# Proving that an $n\times n$ matrix has at most $n$ distinct eigenvalues

$A$ is a $n\times n$ matrix over the field $F$. How can I prove that there are at most $n$ distinct scalars $c$ in $F$ such that $\det(cI - A) = 0$?

Thank you!

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What kind of a function of $c$ comes out, when you expand that determinant? – Jyrki Lahtonen Apr 23 '12 at 9:10
Is it a linear function? – Megan Apr 23 '12 at 9:29
If $n=2$ and $$A=\pmatrix{a_{11}&a_{12}\cr a_{21}&a_{22}\cr},$$ then $$\det(cI-A)=\left|\begin{array}{cc}c-a_{11}&-a_{12}\\-a_{21}&c-a_{22}\end{array}‌​\right|=c^2-c(a_{11}+a_{22})+(a_{11}a_{22}-a_{12}a_{21})$$ looks like a quadratic polynomial in the unknown $c$. What happens when $n>2$? – Jyrki Lahtonen Apr 23 '12 at 9:33
If $F$ is algebraically close, the equation will have $n$ roots, otherwise, less than $n$. – ziyuang Apr 23 '12 at 11:29
Almost correct. It may be less: $(x-1)^2=0$ has only one zero even though it is quadratic :-) So you are seeing the light now, Megan? If so, you may consider writing a summary of what you learned as an answer, so that we can upvote, comment and criticize it! – Jyrki Lahtonen Apr 23 '12 at 13:50

If there are $c_1,\ldots,c_{n+1}$ $n+1$ distinct elements of $F$ such that for all $1\leq j\leq n+1$ we have $\det(c_jI-A)=0$ then $c_jI-A$ is not invertible. Hence we can find a vector $v_j\in F^n$ such that $Av_j=c_jv_j$. The family $\{v_j,1\leq j\leq n+1\}$ is linearly independent, otherwise there would exists a $j_0$ such that $\{v_1,\ldots,v_{j_0}\}$ is linearly independent, but not $\{v_1,\ldots,v_{j_0+1}\}$. So $v_{j_0+1}=\sum_{k=1}^{j_0}\alpha_kv_k$ for $(\alpha_1,\ldots,\alpha_{j_0})\neq (0,\ldots,0)$ and $$Av_{j_0+1}=c_{j_0+1}v_{j_0+1}=\sum_{k=1}^{j_0}\alpha_kc_kv_k.$$ We get $$\sum_{k=1}^{j_0}\alpha_kc_kv_k=\sum_{k=1}^{j_0}\alpha_kc_{j_0+1}v_k$$ so by linear independence $\alpha_k(c_k-c_{j_0+1})=0$ for all $1\leq k\leq j_0$. Since $c_k-c_{j_0+1}\neq 0$ we get $\alpha_k=0$, a contradiction.