tangent of a matrix

Question

If A=$\left[ \begin{array}{ccc} -1 & 4 \\ 2 & 1 \end{array}\right]$ then prove that $3\tan {A}=A\tan{3}$.

I do have a solution with me, but I am not able to understand a part in that.

Why would $\lambda$ satisfy $\tan{A}=aA+bI$?

Answer 1

The most straight forward way is to find a real invertible 2 $\times$ 2 matrix $P$ such that $P^{-1}AP= \text {diag}(3,-3)$. Then $$\sin A=A-\frac{1}{3!}A^3+...$$ $$=P \text {diag}(\sin 3,-\sin 3)P^{-1}$$ and $$\cos A=P\text{diag}(\cos 3,\cos 3)P^{-1}$$ so $$(\cos A)^{-1}=P\text{diag}(\sec 3,\sec 3)P^{-1}$$.Then multiply the sin matrix by the inverse of the cos matrix(Fortunately these matrices commute so the order of multiplication doesn't matter.)We conclude that $$\tan A=P\text {diag}(\tan 3,-\tan 3)P^{-1}.$$ The question would have been more challenging if $A$ had not been diagonalizable.

Answer 2

By holomorphic functional calculus, $\tan A$ may be defined by $$\tan A=\frac1{2\pi i}\int_\Gamma\frac{\tan z}{zI-A}\,dz,$$ where:

• $\Gamma$ is (after correction, thanks to Jean Marie's kind comments) some positively oriented curve delimiting a domain containing $A$'s eigenvalues but containging none of $\tan$'s poles $\left(\Bbb Z+\frac12\right)\pi,$ e.g. the union of the circle $\Gamma_4:|z|=4$ oriented positively and the circle $\Gamma_2:|z|=2$ oriented negatively,
• and by Cayley-Hamilton, $A^2=9I,$ so $$\frac1{zI-A}=\frac1{z^2-9}A+\frac z{z^2-9}I.$$

Plugging this into the definition of $\tan A$ above, we obtain $$\tan A=aA+bI$$ where $a,b$ are the corresponding integrals on $\Gamma=\Gamma_4\cup\Gamma_2.$

Applying exactly the same process to any of the two eigenvalues $\lambda=\pm3$ of $A,$ we obtain $$\tan\lambda=a\lambda+b$$ for the same constants $a,b.$

The end is like in your solution: solving these two equations, we get $b=0$ and $a=\frac{\tan 3}3,$ so $\tan A=\frac{\tan 3}3A.$

Answer 3

Let $A=\left(\begin{array}{rr} -1 & 4 \\ 2 & 1 \end{array}\right) \text{; its eigenvalues are} \ 3, \ -3.$

I will use, as Anne Bauval has done, the Cauchy integral formula for operators :

$$f(A)=\frac{1}{2 i \pi}\int_{\gamma}(zI-A)^{-1}f(A)dz \ \text{with} \ \ f(A)=\tan(A) \tag{1}$$

but in a different way.

Formula (1) is valid

• for a closed path $\gamma$ enclosing all the eigenvalues of $A$

• for a holomorphic function $f$. $f(z)=\tan(z)=\frac{\sin(z)}{\cos(z)}$ is holomorphic only if we restrict it to a region which doesn't contain any of its poles $(2k+1)\frac{\pi}{2}$. Closed path $\gamma$ must be situated in such a region.

Here is such a closed path :

Fig. 1 : Integration path $\gamma$. Poles $-3, 3$ are featured in green. Poles $(2k+1)\frac{\pi}{2}$ of function $\tan$ (in red) are outside path $\gamma$.

Let $I$ be the $2 \times 2$ unit matrix. We can give an explicit form to matrix :

$$(zI-A)^{-1}=\frac{1}{z^2-9}\left(\begin{array}{cc} z-1 & 4 \\ 2 & z+1 \end{array}\right)$$

Therefore, the result is :

$$\tan(A)=\frac{1}{2 i \pi}\left(\begin{array}{cc} \int_{\gamma}\frac{\tan(z)(z-1)}{z^2-9}dz & \int_{\gamma}\frac{4\tan(z)}{z^2-9}dz\\\int_{\gamma}\frac{2\tan(z)}{z^2-9}dz & \int_{\gamma}\frac{\tan(z)(z+1)}{z^2-9}dz \end{array}\right)$$

Each integral is easily computed with the theorem of residues, because poles $-3, 3$ interior to $\gamma$ are simple. If we note $c=\frac{\tan(3)}{3}$ :

$$\tan(A)=\frac{1}{2 i \pi}\left(\begin{array}{rr} -2 i \pi c & 8 i \pi c\\ 4 i \pi c & 2 i \pi c \end{array}\right)=cA$$

A side remark : I have checked my results by using "integral" function of Matlab which can do complex integration along any path. Here is the code for example for the first integral, where the "WayPoints" (integration path represented on Fig. 1) are given in list $p$.

 clear all;
 p=[4+i,-i,-4+i,-4-2i,4-2i,4+i];
 integral(@(z)(tan(z).*(z-1)./(z.^2-9)),p(1),p(1),'WayPoints',p)