According to Wikipedia, an orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors. It also says that this definition is equivalent to saying that an orthogonal matrix $Q$ is a matrix for which $Q^T Q = Q Q^T = I$, where $I$ is the identity matrix. Why are these definitions equivalent?
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3If you write out matrix multiplication, it turns out that you’re computing a bunch of dot products of the columns of $Q$. – Michael Burr Dec 21 '18 at 14:07
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See https://math.stackexchange.com/questions/2028990/orthogonal-matrix-and-orthonormal-columns – Widawensen Dec 21 '18 at 14:46
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Does this answer your question? [Why, if a matrix $Q$ is orthogonal, then $Q^T Q = I$?](https://math.stackexchange.com/questions/1766669/why-if-a-matrix-q-is-orthogonal-then-qt-q-i) – Jul 25 '20 at 22:32
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The $(i,j)$ entry of $Q^{T}Q$ is the dot product of the $i$-th column of $Q$ with the $j$-th column of $Q$. Since $Q$ is orthogonal, these will be $0$ when $i\neq j$, and $1$ when $i=j$ (because the columns are unit vectors). Therefore $Q^{T}Q$ has ones on the diagonal and zeros everywhere else, so it's the identity matrix.
David C. Ullrich
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pwerth
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As pwerth noted, in the product $Q^\top Q$, the $(i,j)$-entry is simply the inner product of the $i$-th column of $Q$ with the $j$-th column. It is worth noting however that this implies that $QQ^\top$ is also the identity. This is because when a square matrix has a left-inverse, this left-inverse is immediately a right-inverse also.
The reason why this is interesting is because this makes a big use of the finite-dimensionality of the vector space we work over (in this case, $\mathbb{R}^n$, with $n$ the size of $Q$). For more details, see this question, where you'll also find a counterexample in the infinite dimensional case.
SvanN
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