# Generalized Eigenspaces Associated to Different Values

Let $$V$$ be some vector space, and $$T \in \mathcal{L}(V)$$. If $$a \neq b$$, then $$G(a,T) \cap G(b,T) = \{0\}$$, where $$G(b,T)$$ denotes the generalized eigenspace.

I am having a lot of trouble with this problem; I've been thinking about it for the past few days with no progress. Here is what I've come up with:

If $$v \in G(\alpha,T) \cap G(\beta,T)$$, then there $$i,j \in \Bbb{N}$$ such that $$(T-aI)^i v =0$$ and $$(T-bI)^j v = 0$$. WLOG, let $$i$$ and $$j$$ be the smallest such integers and suppose $$i < j$$. Then

$$w_1 := (T-aI)^{i-1}v \neq 0$$

and

$$w_2 := (T-aI)^{j-1} \neq 0$$

Note that $$(T-aI)w_1 =0$$ and therefore $$Tw_1 = aw_1$$; similarly, $$Tw_2 = bw_2$$...

I've played with this equations in various ways. My strategy = is to get $$(\mbox{something non-zero}) \cdot v = 0$$, which will obviously force $$v=0$$; but this is proving extremely difficult. My hope is that I can get $$(a-b)v=0$$, but I don't see how to do it.

I could use some help...

Following your set up, let $$v\in G(a,T)\cap G(b,T)$$ and let $$i$$ be the smallest number such that $$(T-aI)^iv=0$$. Let $$w=(T-aI)^{i-1}v\neq 0$$. Then $$(T-aI)w=0$$ and so $$Tw=av$$. As such for $$\lambda \in \mathbb{F}$$ and $$n\in \mathbb{N}$$, $$(T-\lambda I)^nw=(a-\lambda)^n w$$. Now let $$j\in\mathbb{N}$$ such that $$(T-bI)^jv=0$$. Now, $$(T-aI)^{i-1}(T-bI)^jv=0$$. However, both of these are polynomials of $$T$$ and therefore commute (if you are using Linear Algebra Done Right by Axler as I suspect you are this is 5.20 on page 144). As such, $$0=(T-bI)^j(T-aI)^{i-1}v=(T-bI)^jw=(a-b)^jw$$. However, that implies that $$a=b$$ which is a contradiction. As such, no such $$v$$ exists. (If you are using Axler this proof is modelled off of Axler's proof of 8.13).

• This is a much more straightforward proof than the one I gave. Oct 24 '18 at 10:28

I guess the cheapest way is the Bezout identity. The polynomials with coefficients in a field, say the reals or the complexes, form a Euclidean ring. The units in the ring are the nonzero constants, i.e. the nonzero field elements.

Added: after flipping through Dummit and Foote, let me add that the "Bezout" stuff is the theorem that any Euclidean Domain is also a Principal Ideal Domain. In the paragraphs below, this means that the ideal in $$F[x]$$ generated by $$(x-a)^m$$ and $$(x-b)^n$$ must be the entire ring, in particular contain the constant polynomial $$1.$$

For univariate polynomials f and g with coefficients in a field, there exist polynomials a and b such that af + bg = 1 if and only if f and g have no common root in any algebraically closed field (commonly the field of complex numbers).

Given $$a \neq b,$$ we have $$x-a$$ and $$x-b$$ coprime. In particular, $$b-a \neq 0$$ and $$(x-a) - (x-b) = b-a$$ is a constant. Being a bit more careful, $$\frac{x-a}{b-a} - \frac{x-b}{b-a} = 1$$

Anyway, $$(x-a)^m$$ and $$(x-b)^n$$ are coprime as well. Let me check if I know a quick proof for that, there may be a one-liner. (For the moment, I just think that they have no roots in common i any field extension, all roots are accounted for, so no common factor can have any roots, i.e. is a constant). We will have polynomials $$p,q$$ such that $$p(x) (x-a)^m - q(x) (x-b)^n = 1.$$ We switch this back to $$\color{red}{ p(T) (T-aI)^m - q(T) (T-bI)^n = I.}$$ Therefore, if $$(T - aI)^m v = 0$$ and $$(T - bI)^n v = 0,$$ it follows that $$Iv=0$$ and $$v=0.$$

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I did a sample Bezout for $$(x-1)^5$$ and $$(x-2)^3.$$ The final outcome was

$$\left( x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x + 1 \right) \left( \frac{ - 5 x^{2} + 5 x + 19 }{ 27 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ - 5 x^{4} + 4 x^{2} - 11 x + 1 }{ 27 } \right) = \left( 1 \right)$$

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$$\left( x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x + 1 \right)$$

$$\left( x^{3} - 6 x^{2} + 12 x - 8 \right)$$

$$\left( x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x + 1 \right) = \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \cdot \color{magenta}{ \left( x^{2} + x + 4 \right) } + \left( 10 x^{2} - 35 x + 33 \right)$$ $$\left( x^{3} - 6 x^{2} + 12 x - 8 \right) = \left( 10 x^{2} - 35 x + 33 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 5 }{ 20 } \right) } + \left( \frac{ - x + 5 }{ 20 } \right)$$ $$\left( 10 x^{2} - 35 x + 33 \right) = \left( \frac{ - x + 5 }{ 20 } \right) \cdot \color{magenta}{ \left( - 200 x - 300 \right) } + \left( 108 \right)$$ $$\left( \frac{ - x + 5 }{ 20 } \right) = \left( 108 \right) \cdot \color{magenta}{ \left( \frac{ - x + 5 }{ 2160 } \right) } + \left( 0 \right)$$ $$\frac{ 0}{1}$$ $$\frac{ 1}{0}$$ $$\color{magenta}{ \left( x^{2} + x + 4 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{2} + x + 4 \right) }{ \left( 1 \right) }$$ $$\color{magenta}{ \left( \frac{ 2 x - 5 }{ 20 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x^{3} - 3 x^{2} + 3 x }{ 20 } \right) }{ \left( \frac{ 2 x - 5 }{ 20 } \right) }$$ $$\color{magenta}{ \left( - 200 x - 300 \right) } \Longrightarrow \Longrightarrow \frac{ \left( - 20 x^{4} + 16 x^{2} - 44 x + 4 \right) }{ \left( - 20 x^{2} + 20 x + 76 \right) }$$ $$\color{magenta}{ \left( \frac{ - x + 5 }{ 2160 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x + 1 }{ 108 } \right) }{ \left( \frac{ x^{3} - 6 x^{2} + 12 x - 8 }{ 108 } \right) }$$ $$\left( x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x + 1 \right) \left( \frac{ - 5 x^{2} + 5 x + 19 }{ 27 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ - 5 x^{4} + 4 x^{2} - 11 x + 1 }{ 27 } \right) = \left( 1 \right)$$

• Hmm...Unfortunately I think this solution might be a bit beyond the scope of the course I'm enrolled in. E.g., we haven't discussed polynomials being coprime, field extensions, Bezout's Identity. Oct 23 '18 at 21:02

The proof I know of this follows from a bunch of general theory about finitely-generated modules over principal ideal domains. If that means something to you, definitely check it out. Otherwise, I will try to boil out the important points. You do need to know something about polynomial divisibility though, there is no getting around that.

Let $$T: V \to V$$ be a linear transformation of the finite-dimensional vector space $$V$$, and $$k[x]$$ the set of polynomials over the same field $$k$$. Given a polynomial $$p \in k[x]$$, we can create the linear map $$p(T)$$ in the usual way.

Given a polynomial $$p \in k[x]$$ we define $$V_p := \{v \in V \mid p(T)v = 0\}$$, the subspace of $$V$$ of vectors killed by $$p$$. For example, if $$p(x) = (x - \lambda)$$, then $$V_p$$ is the $$\lambda$$-eigenspace of $$T$$. If $$p(x) = (x - \lambda)^n$$ for $$n$$ large enough, then $$V_p$$ will be the generalised eigenspace for $$\lambda$$.

So we can re-state your problem: if $$p(x) = (x - \lambda)^n$$ and $$q(x) = (x - \mu)^m$$ for some $$n, m \geq 1$$ and $$\lambda \neq \mu$$, then why is $$V_p \cap V_q = 0$$? The polynomials $$p, q$$ are coprime, i.e. they share no factors. For any two coprime polynomials $$p, q$$, there exist two other polynomials $$r, s$$ such that $$rp + sq = 1$$. Plugging $$T$$ into this identity, we get $$r(T)p(T) + s(T)q(T) = \mathrm{id}_V$$. Now suppose $$v \in V_p \cap V_q$$, then by the above identity we have $$v = r(T)p(T)v + s(T)q(T)v = 0$$

The only non-obvious part in the above is the assertion that the polynomials $$r, s$$ exist. You might have seen a proof of this before in the case of integers, where the extended Euclidean algorithm is used to create the integers $$r, s$$. The proof for polynomials is exactly the same, and the algorithm still works. But perhaps you can learn about that from somewhere else.