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I am trying to use induction to prove that in the general case that distinct eigenvalues have distinct eigenvectors. This is based off of @Arturo Magidin's answer in the following question:

How to prove that eigenvectors from different eigenvalues are linearly independent

I was hoping someone could tell me if I am going about it in the right way.

N=1 Case

Show it is true for the case where $n =1:$

$a_1v_1 = 0$

$a_1$ must be $0$ here because if $v_1$ is an eigenvector from the corresponding eigenvalue $\lambda_1$, then it cannot be $0$

N=K Case

Assume it is true for the case where $n = k:$

$a_1v_1 + a_2v_2 + ...+ a_kv_k = 0$

If we apply the linear transformation $T$ to the above equation:

$T(a_1v_1 + a_2v_2 + ...+ a_kv_k) = 0$

$=a_1\lambda_1v_1 + a_2\lambda_2v_2 + ...+ a_k\lambda_kv_k=0$

If we multiply

$\lambda_1$

by $a_1v_1 + a_2v_2 + ...+ a_kv_k = 0$

We will get:

$a_1\lambda_1v_1 + a_2\lambda_1v_2 + ...+ a_k\lambda_1v_k = 0$

If we take the difference between the the linear transformation and the above equation, we get the following:

$a_2(\lambda_2-\lambda_1)v_2 + ...+ a_k(\lambda_1-\lambda_k)v_k = 0$

Now, since the eigenvalues are distinct and the eigenvectors cannot equal $0$, then the only solution is t

then $a_i = 0$ for $i = 1,2,..., k$

N+1 Case

Then for the case where $n=k+1:$

$a_1v_1 + a_2v_2 + ...+ a_kv_k + a_{k+1}v_{k+1}= 0$

We know it therefore must be true that the coefficients

$a_i$ for $i = 1,2,..., k+1$ must be $0$

I am not convinced that what I have produced is acceptable so if anyone could comment on it or let me know I would be most grateful.

InvestingScientist
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1 Answers1

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A good way to check if your proof is correct is if you have used all the assumptions (of course, this is no hard and fast rule). You didn't use the fact that eigenvalues are distinct.

Your proof is not quite on the mark. Sure, you can assume linear independence when $n=k$, which guarantees that if $a_1v_1 + ...+a_kv_k = 0$, then $a_1,...,a_k = 0$. However, when $n=k+1$, what you are trying to show is that if $b_1v_1 + ... + b_{k+1}v_{k+1}= 0$, then $b_1, ...b_{k+1} = 0$. Here $b_1 ... b_{k+1}$ are some arbitrary real numbers, not constants "taken" from the $n=k$ case. Hence, you may not assume $b_1,...,b_k=0$.

koifish
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