Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$A$ 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!

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.