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

Question

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} $.

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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?

Answer 1

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

Answer 2

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

Answer 3

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 $e^{iy}=\cos(y)+i\sin(y)$ $$ \exp\left[\begin{array}{cc}x&-y\\y&x\end{array}\right]=\exp M(z)=M(\exp(z))=M(e^x(\cos(y)+i \sin(y))=e^x M(\cos(y)+i\sin(y))= e^x \left[\begin{array}{cc}\cos(y)&-\sin(y)\\\sin(y)&\cos(y)\end{array}\right].$$

Answer 4

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 $$e^{x+iy} \;=\; e^x\,(\cos y +i\sin y)\,.$$

Answer 5

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 .