Show that $\exp \begin{pmatrix} x & -y\\ y & x\end{pmatrix}= \exp(x) \begin{pmatrix} \cos y& -\sin y\\ \sin y& \cos y\end{pmatrix}$

Show that $$\exp \begin{pmatrix} x & -y\\ y & x\end{pmatrix}= \exp(x) \begin{pmatrix} \cos y& -\sin y\\ \sin y& \cos y\end{pmatrix}$$ for all $$x,y \in \mathbb{R}$$.

My thought process is the following:

$$\left(\begin{array}{cc} a & -b\\ b & a\end{array}\right) + \left(\begin{array}{cc}c & -d\\ d& c\end{array}\right) = \left(\begin{array}{cc} a+c & -(b+d)\\ b+d & a+c \end{array}\right)$$

and

$$\left(\begin{array}{cc} a & -b\\ b & a\end{array}\right) \left(\begin{array}{cc}c & -d\\ d& c\end{array}\right) = \left(\begin{array}{cc} ac-bd & -(ad+bc)\\ ad+bc & ac-bd \end{array}\right).$$

Can I use this idea in my proof?

• How did you define $\exp(A)$ where $A$ is a matrix?
– user700480
May 16 at 10:39

The argument of the exponential on the LHS $$\begin{pmatrix} x & -y\\ y & x\end{pmatrix} \;=\; x\,\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix} \;+\; y\,\begin{pmatrix} 0 & -1\\ 1& 0\end{pmatrix}$$ may be identified with the complex number $$\,x+y\,i\,$$ because the matrix summands on the RHS commute and $$\begin{pmatrix} 0 & -1\\ 1& 0\end{pmatrix}^2 \;=\; -\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}\,.$$ Under this correspondence we have $$\:\begin{pmatrix} \cos y& -\sin y\\ \sin y& \cos y\end{pmatrix} \;\longleftrightarrow\;\cos y +i\sin y\,,\,$$ and the identity under consideration is just Euler's identity $$e^{x+iy} \;=\; e^x\,(\cos y +i\sin y)\,.$$

The map $$M:\mathbb C\to \mathbb R^{2\times 2}$$ defined by $$z=x+iy\mapsto \left[\begin{array}{cc}x&-y\\y&x\end{array}\right]$$ satisfies $$M(z+w)=M(z)+M(w)$$, $$M(zw)=M(z)M(w)$$ and $$M(tz)=tM(w)$$ for $$z,w\in\mathbb C$$ and $$t\in\mathbb R$$. This implies that also $$M(\exp(z))=\exp(M(z))$$ for all $$z\in\mathbb C$$. For $$z=x+iy$$ we thus get from Euler's identity $$e^{iy}=\cos(y)+i\sin(y)$$ $$\exp\left[\begin{array}{cc}x&-y\\y&x\end{array}\right]=\exp M(z)=M(\exp(z))=M(e^x(\cos(y)+i \sin(y))=e^x M(\cos(y)+i\sin(y))= e^x \left[\begin{array}{cc}\cos(y)&-\sin(y)\\\sin(y)&\cos(y)\end{array}\right].$$

If the matrix is diagonalize $$\mathbf{A} = \left(\matrix{x & -y \\ y & x}\right)$$ e.g. $$\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$$ To do so we need to compute the eigenvalues $$\det(\lambda \mathbf{I} - \mathbf{A})$$ which is $$(\lambda - x)(\lambda - x) - y \cdot (-y) = \lambda^2 -2x\lambda +x^2 + y^2 = 0$$ thus $$\lambda = \frac{2x \pm \sqrt{4x^2 - 4(x^2 + y^2)}}{2} = \frac{2x \pm2iy}{2} = x \pm iy$$ (if the eigenvalues are degenerate then we can not continue further)

compute the eigenvectors to create $$\mathbf{P}$$ and $$\mathbf{D} = \left(\matrix{x + iy & 0 \\ 0 & x - iy}\right)$$ with this in place $$\exp(\mathbf{A}) = \mathbf{P}\exp(\mathbf{D})\mathbf{P}^{-1}$$ where we have $$\exp(\mathbf{D}) = \left(\matrix{\exp(x + iy) & 0 \\ 0 & \exp(x - iy)}\right)$$ then you need to pre/post multiply with the respective eigenvector matrices $$\mathbf{P},\mathbf{P}^{-1}$$.

notes: Please see details for diagonalization and conditions

My advice is to diagonalise the matrix first, and then it should be easy from there. Expand in power series.

Any analytic function of an $$n\times n$$ matrix is equal to a polynomial of degree $$(n-1)$$ with scalar coefficients: $$\exp \mathbf{A} = \beta_0 \mathbf{I}+ \beta_1 \mathbf{A}$$ The eigenvalues of $$\mathbf{A}$$ are known to be $$\lambda_k=x\pm iy$$ and verify $$\exp(\lambda_k) = \beta_0 + \beta_1 \lambda_k$$

It follows after straightforward manipulations $$\begin{eqnarray} e^x \cos y &=& \beta_0 + \beta_1 x \\ e^x \sin y &=& \beta_1 y \end{eqnarray}$$ from which $$\beta_0, \beta_1$$ are easily found and thus the final relation .