# Union of finite linearly independent subsets of eigenspaces, is a lin-indep subset.

Source: Linear Algebra by Friedberg et al. (4 edn 2002). p. 267.
This is germane to Linear Algebra by Lay (4 edn 2011). p. 285. Section 5.3. Theorem 7c.

$\bbox[,10px,border:4px solid green]{\text{Lemma}} \;$ Let $T$ be a linear operator, and let $\lambda_{1},\ \lambda_{2},\ \ldots,\ \lambda_{k}$ be distinct eigenvalues of T.
For each $i=1,2,\ \ldots,\ k$, let $v_{i}\in E_{\lambda;}$, the eigenspace corresponding to $\lambda_{i}$.
If $v_{1}+v_{2}+\cdots+v_{k}=0,$ then $\mathbf{ v_{i}=0 }$ for all $i$.

$\bbox[,10px,border:4px solid green]{\text{Prove by contradiction}} \;$ Suppose $\mathbf{ v_{i}\neq 0}$. By $\color{red}{\text{renumbering if necessary}}$, suppose that, for $1\leq m\leq k$, we have $\mathbf{ v_{i}\neq 0}$ for $1\leq i\leq m$, and $\mathbf{ v_{i}=0 }$ for $i>m$. Then [...] $v_{1}+v_{2}+\cdots+v_{m}=0.$ But this contradicts Theorem 5.5 which states that these $v_{i}$'s are linearly independent.

$1.$ Is there an easier proof, one without contradiction? Isn't a proof by contrapositive easier?
If $\mathbf{v_i \neq 0}$, then immediately we see that $v_{1}+v_{2}+\cdots+v_{m} \neq \mathbf{0}$?

$2.$ I don't understand $\color{red}{\text{renumbering if necessary}}$. Why'd this be necessary?

$\bbox[,10px,border:4px solid green]{\text{Theorem 5.8}} \,$ For each $i=1$, 2, $\ldots,\ k$, let $S_{i}$ be a finite linearly independent subset of the eigenspace $E(\lambda_i).$ Then $\cup_{1 \le i \le k} S_i$ is a linearly independent subset of $V$.

$\bbox[,10px,border:4px solid green]{\text{Direct Proof}} \;$ Suppose that for each $i$, $S_{i}=\{v_{i1},v_{i2},\ \ldots,v(i,n_i)\}.$ Then $S=\{v_{ij}:1\leq j\leq n_{i}$, and $1\leq i\leq k\}$. Consider any scalar $\{a_{ij}\}$ such that $\sum_{i=1}^{k} \; \color{#C154C1 }{ \sum_{j=1}^{n_{i}}a_{ij}v_{ij} } \mathbf{ = 0} \iff \color{#C154C1 }{ \sum_{j=1}^{n_{i}}a(1, j)v(1, j) + ... + \sum_{j=1}^{n_{i}}a(k, j)v(k, j) } \mathbf{ = 0}$ Then $\color{#C154C1 }{ \sum_{j=1}^{n_{i}}a_{ij}v_{ij} } \in E_{\lambda_{l}}$ for each $i$. By p. 267 Lemma above, $\color{#C154C1 }{ \sum_{j=1}^{n_{i}}a_{ij}v_{ij} } \mathbf{ =0 }$ for all $i$. But each $S_{i}$ is linearly independent, and hence $a_{ij}=0$ for all $j$.

$3.$ Why $\color{#C154C1}{ \sum_{j=1}^{n_{i}}a_{ij}v_{ij} } \in E_{\lambda_{l}}$ ? Because $v(i, j) \in E(\lambda_i)$ and eigenspaces are subspaces, and subspaces contain any linear combination of the vectors inside?

Question 1: Do induction on $k$.

If $k=1$, there's nothing to prove.

Suppose the assertion for $k$ eigenvalues and assume $$v_1+v_2+\dots+v_k+v_{k+1}=0$$ where $v_i$ is an eigenvector of $\lambda_i$ ($i=1,\dots,k+1$) and $\lambda_1,\dots,\lambda_{k+1}$ are pairwise distinct eigenvalues of $T$.

Evaluate the relation with $T$ to get $$\lambda_1v_1+\dots+\lambda_kv_k+\lambda_{k+1}v_{k+1}=0$$

Multiply the relation by $\lambda_{k+1}$ to get $$\lambda_{k+1}v_1+\dots+\lambda_{k+1}v_k+\lambda_{k+1}v_{k+1}=0$$

Subtract the two relations to get $$(\lambda_1-\lambda_k)v_1+\dots+(\lambda_k-\lambda_{k+1})v_k=0$$

By the induction hypothesis you conclude \begin{gather} (\lambda_1-\lambda_k)v_1=0\\ \dots\\ (\lambda_k-\lambda_k)v_k=0 \end{gather} because $v_i\in E_{\lambda_i}$ implies $\alpha v_i\in E_{\lambda_i}$ for all scalars $\alpha$, because $E_{\lambda_i}$ is a subspace. Therefore $$v_1=v_2=\dots=v_k=0$$ because $\lambda_i-\lambda_{k+1}\ne0$ ($i=1,2,\dots,k$), whence also $v_{k+1}=0$.

Having this, the rest is easy.

Question 3: yes, the eigenspaces are vector subspaces.

Question 2: if you have a sum $v_1+\dots+v_k$, you want to assume that some vector is nonzero, by contradiction. You can't know which ones are non zero, so you assume they're the first in the ordering, as the order in which the eigenvalues are presented is irrelevant. So it's not restrictive (by renumbering, if necessary), that $v_1,\dots,v_m\ne0$ and $v_{m+1},\dots,v_k=0$. It's a very common technique and avoids to say something like

suppose $v_{i_1}\ne0,\dots,v_{i_m}\ne0$ with $1\le i_1<i_2<\dots<i_m\le k$ and $v_j=0$ if $j\notin\{i_1,\dots,i_m\}$

that just complicates the proof.

• Thanks, but would you plase explain how this answers my questions 1, 2, 3? I'm not asking about the proof, but about the steps? – Greek - Area 51 Proposal May 16 '14 at 14:57
• @LePressentiment: Why did you edit this answer? – Asaf Karagila May 26 '14 at 14:46
• @AsafKaragila To remove some white space which spaced apart the answer and to elucidate the steps – Greek - Area 51 Proposal May 26 '14 at 14:48
• @LePressentiment: Won't you stipulate to excavate it from thine erupting mind, that it is not necessarily, and in fact quite ostensibly, no impolite to manufacture discontinuity in the style and emblem of another? – Asaf Karagila May 26 '14 at 14:52
• @LePressentiment Please, avoid such edits on my posts in the future. – egreg May 26 '14 at 17:23

I tender some different steps, thanks to egreg.

Question 1: Do induction on $k$. If $k=1$, there's nothing to prove.

Suppose the assertion is true for $k$ eigenvalues.

To prove the $k = n + 1$ case, assume $v_1+v_2+\dots+v_k+v_{k+1}=0 \quad (*)$
where $v_i$ is an eigenvector of $\lambda_i$ ($i=1,\dots,k+1$) and $\lambda_1,\dots,\lambda_{k+1}$ are pairwise distinct eigenvalues of $T$. Now the trick is to apply $T - \lambda_k I$ to (*).

Evaluate the relation with $T$: $\quad \lambda_1v_1+\dots+\lambda_kv_k+\lambda_{k+1}v_{k+1}=0 $$Multiply the relation by \lambda_{k+1}: \quad \lambda_{k+1}v_1+\dots+\lambda_{k+1}v_k+\lambda_{k+1}v_{k+1}=0 Subtract the two relations: \quad (\lambda_1-\lambda_k)v_1+\dots+(\lambda_k-\lambda_{k+1})v_k=0 \qquad (\clubsuit) Because E_{\lambda_i} is a subspace, thus v_i\in E_{\lambda_i} implies \alpha v_i\in E_{\lambda_i} for all scalars \alpha. Therefore by the induction hypothesis,$$ v_1=v_2=\dots=v_k=0$$Because$\lambda_i-\lambda_{k+1}\ne0$($i=1,2,\dots,k)$,$(\clubsuit) \implies \begin{gather} (\lambda_1-\lambda_k)v_1=0\\ \dots\\ (\lambda_k-\lambda_k)v_k=0 \end{gather}$Finally$(*)$becomes:$v_{k+1}=0\$.