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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 $T$.

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    $\begingroup$ 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. $\endgroup$
    – user1317176
    Commented May 1 at 4:58
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    $\begingroup$ 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 . $\endgroup$
    – user1317176
    Commented May 1 at 5:00
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    $\begingroup$ @Arpit what are the Eigen vectors? $\endgroup$
    – Sandeep Tiwari
    Commented May 1 at 5:04
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    $\begingroup$ $T$ 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. $\endgroup$
    – Ben Grossmann
    Commented May 1 at 5:08
  • $\begingroup$ @BenGrossmann I am very lucky that I got a reply form you.In each question your solutions are awesome .Thank you sir. $\endgroup$
    – Sandeep Tiwari
    Commented May 1 at 5:11

2 Answers 2

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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.

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  • $\begingroup$ Excellent solution sir. $\endgroup$
    – Sandeep Tiwari
    Commented May 1 at 5:17
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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 $$\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-2\\0&0&2&0\end{pmatrix}.$$

Eigenvectors, eigenvalues, Jordan form, etc. can then be computed using standard linear algebra techniques.

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