I'm learning linear algebra from Linear Algebra Done Right. The book gives a proof that the degree of the minimal polynomial of an operator on V is at most the dimension of V, but the proof feels unintuitive and doesn't really help me develop intuition for why this is true or what the minimal polynomial is beyond the definition. Is there a more intuitive way to explain it without using terms I haven't learned yet such as characteristic polynomials? You can see the proof given by the book on this link on page 144.
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Will Jagy
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In a general vector space with dimension n you can’t have more than n independent vectors. This is because, in an n-dimensional space, any set of more than n vectors must be linearly dependent. Now consider a linear operator O acting on V. If you apply O repeatedly to V, you generate a sequence of vectors. Since V has dimension n, after n applications these vectors will become linearly dependent which implies that there is a polynomial of degrees at most n that O satisfies.
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2This is not correct. You only get that $p(A) v = 0$ for some polynomial $p$ of degree $\le n$, which doesn't imply that $p(A) = 0$. This can be turned into a complete proof but it requires some more work. – Qiaochu Yuan Aug 19 '24 at 00:09
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Thank you for the clarification. You are correct that the argument I provided shows that for each vector v in V, there exists a polynomial p(x) of degree at most n such that p (T)v =0. However, to conclude that p(T)=0 as an operator on the entire space V, we need to consider that this polynomial can be constructed for any vector in V. Since V is finite-dimensional, we can find a common min polynomial that works for all vectors in V and still has degree at most n. This common polynomial is the min polynomial of the operator. Thank you for pointing out the need to make this distinction clear. – Kurt McKenzie Aug 19 '24 at 00:54
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1@KurtMcKenzie Your assertion that $V$ being finite dimensional implies that there exists a common polynomial which works for all $v$ is essentially the statement of the theorem, so you still haven't provided a proof. – Max Chien Aug 19 '24 at 00:57
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1Thank you, Max! For each vector v in V, there is a polynomial p(x) of degree at most n such that p(T)v =0. Although these p(x)’s might be different for different vectors, since V is finite dimensional the set of all these polynomials forms an ideal in the ring of polynomials. The fact that this Ideal is generated by a single polynomial means there exist one polynomial that works for V. This is what we call the minimal polynomial and it has degree at most n, because it must divide any p(x). So the reason there is a common p(x) isn't just that the space is finite-dimensional. – Kurt McKenzie Aug 19 '24 at 01:32
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@KurtMcKenzie Could you elaborate a bit more on this? I could be misunderstanding, but I think your current statement is a bit incorrect. If we let $$A=\begin{bmatrix}1&0&0\0&1&1\0&0&1\end{bmatrix}$$and take $v=(1,0,0)$, then the polynomial $p(x)=x-1$ works to annihilate $v$, yet the minimal polynomial is $(x-1)^2$. So I'm not sure how it's the case that the minimal polynomial divides all such $p$. I'd be interested in if there is a way to get around this, though. – Max Chien Aug 19 '24 at 04:35
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Thx, Max, for the example. This is a valid point! The polynomial p(x) = x-1 annihilates v, but the minimal polynomial for the entire matrix A is ( x - 1)^2, showing that while individual vectors can have annihilating polynomials of lower degree, the minimal polynomial is the smallest polynomial that works for V. So the minimal polynomial must divide any polynomial that annihilated the operator on the entire space but this doesn't mean that every annihilating polynomial for individual vectors is divisible by the minimal polynomial. – Kurt McKenzie Aug 19 '24 at 05:06
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@KurtMcKenzie Right, that's kind of my point. Because no individual vector gives you a polynomial which we know the minimal polynomial must divide, I don't see how this method proves that the minimal polynomial has degree $\leq n$ without actually exhibiting some polynomial where $p(A)=0$. – Max Chien Aug 19 '24 at 12:29
"A nonzero polynomial has at most as many distinct zeros as its degree (see 4.8). Thus (a) of the previous result, along with the result that the minimal polynomial of an operator on has degree at most dim , gives an alternative proof of 5.12, which states that an operator on has at most dim distinct eigenvalues"
– qqo Aug 18 '24 at 23:36