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Let $T: V\rightarrow V$ be a linear map where $V$ is finite dimensional. Let $\lambda_1,\lambda_2...,\lambda_n$ be distinct eigenvalues. If the characteristic polynomial $C(x)$ is a product of linear factors then show that the sum of the algebraic multiplicities $a_1+...…+a_n=dimV$.

I know that if I Suppose that the characteristic polynomial $C(x)$ can be expressed as $C(x)=(x-B_1)…(x-B_p)$. Then as $B_1...B_p$ are the roots, it follows that $C(x)$ can be expressed as $C(x)=(x-\lambda_1)^{a_1}…(x-\lambda_n)^{a_n}$ $=$ $det(xI-A)$, where $A$ is the matrix of the linear transformation of $T$. As $V$ is finite dimensional, $A$ has $dim(V)$ rows and $dim(V)$ columns. This means that $C(x)$ is a polynomial of degree dim(V). By the fundamental theorem of algebra, $C(x)$ must have at most $dimV$ roots (What type of fields does this statement hold for?). Therefore, since $\lambda_1,...,\lambda_n$ are all the roots of the polynomial, it follows that $a_1+a_2+....+a_n\leq dimV$.

How do I show that $dimV = a_1+a_2+....+a_n$?

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    In general, for any field, a polynomial of degree $d$ has at most $d$ roots. – Michael Burr Nov 20 '19 at 02:23
  • Hint: If you pass to the algebraic closure, you have equality of the algebraic multiplicities. Now, if all the roots are in the base field, then the equality also holds in the base field. – Michael Burr Nov 20 '19 at 02:24
  • @MichaelBurr $\mathbb{F}$ is not assumed to be algebraically closed. –  Nov 20 '19 at 02:25
  • It is not relevant if $\mathbb{F}$ is algebraically closed since $\mathbb{F}$ has an algebraic closure and equality holds there. Since $C(x)$ is a product of linear factors in $\mathbb{F}$, the roots are already in $\mathbb{F}$, so the equality in the algebraic closure descends to $\mathbb{F}$. – Michael Burr Nov 20 '19 at 02:27
  • @MichaelBurr how do you show it without using the notion of algebraic closure? I have not studied field theory or galois theory yet. –  Nov 20 '19 at 02:28
  • Add up the degrees of the factors. If they're equal as polynomials, then they have the same degree. – Michael Burr Nov 20 '19 at 02:29
  • @MichaelBurr may you include a more elementary proof on why this works, please? –  Nov 20 '19 at 02:33

1 Answers1

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You do not need to know anything about fields. All you need to do is the match the degrees of the highest power of $x$. You know that the characteristic polynomial is a polynomial of degree $\dim V$. If you do not know this fact, you can prove it by induction using the cofactor expansion, or by direct observation using the Leibniz formula. The largest power of $(x-\lambda_1)^{a_1}\cdots (x-\lambda_n)^{a_n}$ is $x^{(a_1+\cdots + a_n)}$. Therefore you must have $\dim V = a_1 + \cdots + a_n$ by matching powers.

For your other question, it's a common misconception that the Fundamental Theorem of Algebra says you have $n$ roots for a degree $n$ polynomial. The fact that a polynomial of degree $n$ has at most $n$ roots holds for any field, and is sometimes known as Lagrange's theorem. You can see some proofs here, for example. The non-trivial part of the Fundamental Theorem of Algebra is the "at least" part, i.e., every degree $n$ polynomial has at least $n$ roots, and this is where you need algebraic closure.

EuYu
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  • Alternatively, could I have said that $C(x)$ is of degree dimV because of the matrix of the linear map mentioned above, which has dimV rows and dim$V$ columns? –  Nov 20 '19 at 02:54
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    Sure. The fact that the characteristic polynomial of an $n\times n$ matrix is of degree $n$ is a basic fact of linear algebra. I don't know what your background is, so I mentioned two ways of proving it, but certainly most people would just apply this result without further justification. – EuYu Nov 20 '19 at 03:01