# Let Q be a square n xx n matrix. Let {e_1,e_2,...,e_n} be the n standard basis column vectors of RR^n. Show that the set of vectors {Qe_1,Qe_2,...,Qe_n} also form a set of orthonormal vectors.

Here's the question: Let Q be a square $n×n$ matrix. Let $\left\{{\mathbf{\text{e}}}_{1},{\mathbf{\text{e}}}_{2},...,{\mathbf{\text{e}}}_{n}\right\}$ be the n standard basis column vectors of Rn. Show that the set of vectors $\left\{Q{\mathbf{\text{e}}}_{1},Q{\mathbf{\text{e}}}_{2},...,Q{\mathbf{\text{e}}}_{n}\right\}$ 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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Yuliana Griffith
We have ${e}_{i}^{T}{e}_{j}={\delta }_{ij}$ and want $\left(Q{e}_{i}{\right)}^{T}Q{e}_{j}={\delta }_{ij}$, but the left-hand side is ${e}_{i}^{T}{Q}^{T}Q{e}_{j}={\delta }_{ij}$, which is equivalent to ${Q}^{T}Q=I$. In particular, if ${Q}^{T}Q{e}_{j}$ isn't proportional to ${e}_{j}$, some ${e}_{i},\phantom{\rule{thinmathspace}{0ex}}i\ne j$ won't be orthogonal to it; whereas if ${Q}^{T}Q{e}_{j}\propto {e}_{j}$, we need equality so ${e}_{i}^{T}{Q}^{T}Q{e}_{j}$ is the identity matrix rather than just being diagonal