Let $B$ be a matrix with elements as linear forms of indeterminates. Is there a proper diagonalization procedure for such matrices like those of matrices with real and complex entries?
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$\begingroup$ What do you mean by "diagonalization procedure" ? $\endgroup$– RalphCommented Feb 13, 2012 at 19:43
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$\begingroup$ Say we have a non-singular matrix over complex numbers with unequal eigenvalues, then we can diagonalize such matrices. Can we have a procedure for these matrices as well? I have a bottleneck in a proof. $\endgroup$– TurboCommented Feb 13, 2012 at 19:46
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$\begingroup$ Couldn't $B$ be considered as matrix over a function field $k(x_i\mid i \in I)$ ? $\endgroup$– RalphCommented Feb 13, 2012 at 19:58
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$\begingroup$ Sure but I believe $k[x_{i}|i∈I]$ is more correct since I have only linear forms. $\endgroup$– TurboCommented Feb 13, 2012 at 20:07
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$\begingroup$ Over the function field diagonalization works as over every field. But there is in general no $X \in GL_n(k[x_i|i \in I])$ such that $X^{-1}BX$ is diagonal. As an example take $$B = \begin{pmatrix}x_1 & x_2 \\ x_1 & x_2\end{pmatrix}$$. In this case $X$ is (up to permutation of columns) $$X = \begin{pmatrix}x_2f & g \\ -x_1f & g\end{pmatrix}$$ with polynomials $f,g$. Thus $\det(X) = (x_2 - x_1)fg \notin k[x_i]^\times = k$. // But this is nothing special for linear forms, it's typical for matrices over integral domains. $\endgroup$– RalphCommented Feb 13, 2012 at 21:06
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