Proof that eigenvectors corresponding to different eigenvalues are linearly independent.

Asked Jan 02 '19 at 15:39

Active Jan 02 '19 at 15:54

Viewed 172 times

If eigenvectors $x_1, x_2,...,x_k$ correspond to different eigenvalues $λ_1, λ_2,...,λ_k$, then $x_1, x_2,...,x_k$ are linearly independent.

Proof:

Suppose $$C_1{x_1}+C_2{x_2}+...C_k{x_k}=0 ...(1)$$ $$A(C_1{x_1}+C_2{x_2}+...C_k{x_k})=0$$ $$\Rightarrow C_1{λ_1}{x_1}+C_2{λ_2}{x_2}+...+C_k{λ_k}{x_k}=0...(2)$$ $$λ_k(1)-(2)=C_1{λ_1}{x_1}+C_2{λ_2}{x_2}+...+C_{k-1}{λ_k}{x_{k-1}}-C_1{λ_1}{x_1}-C_2{λ_2}{x_2}-...-C_{k-1}{λ_{k-1}}{x_{k-1}}=0$$ $$\Rightarrow C_1({λ_k}-{λ_1}){x_1}+...+C_{k-1}({λ_k}-{λ_{k-1}}){x_{k-1}}=0$$ And how to do next to show that $C_1=0=C_2=...=C_k$?

edited Jan 02 '19 at 15:54

asked Jan 02 '19 at 15:39

Maggie

All the eigenvalues are different? – Boris Valderrama Jan 02 '19 at 15:46
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So we can never have that $\lambda_i -\lambda_j =0$ – Jan 02 '19 at 15:48
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yes, they all are different. – Maggie Jan 02 '19 at 15:48
@vadim123 it is not general – Boris Valderrama Jan 02 '19 at 15:53
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You should also conclude that eigenvectors are nonzero vectors by definition. – Jan 02 '19 at 15:54
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You can easily finish the argument by induction, start with a base case of two vectors. You sort of already have the inductive step! (hypothesis $x_1, x_2, \dots x_{k-1}$ linearly independent) – Jan 02 '19 at 15:55
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@WesleyStrik I see. I forgot some definition. Thanks a lot! – Maggie Jan 02 '19 at 15:56

Proof that eigenvectors corresponding to different eigenvalues are linearly independent.

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