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I remember from my linear algebra courses that if I have a $n\times n$ matrix with coefficients in a field (denoted as $A$) and I have a polynomial $P $ over the field s.t. $P(A)=0$ and a decompostion $P=f(x)g(x)$ over the field then $f(A)=0$ or $g(A)=0$.

This was used to calculate the minimal polynomial of $A$.

My question is: Is the statement above that $f(A)=0$ or $g(A)=0$ is correct or maybe I remember wrong ? the reason I am asking is that there are non zero matrices $B,C$ s.t. $BC=0$ so I don't see how the conclusion was made

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    $\begingroup$ I think it's wrong, consider matrix $2\times 2$, such that the left top entry equal $1$ and zero otherwise, we have $A^2-A=0$, and $P(x)=x(x-1)$, so $f(x)=x$ and $g(x)=x-1$, but clearly neither $f(A)$ or $g(A)$ is zero. $\endgroup$ – Ajat Adriansyah Sep 15 '12 at 10:22
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This isn't correct. For example, the matrix $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ satisfies $p(A)=0$ where $p(t)=t^2$, however $p(t)=f(t)g(t)$ where $f(t)=g(t)=t$ and $f(A) \ne 0 \ne g(A)$.

There are some similar-ish results that you might be thinking of. I'll list a few. Let $A$ be a matrix with characteristic polynomial $p$ and minimal polynomial $m$ over a given (well-behaved) field.

  • $A$ satisfies its minimal polynomial; that is, $m(A) = 0$.
  • If $f$ is a polynomial then $f(A)=0$ if and only if $m\, |\, f$; in particular, $p(A)=0$.
  • If $m(t)=f(t)g(t)$ and $f(A)=0$ then $g$ is constant.
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  • $\begingroup$ Thanks, I was thinking on the second point $\endgroup$ – Belgi Sep 15 '12 at 10:28
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It is not true.

The minimal polynomial doesn't have to be irreducible.For instance if you consider the $2 \times 2$ diagonal matrix with $1$ and $2$ on the main diagonal, then it's minimal polynomial would be $X^2-3X+2$, but $X-1$ and $X-2$ won't vanish the matrix.

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