# Determining Jordan canonical form(JCF) of an operator given by complex differentiation.

Let $$W$$ be the subspace of $$\Bbb C$$ linear combination of the following functions:$$f_1(z)=\sin z,\qquad f_2(z)=\cos z,\qquad f_3(z)=\sin2z,\qquad f_4(z)=\cos2z.$$ Let $$T$$ be the linear opeartor on $$W$$ given by complex differentition.Which of the following statements are true?

$$1$$.Dimension of $$W$$ is $$3$$.
$$2$$.The span of $$f_1$$ and $$f_2$$ is a Jordan block of $$T$$.
$$3$$.$$T$$ has two Jordan blocks.
$$4$$. $$T$$ has four Jordan blocks.

Since $$f_1(z)=\sin z$$,$$f_2(z)=\cos z$$,$$f_3(z)=\sin2z$$ and $$f_4(z)=\cos2z$$ are linearly independent,hence dimension of $$W$$ is $$4$$.So option (1) is incorrect.

Also $$W$$ is $$T$$ invariant (as $$f'_1(z)=\cos z$$, $$f'_2(z)=-\sin z$$, $$f'_3(z)=2\cos2z$$, $$f'_4(z)=-2\sin2z$$)
Now for other options one should find Jordan canonical form of $$T$$ and for Jordan canonical form one must have eigenvalues but I am unable to find eigenvalues of $$T$$.

• T acts differently on f_1(z) and f_2(z), indicating that it may not have a diagonalizable form and consequently may not have distinct eigenvalues.
– user1317176
Commented May 1 at 4:58
• let's consider the Jordan canonical form directly by observing the chains of generalized eigenvectors associated with each eigenvalue. We can see that T has two distinct eigenvalues: lemda_1 = i and lamda_2 = -i .
– user1317176
Commented May 1 at 5:00
• @Arpit what are the Eigen vectors? Commented May 1 at 5:04
• $T$ has the four eigenvalues $\pm i$ and $\pm 2i$, so it is diagonalizable. Each eigenvector is of the form $e^{\lambda z }$ where $\lambda$ is the associated eigenvalue. Statement 4 is correct. Commented May 1 at 5:08
• @BenGrossmann I am very lucky that I got a reply form you.In each question your solutions are awesome .Thank you sir. Commented May 1 at 5:11

An eigenvector for $$T$$ satisfies $$g'=\lambda g$$. Then $$g(z)=e^{\lambda z}$$. You have $$e^{\lambda z}=\cos(-i\lambda z)+i \sin(-i\lambda z),$$ So $$g\in W$$ when either $$-i\lambda =\pm 1$$ or $$-i\lambda=\pm 2$$. That is, the eigenvalues of $$T$$ are $$i,-i,2i,-2i$$. Then $$T$$ has a basis of eigenvectors, namely $$e^{iz}$$, $$e^{-iz}$$, $$e^{2iz}$$, $$e^{-2iz}$$, and it has four Jordan blocks.
This problem can also be solved using linear algebra and minimal calculus by expressing $$T$$ as a matrix. If we take the derivatives, we have $$Tf_1=f_2,Tf_2=-f_1,Tf_3=2f_4,Tf_4=-2f_3$$ so in the $$f_i$$ basis $$T$$ corresponds to the matrix $$\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-2\\0&0&2&0\end{pmatrix}.$$