Could you please give some hint how to prove this statement without using Jordan form:
If $A\in M_{nxn}^C$ is not diagonalizable matrix then exists polynomial $Q(t)=C_n[t]$ whose degree are smaller than $n$ and ${[Q(A)]}^2=0$.
Thanks.
A question about not not diagonalizable matrix
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linear-algebra
polynomials
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0This question is similar to : http://math.stackexchange.com/questions/624962/nondiagonalizable-matrix-and-polynomials – 2017-01-22
2 Answers
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A matrix over any field is non-diagonalizable iff its minimal polynomial is divisible by a linear factor to a power $\;\ge2\;$ , so $\;A\;$ is non diagonalizable iff
$$\min_A(x)=(x-a_1)^{n_1}g(x)\;,\;\;2\le n_1\in\Bbb N$$
and thus we can take $\;Q(x)=(x-a_1)g(x)\;$ and, of course, $\;Q(A)^2=0\;$
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Hint : Let $P_A(t)$ be the characteristic polynomial of $A$. $P_A(t)$ can be written as a product of $n$ linear factors $(t-\lambda_i)$; since $A$ is not diagonalizable one of the factors has to appear twice.