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Given a field $F$ of characteristic 2, I need to give an example of symmetric bilinear form $f$ on a finite dimentional $F$ vector space such that $f$ is not diagonalizable. I will appreciate any hint.

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Try to find an example for the simplest possible field $\mathbb{F}$ and vector space $V$ over $\mathbb{F}$. Take $\mathbb{F} = \mathbb{F}_2$, the field with two elements. If $\dim_{\mathbb{F}} V = 1$, then clearly every bilinear form is diagonalizable, so try the next non-trivial case $\dim_{\mathbb{F}} V = 2$ (for example $V = \mathbb{F}_2^2$). A symmetric bilinear form $f \colon V \times V \rightarrow \mathbb{F}$ can be represented with respect to a certain basis (say, the stanard one of $\mathbb{F}_2^2$) by a $2 \times 2$ symmetric matrix $A \in \mathrm{M}_2(\mathbb{F}_2)$. There are only eight matrices in $\mathrm{M}_2(\mathbb{F}_2)$. You don't want a diagonal $A$, so that leaves only four possible matrices. Play with the remaining ones and you will find what you are looking for.

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