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I am having a terribly difficult time with the following problem. Any help will be appreciated. Thank you!

Let $A$ be a real symmetric $n × n$ matrix, and suppose $A$ has $n$ real, distinct eigenvalues,$λ_1,...,λ_n$ with corresponding eigenvectors $φ_1,...,φ_n$. Prove that $Φ = \{φ_1, . . . , φ_n\}$ is an orthogonal basis of $\mathbb{R}^n$.

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marked as duplicate by vonbrand, Amzoti, Paul, Micah, muzzlator Apr 8 '13 at 4:24

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Choose two $\phi_k$. At least one will correspond to a non-zero eigenvalue, so you can write $\phi_k = \frac{1}{\lambda_k} A \phi_k$. Now take the inner product with the other vector. –  copper.hat Apr 8 '13 at 0:59
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1 Answer 1

I like this proof that was presented in Paul Garrett's algebra textbook:

Suppose $a_1 \phi_1+\cdots+a_n\phi_n=0$ (and all of the $a_i$ are nonzero) is the shortest dependency relation.

Now consider $0=(A-\lambda_1 I)(a_1 \phi_1+\cdots+a_n\phi_n)=0+a_2(\lambda_2-\lambda_1)v_2+\cdots+a_n(\lambda_n-\lambda_1)v_n$. Since we assumed that the first relation was the shortest, we must have that all of these $a_i$ are zero.

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