Proving the uniqueness of eigenspaces

[Eclair_de_XII]

TL;DR

Let $A$ be a linear transformation that maps some vector space to itself. Let $\lambda,\mu$ be two distinct scalars and define $T_k:=A-k\cdot I$, where $I$ denotes the identity transformation. Show that the following hold:

\begin{eqnarray}
\ker(T_\lambda^2)\cap\ker(T_\mu^2)=0
\end{eqnarray}

Let $x\in\ker(T_\lambda^2)\cap\ker(T_\mu^2)$. Then the following must hold:

\begin{eqnarray}
(A^2-2\lambda\cdot A+\lambda^2I)x=0\\
(A^2-2\mu\cdot A+\mu^2I)x=0
\end{eqnarray}

Subtracting the latter equation from the former gives us:

\begin{eqnarray}
0-0&=&0\\
&=&(-2\lambda\cdot Ax+\lambda^2x)-(-2\mu\cdot Ax+\mu^2x)\\
0&=&-2(\lambda-\mu)Ax+(\lambda-\mu)(\lambda+\mu)x
\end{eqnarray}

Bearing in mind that performing row operations on any system of linear equations does not change its solution set, we proceed by dividing both sides of the equality by $\lambda-\mu$:

\begin{eqnarray}
0&=&-2Ax+(\lambda+\mu)x\\
&=&-2Ax+(\lambda+\lambda+(\mu-\lambda)x\\
&=&-2Ax+2\lambda\cdot x+(\mu-\lambda)x\\
&=&2(\lambda\cdot I-A)x+(\mu-\lambda)x\\
&=&2T_\lambda x+(\mu-\lambda)x
\end{eqnarray}

By definition, $x\in\ker(T_\lambda+\frac{1}{2}(\lambda-\mu)I)$ whenever $I$ is restricted to $\ker(T_\lambda^2)\cap\ker(T_\mu^2)$. It can be shown through similar means that $x\in\ker(T_\mu+\frac{1}{2}(-\lambda+\mu)I)$.

(I am still figuring out how to finish this proof, if it can be finished.)

 

[mathwonk]

the problem is unnecessarily complicated. denote A-lambda.Id by B, and then denote A-mu.Id by B-c.Id, where c= mu-lambda ≠0. then try to show that if B^2 x=0 and also (B-c)^2 x = 0, where c≠0, then x = 0.

 

[Eclair_de_XII]

Sure. I show first that $(B-c)^2$ can be expanded like a polynomial.

\begin{eqnarray}
(B-cI)^2&=&[B^2-B(cI)-(cI)B+c^2I]\\
&=&[B^2-c(BI)-c(BI)+c^2I]\\
&=&[B^2-2c(BI)+c^2I]\\
\end{eqnarray}

Let $x\in\ker(B^2)\cap\ker[(B-c)^2]$.

\begin{eqnarray}
0&=&B^2x\\
&=&(B-cI)^2x\\
&=&0-0\\
&=&(2cB-c^2I)x\\
&=&(2B-cI)x\\
&=&(B+B-cI)x\\
-Bx&=&(B-cI)x
\end{eqnarray}

By uniqueness of the inverse, either $B-cI=-B$ or $x=0$. The former fails to hold if $B\neq\frac{1}{2}cI$, in which case, $x=0$.

If this solution were correct, I am wondering how I would show this is true (if it is true) for higher powers. I don't like thinking about the amount of algebra I'd need to do to prove it in the general case like how I did it for the case when the power is just two.

 

[mathwonk]

or, applying B again at step 19 implies cBx= 0, so Bx = 0, so from step 18, also c^2 x = 0, so x = 0.

a basic algebra fact is that in the integers and in the polynomial ring over a field, and in any euclidean domain, the greatest common divisor of two elements can be written as a linear combination of those elements. Hence if P,Q are relatively prime polynomials, so their gcd is 1, there are polynomials f, g, such that fP+gQ = 1. Then for any map A, plugging A into both sides of this equation, gives f(A)P(A) + g(A)Q(A) = Id. Now assuming P(A)x = Q(A)x = 0, gives that 0 = x.

In your original problem, P = (X-lambda)^2, and Q = (X-mu)^2.

 

[Eclair_de_XII]

Thank you. That is very helpful.

applying B again at step 19 implies cBx= 0

Will that not change the solution set, though, if for example, $B$ is singular? Wait, never mind. I think that the new solution set obtained from applying $B$ to both sides would just be a superset of the original solution set. And since the new solution set is trivial, it must follow that the original solution set is also trivial.

 

[mathwonk]

using this idea on your original problem directly, note that if P = X-a, and Q = X-b, with a ≠b, then P-Q = b-a ≠0, so we can take f = 1/(b-a) and g = -1/(b-a), to get fP+gQ = 1.

Then to do the case of P^2 and Q^2, just cube both sides of this equation. Every term on the left will be divisible by either P^2 or Q^2, and the right side will still equal 1. So we get some equation of form FP^2 + GQ^2 = 1. the same trick works for all higher powers. I.e. to deal with P^n and Q^n, raise the equation to the power 2n-1, I guess.

 

[mathwonk]

i don't understand your concern in post #5, since if B is not singular, you are done immediately, so we may assume if we wish that B is singular, but that is irrelevant to the argument. I am just trying to prove that x = 0. from step 19 we know that 2Bx = cx, so applying B again gives 2B^2x = cBx, but B^2 x = 0, by hypothesis, so cBx = 0. but c≠0 so Bx = 0. now either step 18 or 19 gives that 0 = cx or 0 = c^2x, either way x = 0.