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 $\tan{A}=aA+bI$?
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- $\begingroup$ I'm skeptical: for an analytic function $f(x) $ where the power series at zero has large enough radius of convergence, we can define $f(A)$ for matrix $A.$ If that works, the Cayley-Hamilton theorem says that we can calculate $f(A)$ in terms of just the first few powers of $A.$ However, the Taylor series for tangent has radius of convergence $\pi / 2.$ You are not giving any evidence that you have a way around divergence, so I would regard the conclusion as meaningless. What is your background? $\endgroup$Will Jagy– Will Jagy2023-01-19 22:31:33 +00:00Commented Jan 19, 2023 at 22:31
- $\begingroup$ Well, you are correct in your statement and I have not presented any evidence-right. But, considering the existence of the required premise and writing the Taylor function to be explicit-how would the eigenvalue satisfy the explicit relation? (tanA=kA is the final deduction) $\endgroup$Arsenic– Arsenic2023-01-19 22:46:17 +00:00Commented Jan 19, 2023 at 22:46
- $\begingroup$ Tell you what, the sine function has Taylor series with infinite radius of convergence. There is an unambiguous matrix value for $\sin A.$ Using $A^2 = 9I,$ write out the first few, $A - \frac{A^3}{6} + \frac{A^5}{120} -+$ see if you can calculate $\sin A$ in its entirety. Comes out similar to your given problem. $\endgroup$Will Jagy– Will Jagy2023-01-19 23:07:06 +00:00Commented Jan 19, 2023 at 23:07
- 1$\begingroup$ @JeanMarie Hi! I was not aware that there was a general treatment possible. I would guess that, if $A,B$ don't commute, $\tan(A+B)$ is not so simple in terms of $\tan A, \tan B.$ Actually, I guess sine and cosine of a sum of matrices don't come out right, despite convergent power series as in $e^A$ $\endgroup$Will Jagy– Will Jagy2023-01-23 18:15:14 +00:00Commented Jan 23, 2023 at 18:15
- 2$\begingroup$ Cf. math.stackexchange.com/questions/1634488/… $\endgroup$Travis Willse– Travis Willse2023-01-23 22:35:40 +00:00Commented Jan 23, 2023 at 22:35
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3 Answers
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2 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.
- $\begingroup$ I fully agree with you but this does not answer the question, i.e. it does not explain where this "$\tan A=aA+bI$ (1)" come from and why "equation (1) is satisfied by eigen values". I am preparing an answer... $\endgroup$Anne Bauval– Anne Bauval2023-01-20 12:18:42 +00:00Commented Jan 20, 2023 at 12:18
- $\begingroup$ [+1] But what prevented you to write directly $\tan A=P\text {diag}(\tan 3,-\tan 3)P^{-1}$ ? Is it because, going from $-3$ to $3$ you cross poles of $\tan$ on the $x$ axis ? $\endgroup$Jean Marie– Jean Marie2023-02-12 08:12:20 +00:00Commented Feb 12, 2023 at 8:12
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3 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 $\tan A=\frac{\tan 3}3A.$
- 1$\begingroup$ [+1] Excellent answer. $\endgroup$Jean Marie– Jean Marie2023-01-23 14:19:03 +00:00Commented Jan 23, 2023 at 14:19
- $\begingroup$ Have you heard about Schur-Parlett algorithm ? $\endgroup$Jean Marie– Jean Marie2023-02-12 08:07:57 +00:00Commented Feb 12, 2023 at 8:07
- $\begingroup$ No, I hadn't. Vu de loin ce papier a l'air intéressant mais je suis trop paresseuse pour m'y plonger, et l'analyse numérique ne m'attire guère. Merci quand même ! $\endgroup$Anne Bauval– Anne Bauval2023-02-12 11:33:01 +00:00Commented Feb 12, 2023 at 11:33
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0 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) 