{pmatrix} \cos(1)\cos(2+i-2)-\sin(1)\sin(2+i-2) & \cos(1)\cos(2+i-1)-\sin(1)\sin(2+i-1) & \cdots & \cos(1)\cos(2+i-2+n-1)-\sin(1)\sin(2+i-2+n-1) \\ \end{pmatrix} $$ Thus $V_i-{{V_i-V_{i-2}} \over 2}$ equals to : $$ V_i=\begin{pmatrix} \cos(1)\cos(2+i-2) & \cos(1)\cos(2+i-1) & \cdots & \cos(1)\cos(2+i-2+n-1) \\ \end{pmatrix} $$ Now it is easy to see that : $$ V_i-{{V_i-V_{i-2}} \over 2}-V_{i-1}\cos(1)=0 $$

It is show that all the $V_i, 3\le i\le n$ genrated by first two row thus $rank(A)=2$.

Answer 2

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 $A$.