I think I got this problem, but just want to make sure:

Let $M$ be a $n\times n$ complex matrix with 3 distinct eigenvalues $\lambda_1$,$\lambda_2$, $\lambda_3$. Suppose $\{A_1,A_2\}$ are linearly independent e-vectors associated to $\lambda_1$, $\{B_1,B_2\}$ are linearly independent e-vectors associated to $\lambda_2$ and $\{C_1,C_2\}$ are linearly independent e-vectors associated to $\lambda_3$. Using the fact that "eigenvectors associated to distinct eigenvalues are linearly independent", prove that $\{A_1,A_2,B_1,B_2,C_1,C_2\}$ are linearly independent.

My attempt: Since we have that "eigenvectors associated to distinct eigenvalues are linearly independent", $A_i$ $B_j$ and $C_k$ are linearly independent for all $i,j$ and $k$ and since each set of of eigenvectors $\{A_1,A_2\}$, $\{B_1,B_2\}$ and $\{C_1,C_2\}$ are linearly independent by assumption, then $$\{A_1,A_2,B_1,B_2,C_1,C_2\}$$ is also linearly independent.

Is this correct? I feel like this can't be that easy. Any help will be appreciated.


1 Answer 1


What you are stating is not enough to conclude that $\{A_1, A_2, B_1, B_2, C_1, C_2\}$ is independent set. (Also notice, vectors cannot be independent by itself, it is a set that is either linearly independent / dependent). Consider the following example: $$A_1=(1,0,0,0), A_2=(1,1,0,0) \\ B_1 = (0,1,0,0), B_2 = (0,1,1,0) \\ C_1 = (0,0,1,0), C_2 = (0,0,1,1)$$ Here $\forall i,j,k \in \{1,2\}$ the set $\{A_i, B_j, C_k\}$ is linearly independent, which also holds for $\{A_1, A_2\}$, $\{B_1, B_2\}$ and $\{C_1, C_2\}$. But on the contrary, $\{A_1, A_2, B_1, B_2, C_1, C_2\}$ is not linearly independent.

Correct proof:

Let $\alpha_1A_1+\alpha_2A_2+\beta_1B_1+\beta_2B_2+\gamma_1C_1+\gamma_2C_2 = 0$. We want to prove that only combination of scalars satisfying this equation is the trivial combination ($\alpha_i=\beta_i=\gamma_i=0, $ for $i\in\{1,2\}$).

Let's define $$A:=\alpha_1A_1+\alpha_2A_2 \\ B:=\beta_1B_1 + \beta_2B_2 \\ C:=\gamma_1C_1+\gamma_2C_2$$

Now we want to prove that $A=B=C=0$. Let's suppose that some of these three vectors are not equals to 0. For easier notation, I will assume the worst case scenario - all of them being non-zero. Notice that $A,B,C$ are laying in the corresponding characteristic subspaces (for example, $A\in[\{A_1, A_2\}]$) which means that they are also the eigenvectors for different eigenvalues. Thus, the set $\{A,B,C\}$ is linearly independent. The above equation is now equivalent to $A+B+C=0$, but this is a contradiction to $\{A,B,C\}$ being independent.

Now we have that $A=B=C=0$, which leads to all scalars being 0 in the first equation. (For example $A=0=\alpha_1A_1+\alpha_2A_2 \implies \alpha_i=0$ since $\{A_1, A_2\}$ is linearly independent). QED.

  • $\begingroup$ Ok I definitely see that with this counter example. So am I interpreting the question wrong then? With this counter example, the statement of the problem cannot be proved. Also I fixed a typo I saw in the problem statement, $\{B_1,B_2\}$ are associated to $\lambda_2$ $\endgroup$ Oct 15, 2020 at 23:08
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    $\begingroup$ I don't think it is a valid counterexample, as it didn't guarantee that $\{A_1, A_2\}, \{B_1, B_2\}, \{C_1, C_2\}$ are eigenvectors. In fact, the conclusion of OP's question is correct. To prove it, first note $V_{\lambda_1} + V_{\lambda_2} + V_{\lambda_3}$ is a direct sum since eigenvalues are distinct. Then use the independence assumption within each $V_{\lambda_i}$ to conclude. $\endgroup$
    – Zhanxiong
    Oct 15, 2020 at 23:29
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    $\begingroup$ I added the correct proof of the statement to my answer. The counterexample is with respect to arguments used to prove the statement, not the statement itself (i.e. you could use the same arguments to prove that the above counterexample is linearly independent, which is not). $\endgroup$ Oct 15, 2020 at 23:42

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