I am considering the following matrix: $$ M_m = \begin{bmatrix} \cos\bigl(\tfrac{0\cdot 0}{m}\pi\bigr) & \cos\bigl(\tfrac{0\cdot 1}{m}\pi\bigr) & \dots & \cos\bigl(\tfrac{0\cdot m}{m}\pi\bigr) \\ \cos\bigl(\tfrac{1\cdot 0}{m}\pi\bigr) & \cos\bigl(\tfrac{1\cdot 1}{m}\pi\bigr) & \dots & \cos\bigl(\tfrac{1\cdot m}{m}\pi\bigr) \\ \vdots & \vdots & \ddots & \vdots \\ \cos\bigl(\tfrac{m\cdot 0}{m}\pi\bigr) & \cos\bigl(\tfrac{m\cdot 1}{m}\pi\bigr) & \dots & \cos\bigl(\tfrac{m\cdot m}{m}\pi\bigr) \\ \end{bmatrix} \in \mathbb R^{(m+1)\times(m+1)} $$
A short notation would be $$M_m = \Bigl(\cos\bigl(\tfrac{ij}{m}\pi\bigr)\Bigr)_{i,j=0}^m.$$
I want to calculate the spectral norm $\|M_m\|_2$.
Since $M_m$ is symmetric, one could investigate its Eigenvalues, but I could not come up with anything.
Maybe one can come up with a clever vector $v$ with $\|v\|_2 = 1$, where it can be seen that it maximizes $\|M_m\cdot v\|_2$?
I already saw that numerically, the norm $\|M_m\|_2$ seems to grow like $\sqrt m$, so I suppose in the algebraic expression for $\|M_m\|_2$ there should be something like $\sqrt m$. I also saw that I may be a good idea to consider $m=2k$ and $m=2k+1$ differently, since it seems the norm is jumping around between two slightly different functions.
I would conjecture that for $$ \alpha(m) = \begin{cases}\frac1{\sqrt2}\sqrt{2m+1 + \sqrt{4m+1}},&m\text{ is even}\\ \frac12\sqrt{4m+1 + \sqrt{8m+1}},&m\text{ is odd}\\\end{cases} $$ it holds $$ \alpha(m) = \|M_m\|_2. $$
I know that it is true for $m=2,\dots,8$, because of this Mathematica Code:
mM[m_] := Table[Cos[(i j)/m Pi], {i, 0, m}, {j, 0, m}]
alpha[m_] := If[EvenQ[m],
Sqrt[(2 m + 1 + Sqrt[4 m + 1])/2],
Sqrt[ 4 m + 1 + Sqrt[8 m + 1]]/2 ]
Table[Norm[mM[m]] - alpha[m], {m, 2, 8}]
The conjecture can be formulated a bit different and easier: For $$ \beta(m) = \begin{cases}\frac12(1 + \sqrt{4m+1}),&m\text{ is even}\\ \frac{\sqrt{2}}4(1 + \sqrt{8m+1}),&m\text{ is odd}\\\end{cases} $$ it (probably) holds $$ \beta(m) = \|M_m\|_2. $$ Again, validation for $m=2,\dots,8$ via Mathematica:
mM[m_] := Table[Cos[(i j)/m Pi], {i, 0, m}, {j, 0, m}]
beta[m_] :=
If [EvenQ[m], (1 + Sqrt[4 m + 1])/2, (1 + Sqrt[8 m + 1]) Sqrt[2]/4]
Table[Norm[mM[m]] - beta[m], {m, 2, 8}] // FullSimplify