Suppose $m_A(\lambda)$ is the minimal polynomial of $A_{n\times n}$.

  1. Show that if $\lambda$ is an eigenvalue of $A$, then $m_A(\lambda)=0$

  2. Show that $m_A(x)$ of a diagonalizable matrix $A$ divides the characteristic polynomial $f_A(x)$.

For the second question, I was thinking we could use the fact that if $p(x)$ is a polynomial such that $p(A)=0$, then $m_A(x)$ divides $p(x)$. I think this requires a proof of number 1, though, since a diagonalizable matrix $A$ will have the eigenvalues on the diagonal, but I'm not really sure.

  • $\begingroup$ The second one is actually true in general, by Cayley-Hamilton. But for a diagonalizable matrix, the characteristic polynomial and minimal polynomial are easy to find, so you can check it directly. For 1, what happens when you apply $m_A(A)$ to an eigenvector? $\endgroup$ – Nishant Dec 3 '14 at 3:59

Hints: For the first question: let $x$ be an eigenvector. Consider $m_A(A)x$. For the second: yes, that will work.

  • $\begingroup$ What does $m_A(A)x$ represent? $\endgroup$ – Carley Dec 3 '14 at 4:10
  • $\begingroup$ $m_A(A)$ is the matrix that comes out of plugging $A$ into the polynomial $m_A$. Take that matrix, and multiply it by $x$. $\endgroup$ – Omnomnomnom Dec 3 '14 at 4:26
  • $\begingroup$ OK, I think I got #1. For #2, what is $m_A(x)$ of a diagonal matrix $A$? Why is it zero? $\endgroup$ – Carley Dec 3 '14 at 4:36
  • $\begingroup$ Note that for any diagonal matrix $$ D = \pmatrix{d_1\\&\ddots\\&&d_n} $$ and any polynomial $p$, we have $$ p(D) = \pmatrix{p(d_1)\\&\ddots\\&&p(d_n)} $$ $\endgroup$ – Omnomnomnom Dec 3 '14 at 4:41
  • $\begingroup$ When we say that "matrix $A$ is diagonalizable," are we really saying that $A$ can be written as a diagonal matrix where its eigenvalues are on the diagonal? I thought it just meant that $A$ could be factored into $SBS^{-1}$, where $B$ is the diagonal matrix with eigenvalues on the diagonal. $\endgroup$ – Carley Dec 3 '14 at 4:44

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