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Let $L: M_{n \times n}(\mathbb{R})\to M_{n \times n}(\mathbb{R})$ be a linear transformation defined by $L(A)=(A+A^t)/2$. Then how to find the characteristic polynomial of L?

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    click on edit to see the latex I typed. Did you try with $n=2$ ? You can see $M_{n \times n}(\mathbb{R})$ as $M_{4}(\mathbb{R})$ and find the matrix corresponding to $L$ – 2017-02-08
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    First find the matrix of this linear transformation w.r.t. some basis. Which basis of $M_{n\times n}(\mathbb{R})$ do you think will be fairly easy? – 2017-02-08
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    @Mathematician42 are you suggesting to see $L$ as a proj**** ? – 2017-02-08
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    I was not actually. – 2017-02-08
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    In fact using the ordered basis $\left\{e_{11}, e_{22}, \dots ,e_{nn}, e_{12}, e_{21}, e_{13}, e_{31}, \dots , e_{1n},e_{n1}, e_{23}, e_{32}, e_{24}, e_{42}, \dots\right\}$ seems to work nicely. Doing this you see you will get a factor $(1-X)$ for each diagonal element. For each pair of off-diagonal elements which are each others transposes, you will get one factor $X$ and one additional factor which you can easily determine in general. – 2017-02-08
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    @Mathematician42 If you want to write an answer, then $L$ is really $\begin{pmatrix} I & 0 \\ 0 & 0 \end{pmatrix}$ in some basis – 2017-02-08
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    True, the clue is that you have to take those pairs together in a good way. – 2017-02-08

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Consider the following basis of $M_{n\times n}(\mathbb{R})$: $$\left\{e_{11}, \dots , e_{nn}, e_{12}+e_{21}, \dots , e_{1n}+e_{n1},e_{23}+e_{32}, \dots , e_{12}-e_{21}, \dots, e_{1n}-e_{n1}, \dots\right\}.$$ There are a lot of dots to fill in in this basis, but I think the idea should be clear.

Then the matrix of $L$ w.r.t. this basis is

\begin{pmatrix} I_{n+\frac{n^2-n}{2}}&0\\ 0&0 \end{pmatrix}

Thus the characteristic polynomial is $(1-X)^{\frac{n^2+n}{2}}X^{\frac{n^2-n}{2}}.$