Let $A$ and $B$ be two real symmetric matrices such that $A^{2018} = B^{2018}$. Show that $\cos A = \cos B$.

Question

Let $A$ and $B$ be two real symmetric matrices such that $A^{2018} = B^{2018}$. Show that $\cos A = \cos B$.

(And is it true that $\sin A = \sin B$ ?)

We can write $A = PDP^T$ and $B = QD'Q^T$ by the spectral theorem ;

Hence $PD^{2018}P^T = QD'^{2018}Q^T$

Do you have a hint to solve this?

Answer 1

$A^2$ and $B^2$ are positive definite matrices satisfying $$ (A^2)^{1009} = (B^2)^{1009} $$ It follows that $A^2 = B^2$.

Now, note that the power series for $\cos(A)$ may be written as a power series on $A^2$.

It is easy to come up with a counterexample for $\sin$. If you like, take $A = \pi/2$, $B = -\pi/2$ (both $1 \times 1$ matrices).

Answer 2

First note that $A^{2018}$ and $B^{2018}$ are positive.

The function ``$2018$-th root ($\sqrt[2018]{\ast}$)'' is defined for all positive matrices, and it takes "a positive matrix $X$ to the only positive matrix $B$ in the algebra generated by $A$ such that $B^{2018}=X$"

Moreover, if $X$ is symmetric then $\sqrt[2018]{X^{2018}}=|X|$, where $|X|=\sqrt{X^2}$.

From this, we get $|A|=|B|$. Take cosine on both sides to get $\cos|A|=\cos|B|$, and again use symmetry to conclude that $$\cos A=\cos|A|=\cos|B|=\cos B$$

All of this works in any C*-algebra and $A,B$ self-adjoint elements and if $\cos$ is replaced by any even continuous function (note that in this case the trick with the Taylor expansion of $\cos$ mentioned by others doesn't work anymore).