# If $\lambda$ is an eigenvalue of $A$, then $m_A(\lambda)=0$

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.

• 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? – 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.
• What does $m_A(A)x$ represent? – Carley Dec 3 '14 at 4:10
• $m_A(A)$ is the matrix that comes out of plugging $A$ into the polynomial $m_A$. Take that matrix, and multiply it by $x$. – Omnomnomnom Dec 3 '14 at 4:26
• OK, I think I got #1. For #2, what is $m_A(x)$ of a diagonal matrix $A$? Why is it zero? – Carley Dec 3 '14 at 4:36
• Note that for any diagonal matrix $$D = \pmatrix{d_1\\&\ddots\\&&d_n}$$ and any polynomial $p$, we have $$p(D) = \pmatrix{p(d_1)\\&\ddots\\&&p(d_n)}$$ – Omnomnomnom Dec 3 '14 at 4:41
• 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. – Carley Dec 3 '14 at 4:44