Trouble with eigenvalue proof Hello I am having a lot of trouble trying to prove a statement in linear algebra,
il post it and what I have tried etc,
Let $A \in \mathbb M_{nxn}$ and let $\lambda_1,...,\lambda_r$ be distinct eigenvalues of A,
Let $S_1=\{v_1^1,...,v_{k1}^1\}, ..., S_r=\{v_1^r,...,v_{kr}^r\}$ and suppose that the $S_i$ are linearly independent for $i=1,..,r$ and that $S_i \subset E_{\lambda i}$ (eigenspace)
Show that the set $\{v_1^1,...,v_{k1}^1,....,v_1^r,...v_{kr}^r\}$ is linearly independent.
Overall I am really confused, I do know that and how to prove that eigenvectors are linearly independent if they all have distinct eigenvalues. But I am not sure how to compare it with this, secondly, all the notation is a little confusing and I am having trouble even understand what I need to show. For example what is the diffirence between $v_1^1$ and $v_1^r$?
I think it can maybe be done with showing all the coefficients are zero.
Any suggestions or hints to this?
Thanks
 A: It's easier if you split the proof into two parts.
Theorem. If $x_1\in E_{\lambda_1},x_2\in E_{\lambda_2},\dots,x_k\in E_{\lambda_k}$ for pairwise distinct eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_k$ and $x_1+x_2+\dots+x_k=0$, then $x_1=x_2=\dots=x_k=0$.
Proof. Induction on $k$. The case $k=1$ is obvious. Suppose the result is true for $k-1$ vectors. Then we have
\begin{gather}
0=A(x_1+x_2+\dots+x_k)=
\lambda_1x_1+\lambda_2x_2+\dots+\lambda_kx_k
\\
0=\lambda_k(x_1+x_2+\dots+x_k)=
\lambda_kx_1+\lambda_kx_2+\dots+\lambda_kx_k
\end{gather}
Subtracting the two relations, we get
$$
0=(\lambda_1-\lambda_k)x_1+(\lambda_2-\lambda_k)x_2+\dots+(\lambda_{k-1}-\lambda_k)x_{k-1}
$$
and, by the induction hypothesis, we get
$$
(\lambda_1-\lambda_k)x_1=0\\
(\lambda_2-\lambda_k)x_2=0\\
\dots\\
(\lambda_{k-1}-\lambda_k)x_{k-1}=0
$$
So $x_1=x_2=\dots=x_{k-1}=0$ and hence also $x_k=0$. QED
Now you should be able to conclude, by invoking the theorem above. Instead of the complicated notation, write
\begin{align}
S_1&=\{v_{1,1},\dots,v_{1,k_1}\},\\
\dots,\\
S_r&=\{v_{r,1},\dots,v_{r,k_r}\}
\end{align}
A zero linear combination can be written
$$
\underbrace{\alpha_{1,1}v_{1,1}+\dots+\alpha_{1,k_1}v_{1,k_1}}_{\in E_{\lambda_1}}
+
\underbrace{\alpha_{2,1}v_{2,1}+\dots+\alpha_{2,k_2}v_{2,k_2}}_{\in E_{\lambda_1}}
+
\dots
+
\underbrace{\alpha_{r,1}v_{r,1}+\dots+\alpha_{r,k_r}v_{r,k_r}}_{\in E_{\lambda_r}}
=0
$$
and you can set
\begin{align}
x_1&=\alpha_{1,1}v_{1,1}+\dots+\alpha_{1,k_1}v_{1,k_1}\\
x_2&=\alpha_{2,1}v_{2,1}+\dots+\alpha_{2,k_2}v_{2,k_2}\\
\dots\\
x_r&=\alpha_{r,1}v_{r,1}+\dots+\alpha_{r,k_r}v_{r,k_r}
\end{align}
so invoking the above theorem and then the linear independence of the subsets you're done.
A: Suppose $\sum_{k,j} r_{kj}v_{kj} = 0$. Rewrite this sum as
\begin{equation*}
  \sum_j\left(\sum_k r_{kj}v_{kj}\right).
\end{equation*}
Each term in the outer sum is a sum of elements of $S_j$, so it is in $S_j$. But the $S_j$ are linearly independent subspaces since they correspond to different eigenvalues, so each sum is zero. Now use linear independence of $v_{1j}, \dotsc, v_{kj}$.
A: Hint
$\alpha^i_1v^i_1+\cdots +\alpha^i_{k_i}v^i_{k_i}$ is a $\lambda_i$-eigenvalue of $A$ for any $\alpha^i_1, \cdots,\alpha^i_{k_i}\in \mathbb{R}.$
