# How to prove this set of vectors is a set comprised entirely of orthonormal vectors?

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?

• I guess there's a condition here on $Q$. Probably orthogonal matrix? Note that $Qe_i$ is just the $i$th column. – Berci May 2 at 14:00

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.