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I've found this proof of the independence of eigenspaces in a textbook and I can't reproduce it myself.

Theorem. Let $\varphi: L \rightarrow L$ be a linear operator on a linear space L. If numbers ${\lambda}_{1},...,{\lambda}_{k}$ are different, then the subspaces $\ker(\varphi - {\lambda}_{i}\varepsilon)$ are independent.

Proof.

  1. Suppose that nonzero vectors ${x}_{i} \in \ker(\varphi - {\lambda}_{i}\varepsilon)$ satisfy ${x}_{1} + ... + {x}_{k}=0$.
  2. Let us apply the operators $(\varphi - {\lambda}_{i}\varepsilon)$ to the identity above for $i = 2, 3,...,k$.
  3. Then we get $({\lambda}_{1}-{\lambda}_{2})\cdot({\lambda}_{1}-{\lambda}_{3})\cdot...\cdot({\lambda}_{1}-{\lambda}_{k}){x}_{1}$
  4. Since the numbers ${\lambda}_{i}$ are different from the identity above we get ${x}_{1} = 0$. That is a contradiction, hence the subspaces are independent.

Well, first of all I don't quite get what the second step means. Does it mean a composition of all these operators is applied to the equation, i.e. $(\varphi - {\lambda}_{k}\varepsilon)\circ...\circ(\varphi - {\lambda}_{2}\varepsilon)({x}_{1}+...+{x}_{k})=0$? If so, how do we get ${\lambda}_{1}$ in the third step if we don't apply $(\varphi - {\lambda}_{1}\varepsilon)$? Thank you in advance.

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There's something not really precise in point 1, which should be

Suppose that $x_1+x_2+\dots+x_k=0$ where at least one of the vectors $x_1,x_2,\dots,x_k$ is nonzero; without loss of generality we can assume $x_1\ne0$.

Now follow the hint; if we apply $\varphi-\lambda_2\varepsilon$ we get $$ \varphi(x_1)-\lambda_2x_1+ \varphi(x_2)-\lambda_2x_2+ \varphi(x_3)-\lambda_2x_3+ \dots+ \varphi(x_k)-\lambda_2x_k =0 $$ that, in view of the properties of the vectors, becomes $$ \lambda_1x_1-\lambda_2x_1+ \lambda_3x_3-\lambda_2x_3+ \dots+ \lambda_kx_k-\lambda_2x_k =0 $$ or $$ (\lambda_1-\lambda_2)x_1+(\lambda_3-\lambda_2)x_3+\dots+ (\lambda_k-\lambda_2)x_k=0 $$ Upon applying $\varphi-\lambda_3\varepsilon$, the term with $x_3$ vanishes and the term with $x_1$ is multiplied by $\lambda_1-\lambda_3$.

Proceeding in the same way, we're indeed left with $$ (\lambda_1-\lambda_k)\dots(\lambda_2-\lambda_1)x_1=0 $$ and this implies $x_1=0$, which is a contradiction.


I prefer a different kind of proof, by induction on $k$. The result is obvious for $k=1$, so assume we know it for $k-1$ eigenvalues. Suppose $$ x_1+\dots+x_{k-1}+x_k=0 $$ with $x_i\in\ker(\varphi-\lambda_i\varepsilon)$ for $i=1,2,\dots,k$. Apply $\varphi$ to the relation, getting $$ \lambda_1x_1+\dots+\lambda_{k-1}x_{k-1}+\lambda_kx_k=0 $$ and also multiply the relation by $\lambda_k$: $$ \lambda_kx_1+\dots+\lambda_kx_{k-1}+\lambda_kx_k=0 $$ Now subtract getting $$ (\lambda_1-\lambda_k)x_1+\dots+(\lambda_{k-1}-\lambda_k)x_{k-1}=0 $$ By the induction hypothesis, $$ (\lambda_1-\lambda_k)x_1=0,\quad \dots,\quad (\lambda_{k-1}-\lambda_k)x_{k-1}=0 $$ and since $\lambda_i-\lambda_k\ne0$ for $i=1,2,\dots,k-1$, we get $$ x_1=0,\quad \dots,\quad x_{k-1}=0 $$ which also implies $x_k=0$.

The proof is direct, as you see.

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  • $\begingroup$ Thank you for this detailed explanation. $\endgroup$ – Eli Korvigo Jun 13 '15 at 21:00
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Yes the second step is what you think. Then to understand step 3) just keep in mind that $x_1$ was assumed to be in $\ker(\varphi - \lambda_1 \epsilon)$ i.e. $\varphi (x_1) = \lambda_1 x_1$.

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  • $\begingroup$ Thank you for your answer. Your final note helped me sort the idea in my mind. I accepted the answer by @egreg, because he provided a detailed explanation. $\endgroup$ – Eli Korvigo Jun 13 '15 at 20:59
  • $\begingroup$ Indeed @egreg answer put in evidence something not really precise so you did well in accepting his answer. There is also a nice proof of the independence by using the Vandermonde matrix en.wikipedia.org/wiki/Vandermonde_matrix $\endgroup$ – Holonomia Jun 13 '15 at 21:06

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