Hyperbolic cosine of a matrix

Question

Define the hyperbolic of a matrix $A$ as $\cosh(A)=I+\frac{A^2}{2!}+\frac{A^4}{4!}+\cdots$. Does there exist a complex matrix of order $2$ s.t. $\cosh(A)=\left(\begin{array}{cc}1&2013\\0&1\end{array}\right)$ ?

Answer 1

Suppose that there is a matrix $A$ such that $\cosh(A)=\left(\begin{array}{cc}1&2013\\0&1\end{array}\right)$. Then $A$ has eigenvalues $0$. Clearly $A$ can't havethe Jordan form $\left(\begin{matrix}0&0\\0&0\end{matrix}\right)$. Suppose $A$ hasthe Jordan form $\left(\begin{matrix}0&1\\0&0\end{matrix}\right)$. Without loss of generality, let $A=\left(\begin{matrix}0&1\\0&0\end{matrix}\right)$. Then $A^n=\left(\begin{matrix}0&0\\0&0\end{matrix}\right)$ for $n\ge 2$ and hence $$ \cosh(A)=I+\frac{A^2}{2!}+\frac{A^4}{4!}+\cdots=I.$$ Thus it is impossible for such matrix.

Answer 2

The answer is no.

Note that if $\lambda$ is an eigenvalue of $A$, then $\cosh(\lambda)$ is an eigenvalue of $\cosh(A)$. So, if $A$ were such a matrix, any eigenvalue $\lambda$ of $A$ must satisfy $\cosh(\lambda) = 1$, which means that $\lambda = 0$. Thus, $A$ would have characteristic polynomial $x^2$. By the Cayley-Hamilton theorem, $A^2 = 0$. It would follow that $\cosh(A) = I$.

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As the comment points out, we should really address the case in which $\lambda = 2 \pi i n$ for $n \in \Bbb Z$, which is also a solution to $\cosh(\lambda) = 1$. If $A$ is diagonalizable, then $\cosh(A)$ must be diagonalizable. So, we are left with the case in which $A$ has $\lambda = 2 \pi i n$ as its only eigenvalue, and $A \neq \lambda I$.

In fact, it suffices to apply the identity $$ \cosh(X + Y) = \cosh(X)\cosh(Y) + \sinh(X)\sinh(Y) $$ With $X = (A - \lambda I)$ and $Y = \lambda I$. We find (applying our earlier analysis) that $$ \cosh(A) = (I)(I) + (A-\lambda I)(0) = I $$ so once more, $A$ fails to satisfy the desired equation.