Cindy Noble

2022-09-06

Here's the question: Let Q be a square $n\times n$ matrix. Let $\{{\mathbf{\text{e}}}_{1},{\mathbf{\text{e}}}_{2},...,{\mathbf{\text{e}}}_{n}\}$ be the n standard basis column vectors of Rn. Show that the set of vectors $\{Q{\mathbf{\text{e}}}_{1},Q{\mathbf{\text{e}}}_{2},...,Q{\mathbf{\text{e}}}_{n}\}$ 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?

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?

Yuliana Griffith

Beginner2022-09-07Added 6 answers

We have ${e}_{i}^{T}{e}_{j}={\delta}_{ij}$ and want $(Q{e}_{i}{)}^{T}Q{e}_{j}={\delta}_{ij}$, 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

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