# Show that the set $\{v_1, \dots, v_n\}$ is linearly independent

I have the following question

Suppose an $$n \times n$$ matrix $$A$$ has $$n-1$$ distinct eigenvalues $$\{\lambda_1, \lambda_2, \dots, \lambda_{n-1}\}$$ and that the eigenvalue $$\lambda_{n-1}$$ has algebraic multiplicity $$2$$ and geometric multiplicity $$1$$. The corresponding eigenvector set is $$\{v_1, v_2, \dots, v_{n-1}\}$$. Suppose now there is a vector $$v_n$$ such that $$(A - \lambda_{n-1}I)v_n = v_{n-1}$$. Show that the set $$\{v_1, \dots, v_n\}$$ is linearly independent.

I have no idea on how to proceed on this. Any suggestions would be helpful. Thanks in advance!!!

EDIT I know how to prove $$\{v_1, \dots, v_{n-1}\}$$ are linearly independent but having this knowledge of linear independence of $$\{v_1, \dots, v_{n-1}\}$$, how do I incorporate $$v_n$$ into the set?

Here's a proof of the fact that $$\{v_1, \dots, v_{n-1}\}$$ are linearly independent.

All we need to prove is that eigenvectors corresponding to different eigenvalues are linearly independent. Let $$P = \begin{bmatrix} \mathbb{v}_1 \mathbb{v}_2 \dots \mathbb{v}_n\end{bmatrix}_{n \times n}$$ be a matrix of eigenvectors. Let the eigen decomposition of $$A$$ be $$A = P\Sigma P^{-1}$$ and let WLOG, $$\lambda_1 > \lambda_2 > \dots > \lambda_{n-1} = \lambda_n$$. So, $$\Sigma = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_{n-1}, \lambda_n)$$. Let $$\mathbb{c} = \begin{bmatrix} c_1 & c_2 & \dots & c_{n-1} & c_n \end{bmatrix}^T$$. Since $$\lambda_{n-1}$$ has geometric multiplicity of $$1$$, we have $$v_n = \alpha \mathbb{v}_{n-1}$$, $$\alpha \in \mathbb{R}\setminus \{0\}$$, We need to show, \begin{align*} c_1\mathbb{v_1} + c_2\mathbb{v_2} + \dots + c'_{n-1}\mathbb{v_{n-1}} + c'_n\mathbb{v_n} \\ = c_1\mathbb{v_1} + c_2\mathbb{v_2} + \dots + c'_{n-1}\mathbb{v_{n-1}} + c'_n(\alpha v_{n-1}) \\ = c_1\mathbb{v_1} + c_2\mathbb{v_2} + \dots + (c'_{n-1} + \alpha c'_n)\mathbb{v_{n-1}} \\ = c_1\mathbb{v_1} + c_2\mathbb{v_2} + \dots + c_{n-1}\mathbb{v_{n-1}} = 0 \\ \Rightarrow c_1 = c_2 = \dots = c_{n-1} = 0 \end{align*} Applying $$A\mathbb{v_i} = \lambda_i\mathbb{v_i}$$ we get, $$$$\label{eq3} c_1\lambda_1\mathbb{v_1} + \dots + c_n\lambda_n\mathbb{v_n} = 0$$$$ We can re-write the above equation in matrix form as $$$$P\Sigma \mathbb{c} = \mathbb{0}$$$$ but since $$A$$ can be diagonalized to $$\Sigma$$, we know $$P \Sigma = AP$$ and thus the above equation becomes $$$$\label{eq5} AP\mathbb{c} = \mathbb{0}$$$$ but since $$AP \neq 0$$, it must be the case that $$\mathbb{c} = \mathbb{0}$$

• Notice that in the Math Stack Exchange page How to ask a good question there's an specific section Avoid "no clue" questions, that begins with: "Too many questions begin or end with "I don't even know how to begin with this problem". While this may be true (you may genuinely have no idea how to approach the problem), it is still not a valid reason to limit your post to the statement of the problem without any mention of your own thoughts." Commented Nov 25, 2023 at 2:39
• Start with simple cases. Can you do the case $n=2$? Edit your question to include this. Now try $n=3$. Commented Nov 25, 2023 at 2:46
• @jjagmath Now that I've added what I know, can you remove your downvote? Commented Nov 25, 2023 at 3:27
• Given your edit, my question stands. Just do the case with $n=2$. Commented Nov 25, 2023 at 3:36

The set $$\{v_1,\ldots, v_{n-1}\}$$ is L.I. as you have shown.

Suppose, including $$v_n(\neq 0)$$ to this set makes it L.D. which is possible $$\iff$$ there exists coefficients $$a_{i}$$'s (not all zero) such that $$v_n=\sum_{i=1}^{n-1}a_iv_i\\ \Rightarrow (A-\lambda_{n-1}I)v_n=\sum_{i=1}^{n-1}a_i(A-\lambda_{n-1}I)v_i$$

By definition of eigen vector, $$Av_i=\lambda_iv_i$$ for each $$i$$.

Given, $$(A-\lambda_{n-1}I)v_{n}=v_{n-1}$$.

Simplify and you have:

$$v_{n-1}=\sum_{i=1}^{n-2} a_i(\lambda_i-\lambda_{n-1})v_i$$

This contradicts that $$\{v_1,\ldots,v_{n-1}\}$$ is L.I. $$\blacksquare$$