Eigenvalue decomposition: why are the eigenvectors columns of $Q$, and not some other vectors?

Taken from Wikipedia, https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix

Let $$A$$ be a square $$n × n$$ matrix with $$n$$ linearly independent eigenvectors $$q_i$$ (where $$i = 1, ..., n$$). Then $$A$$ can be factorized as

$${A}= {Q}{\Lambda}{Q}^{-1}$$ where $$Q$$ is the square $$n × n$$ matrix whose $$i$$th column is the eigenvector $$q_i$$ of $$A$$, and $$Λ$$ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues, $$Λ_{ii} = λ_i$$.

But why is $$Q$$ made up of the eigenvectors?

Recall that a general theorem says, given any real symmetric matrix $$A$$, there exists some orthogonal matrix $$Q$$ such that $${A} ={Q} {\Lambda } {Q} ^{-1}$$. Emphasis on exists.

So why can't $$Q$$ be some other orthogonal matrix not associated with the eigenvectors of $$A$$?

Let us write $$Q = (q_1 \, | \, \dots \, | \, q_n)$$ where $$q_i$$ are the columns of $$Q$$. Since $$Q$$ is invertible, we have
$$A = Q \Lambda Q^{-1} \iff AQ = Q \Lambda$$
so the columns of $$AQ$$ and $$Q\Lambda$$ must be identical. By the definition of matrix multiplication and the fact that $$\Lambda$$ is diagonal, the $$i$$-th column of $$Q\Lambda$$ is $$\lambda_i q_i$$ while the $$i$$-th column of $$AQ$$ is $$Aq_i$$. Hence, $$A = Q \Lambda Q^{-1}$$ iff $$Aq_i = \lambda_i q_i$$ for all $$1 \leq i \leq n$$ so for the identify $$A = Q \Lambda Q^{-1}$$ to hold, the columns of $$Q$$ must be eigenvectors of $$A$$ (and the diagonal entries of $$\Lambda$$ must be eigenvales of $$A$$).
• (If $Q$ is orthogonal, then not only the columns of $Q$ are a basis of eigenvectors, they are also an orthonormal basis but this has nothing to do with whether $Q$ is orthogonal or not). Sep 27, 2020 at 17:36