# Show that every integer eigenvalue of $A$ divides the determinant of $A$.

Let $A$ be a square matrix of order $n$ whose entries are all integers. Show that every integer eigenvalue of $A$ divides the determinant of $A$.

I am not able to understand how to show this. We know that $\det A$ is the product of eigen values and so every eigen value must divide $\det A$.

But if a matrix has eigen values $3$ and $\frac{4}{3}$ then if I do the product then the factor $3$ gets neutralized if I do the product then how does it appear as a factor of $\det A$?

Please help.

• It says to show that every integer eigenvalue divides the determinant. If there are other eigenvalues which are rational or real, don't worry about them. – Joppy Dec 15 '17 at 13:07
• @Joppy; what is there to prove then? – Learnmore Dec 15 '17 at 13:09
• In your example where a $2 \times 2$ matrix has eigenvalues $3$ and $\frac{3}{4}$, the determinant will be $1$, and $4$ does not divide $1$. You need to show that this cannot happen. – Joppy Dec 15 '17 at 13:11

## 1 Answer

Let $P(x)=\sum_{k=0}^na_kx^k$ be the charasteristic polynomial. Then $|P(0)|=|\det A|$. Moreover, the coefficients of $P$ are integer. If $\lambda$ is an integer eigenvalue, then $$0=P(\lambda)=P(0)+\sum_{k=1}^na_k\lambda^k$$ Therefore $$|\det A|=|P(0)|=|\lambda|\cdot\left|\sum_{k=1}^na_k\lambda^{k-1}\right|$$ Hence $\lambda$ divides $\det A$.

• So you mean that a matrix with integer entries always has eigen values as integers? – Learnmore Dec 15 '17 at 13:17
• No, it can have irrational or complex eigenvalues. – ajotatxe Dec 15 '17 at 13:19
• @Find_X: But since the characteristic polynomial is monic, the eigenvalues are algebraic integers and if they are rational, they are integer. – robjohn Dec 15 '17 at 13:33
• I don't understand one thing. $\det A$ is the product of eigen values so does it not follow trivially – Learnmore Dec 15 '17 at 13:34
• @Find_X: it is also the constant coefficient of the characteristic polynomial, and that is what we are looking at here. That is, for any eigenvalue, $\lambda$, $$0=\det(A)+\sum_{k=1}^na_k\lambda^k$$ therefore $$\det(A)=-\lambda\sum_{k=1}^na_k\lambda^{k-1}$$ – robjohn Dec 15 '17 at 13:35