# eigen vectors and linear independence

I know that it is the case that if we have eigenvectors $$v_1....v_r$$ that correspond to distinct eigen values $$λ_1....λ_r$$ of an $$n$$ x $$n$$ matrix, then {$$v_1....v_r$$} is linearly independent.

But is is it possible to have $$a_1$$ linearly independent eigen vectors corresponding to $$λ_1$$, $$a_2$$ linearly independent eigen vectors corresponding to $$λ_2$$... $$a_p$$ linearly independent eigen vectors corresponding to $$λ_n$$, and the set of all $$r$$ eigenvectors where $$r=a_1+a_2+...+a_p$$ be linearly dependent?

For example: consider a $$3x3$$ matrix. Imagine we have one vector(call it $$v_1$$) forming a basis for the eigenspace of a one eigen value, but two vectors(call them $$v_2$$ and $$v_3$$) forming a basis for different eigen value. are we guaranteed that {$$v_1,v_2,v_3$$} are linearly independent?

• No, this is not possible. Eigenvectors from distinct eigenspaces are independent. Nov 3 '20 at 19:24
• NO. Your first paragraph gives the explanation. Nov 3 '20 at 19:25
• @TitoEliatron No, the first paragraph does not explain the case they are asking about. But this independence is still a standard result in every textbook. Nov 3 '20 at 19:25
• Yes it does. Take $r=2$. and you'll get that ANY eigenvector associated to $\lambda_1$ is independent of ANY eigenvector associated to $\lambda_2\ne\lambda_1$. In fact, is what you say in your first commnet. Nov 3 '20 at 19:27

Suppose 𝔽 is any field. For example, it could be ℝ or ℂ. Suppose $$M$$ is an n×n 𝔽-matrix. Suppose $$\lambda_1, \dots, \lambda_r$$ is a list of distinct eigenvalues of $$M$$.

Suppose:

• $$v_{1, 1}, \dots, v_{1, k_1}$$ is a linearly independent list of eigenvectors of $$M$$ corresponding to $$\lambda_1$$.
• $$v_{2, 1}, \dots, v_{2, k_2}$$ is a linearly independent list of eigenvectors of $$M$$ corresponding to $$\lambda_2$$.
• $$\vdots$$
• $$v_{r, 1}, \dots, v_{r, k_r}$$ is a linearly independent list of eigenvectors of $$M$$ corresponding to $$\lambda_r$$.

Suppose $$\alpha_{1, 1}, \dots, \alpha_{1, k_1}, \alpha_{2, 1}, \dots, \alpha_{2, k_2}, \dots, \alpha_{r,1}, \dots, \alpha_{r, k_r} \in \mathbb{F}$$ are scalars such that $$(\alpha_{1, 1} v_{1,1} + \dots + \alpha_{1, k_1} v_{1, k_1}) + (\alpha_{2, 1} v_{2,1} + \dots + \alpha_{2, k_2} v_{2, k_2}) + \dots + (\alpha_{r,1} v_{r,1} + \dots + \alpha_{r, k_r} v_{r, k_r}) = 0.$$

Define $$u_1 = (\alpha_{1, 1} v_{1,1} + \dots + \alpha_{1, k_1} v_{1, k_1}), \dots, u_r=(\alpha_{r,1} v_{r,1} + \dots + \alpha_{r, k_r} v_{r, k_r})$$.

Clearly, $$u_1$$ is either the zero vector or a $$\lambda_1$$-eigenvector of $$M$$. Analagously, $$u_2$$ is either the zero vector or a $$\lambda_2$$-eigenvector of $$M$$. And so on for $$u_3, \dots, u_r$$.

By the theorem you say you know, it must be the case that for each $$i \in \{1, \dots, r\}$$, $$u_i$$ is actually the zero vector. But then for each $$i \in \{1, \dots, r\}$$, all the coefficients $$\alpha_{i,1}, \dots, \alpha_{i, k_i}$$ must be zero, because the vectors $$v_{i,1}, \dots, v_{i, k_i}$$ are linearly independent.

We have shown that any linear combination of $$v_{1, 1}, \dots, v_{1, k_1}, \dots, v_{r, 1}, \dots, v_{r, k_r}$$ equal to zero must actually be the trivial combination (i.e., with all coefficients zero). ∎

• This answer is very helpful but my one I still don't understand why "it must be the case that for each i∈{1,…,r}, ui is actually the zero vector."
– bfff
Nov 12 '20 at 17:07
• @bfff $u_1, \dots, u_r$ belong to eigenspaces of distinct eigenvalues. Suppose some of them are nonzero. By the theorem OP stated, those of them which are nonzero are linearly independent. Therefore their sum can give zero only if they are actually all zero. But their sum IS zero by my assumption. Contradiction. Nov 12 '20 at 21:38

We claim that the set of all $$r$$ eigenvectors is linearly independent. The main proof ideas are already provided in the comments. For each $$i=1,\dots, p$$ let $$v_{i1},\dots, v_{ia_i}$$ be linearly independent system of eigenvectors corresponding to the eigenvalue $$\lambda_i$$ and $$\mu_{i1},\dots, \mu_{ia_i}$$ be scalars such that $$\sum_{i=1}^p \sum_{j=1}^{a_i} \mu_{ij} v_{ij}=0.$$

For each $$i$$ put $$v_i=\sum_{j=1}^{a_i} \mu_{ij} v_{ij}$$. Then $$v_i$$ is an eigenvectors corresponding to $$\lambda_i$$ and

$$\sum_{i=1}^p v_i=\sum_{i=1}^p \sum_{j=1}^{a_i} \mu_{ij} v_{ij}=0.$$

The linear independence property formulated in the first paragraph of the question implies that $$0=v_i=\sum_{j=1}^{a_i} \mu_{ij} v_{ij}$$ for each $$i$$. Then the linear independence property formulated in the second paragraph of the question implies that $$\mu_{ij}=0$$ for each $$j$$.