If $T$ has $k$ distinct eigenvalues and $S_i$ (set of eigenvectors) is linearly independent, then $S_1 \cup \ldots \cup S_k$ is linearly independent

Question

Consider the following proof:

Theorem 5.5. Let $\mathsf{T}$ be a linear operator on a vector space, and let $\lambda_1, \lambda_2, \ldots, \lambda_k$ be distinct eigenvalues of $\mathsf{T}$. For each $i = 1, 2, \ldots, k$, let $S_i$ be a set of eigenvectors of $\mathsf{T}$ corresponding to $\lambda_i$. If each $S_i \, (i = 1, 2, \ldots, k)$ is linearly independent, then $S_1 \cup S_2 \cup \cdots \cup S_k$ is linearly independent.

Proof. The proof is by mathematical induction on $k$. If $k = 1$, there is nothing to prove. So assume that the theorem holds for $k - 1$ distinct eigenvalues, where $k - 1 \geq 1$, and that we have $k$ distinct eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_k$ of $\mathsf{T}$. For each $i = 1, 2, \ldots, k$, let $S_i = \{v_{i1}, v_{i2}, \ldots, v_{in_i} \}$ be a linearly independent set of eigenvectors of $\mathsf{T}$ corresponding to $\lambda_i$. We wish to show that $S_1 \cup S_2 \cup \cdots \cup S_k$ is linearly independent.
Consider any scalars $\{a_{ij}\}$, where $i = 1, 2, \ldots, k$ and $j = 1, 2, \ldots, n_i$, such that $$ \sum_{i=1}^{k} \sum_{j=1}^{n_i} a_{ij} v_{ij} = 0. \tag{1} $$ Because $v_{ij}$ is an eigenvector of $\mathsf{T}$ corresponding to $\lambda_i$, applying $\mathsf{T} - \lambda_k I$ to both sides of $(1)$ yields $$\sum_{i=1}^{k-1} \sum_{j=1}^{n_i} a_{ij} (\lambda_i - \lambda_k) v_{ij} = 0. \tag{2} $$ But $S_1 \cup S_2 \cup \cdots \cup S_{k-1}$ is linearly independent by the induction hypothesis, so that $(2)$ implies $a_{ij} (\lambda_i - \lambda_k) = 0$ for $i = 1, 2, \ldots, k - 1$ and $j = 1, 2, \ldots, n_i$. Since $\lambda_1, \lambda_2, \ldots, \lambda_k$ are distinct, it follows that $\lambda_i - \lambda_k \neq 0$ for $1 \leq i \leq k-1$. Hence $a_{ij} = 0$ for $i = 1, 2, \ldots, k - 1$ and $j = 1, 2, \ldots, n_i$, and therefore $(1)$ reduces to $\sum_{j=1}^{n_k} a_{kj} v_{kj} = 0$. But $S_k$ is also linearly independent, and so $a_{kj} = 0$ for $j = 1, 2, \ldots, n_k$. Consequently $a_{ij} = 0$ for $i = 1, 2, \ldots, k$ and $j = 1, 2, \ldots, n_i$, proving that $S$ is linearly independent.

I've been looking at this proof for too long and don't get how the author went from Equation $(1)$

$$\sum_{i=1}^k \sum_{j=1}^{n_i} a_{ij}v_{ij} = 0 \tag{1}$$

to Equation $(2)$

$$\sum_{i=1}^{k-1} \sum_{j=1}^{n_i} a_{ij}(\lambda_i - \lambda_k) v_{ij} = 0 \tag{2}.$$

I think I just don't get what applying $\mathsf{T} - \lambda_k I$ on the sum means.

Answer 1

By definition of an eigenvector/eigenvalue pair,

$$ Tv_{ij}=\lambda_i v_{ij}$$

so $(T-\lambda_k I)v_{ij}=\lambda_i v_{ij}-\lambda_k v_{ij}$

Linearity allows us to assert that applying $(T-\lambda_k I)$ to each term is the same as applying it to the sum as a whole, while any matrix left-multiplied by the 0 matrix on the RHS yields 0.

Answer 2

The full expansion (hopefully correct) after seeing the correct answer from A.M. I wanted to share since I struggled with it \begin{align*} (T-\lambda_kI)(\sum_{i=1}^k \sum_{j=1}^{n_i} a_{ij}v_{ij}) &= 0 \\ \sum_{i=1}^k \sum_{j=1}^{n_i} T-\lambda_kI(a_{ij}v_{ij}) &= 0 \quad \text{ (by the linearity of $T$)}\\ \sum_{i=1}^k \sum_{j=1}^{n_i} T(a_{ij}v_{ij}) - \lambda_k a_{ij} v_{ij} &= 0 \\ \sum_{i=1}^k \sum_{j=1}^{n_i} \lambda_i a_{ij}v_{ij} - \lambda_k a_{ij} v_{ij} &= 0 \quad \text{ ($v_{ij}$ is an eigenvector)}\\ \sum_{i=1}^k \sum_{j=1}^{n_i} a_{ij} (\lambda_i - \lambda_k) v_{ij} &= 0\\ \sum_{j=1}^{n_k} a_{kj} (\lambda_k - \lambda_k) v_{kj} + \sum_{i=1}^{k-1} \sum_{j=1}^{n_i} a_{ij} (\lambda_i - \lambda_k) v_{ij} &= 0\\ \sum_{i=1}^{k-1} \sum_{j=1}^{n_i} a_{ij} (\lambda_i - \lambda_k) v_{ij} &= 0\\ \end{align*}