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Here's the question: Let $Q$ be a square $n\times n$ matrix. Let $\{ \textbf{e}_1,\textbf{e}_2,...,\textbf{e}_n\}$ be the $n$ standard basis column vectors of $\mathbb{R}^n$. Show that the set of vectors $\{ Q\textbf{e}_1,Q\textbf{e}_2,...,Q\textbf{e}_n\}$ also form a set of orthonormal vectors.

In terms of my attempts, I've proven that each column vector of $Q$ forms a set of orthonormal vectors in $\mathbb{R}^n$. I feel like this may be very close but I'm struggling to picture where to go from here. If this method is correct, where do I go from here? If this method is not correct, what would be the best way to prove this?

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    $\begingroup$ I guess there's a condition here on $Q$. Probably orthogonal matrix? Note that $Qe_i$ is just the $i$th column. $\endgroup$ – Berci May 2 at 14:00
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We have $e_i^Te_j=\delta_{ij}$ and want $(Qe_i)^TQe_j=\delta_{ij}$, but the left-hand side is $e_i^TQ^TQe_j=\delta_{ij}$, which is equivalent to $Q^TQ=I$. In particular, if $Q^TQe_j$ isn't proportional to $e_j$, some $e_i,\,i\ne j$ won't be orthogonal to it; whereas if $Q^TQe_j\propto e_j$, we need equality so $e_i^TQ^TQe_j$ is the identity matrix rather than just being diagonal.

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