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$A$ 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?2012-04-23
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    Is it a linear function?2012-04-23
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    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 $n>2$?2012-04-23
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    If $F$ is algebraically close, the equation will have $n$ roots, otherwise, less than $n$.2012-04-23
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    Unrelated note: I have edited this to make the title a little more useful. With foresight, I can imagine similar questions frequenting the site. So, this title will help locating the question than the previous one. I apologise to OP for making this draconic edit.2012-04-23
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    @No need to apologize. Thanks for the help.2012-04-23
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    @Jyrki Lahtonen Yes, should be polynomials of order n.2012-04-23
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    Correct, Megan. Though more commonly the highest power is referred to as the *degree* of the polynomial. What do you know about the number of zeros of a polynomial of degree $n$? In a field.2012-04-23
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    @Jyrki Lahtonen Polynomials of degree n has n zeros or roots.2012-04-23
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    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!2012-04-23

2 Answers 2

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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.

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The charecteristic polynomial of $A\in M_{n x n}(\mathbb R)$ is a polynomial of degree n with leading coefficient $(-1)^n$ and eigenvalues are zeros of this charesterictic polynomial so by Fundamental Theorem of Algebra n degree polynomial can have at most n roots.