- #1

Eclair_de_XII

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- TL;DR Summary
- 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.)

\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.)

Last edited: