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Let $Q$ be an $N \times N$ unitary matrix (its columns are orthonormal). Since $Q$ is unitary, it would preserve the norm of any vector $X$, i.e., $\|QX\|^2 = \|X\|^2$.

My confusion comes when the columns of $Q$ are orthogonal, but not orthonormal, i.e., if the columns are weighted by weights $w_1,\dots,w_N$, the dot product of any two different columns would still be zero, but $Q^H Q \neq I$ anymore.

  1. What are these matrices called? The literature always refers to matrices with orthonormal columns as orthogonal, however I think that's not quite accurate.

  2. Would a square matrix with orthogonal columns, but not orthonormal, change the norm of a vector?

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  • $\begingroup$ "What are these matrices called?" There appears to be no established name for them. "Would a square matrix with orthogonal columns, but not orthonormal, change the norm of a vector?" Why don't you try a simple $2\times 2$ example and see for yourself? $\endgroup$ – user856 Feb 27 '13 at 20:15
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    $\begingroup$ Thank you. I tried a simple 2 by 2 diagonal matrix and it does change the norm. $\endgroup$ – user63552 Feb 27 '13 at 20:24
  • $\begingroup$ According to wikipedia, en.wikipedia.org/wiki/Orthogonal_matrix, all orthogonal matrices are orthonormal, too: "An orthogonal matrix is a square matrix whose columns and rows are orthogonal unit vectors (i.e., orthonormal vectors)". Is wikipedia wrong? $\endgroup$ – makansij Jul 2 '18 at 2:58
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If $Q=(x_1,\ldots,x_n)$ is a matrix with orthogonal columns ($x_i^Hx_j=0$), then provided that its columns $x_1,\ldots,x_n$ are nonzero, we have $$ Q=\left(\frac{x_1}{\|x_1\|},\ldots,\frac{x_n}{\|x_n\|}\right)\begin{pmatrix}\|x_1\|\\ &\ddots\\ &&\|x_n\|\end{pmatrix}=UD. $$ Hence $Q$ is the product of a unitary matrix $U$ with a diagonal matrix $D$. The unitary matrix $U$ preserves norm, but the diagonal matrix $D$ in general doesn't.

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For the second question: yes, the columns of the matrix are the image of the canonical base. So if a column has norm different from one, the corresponding vector of the base (which has norm 1) changes its norm.

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    $\begingroup$ Thank you, although I don't know what you meant by "image of canonical base". $\endgroup$ – user63552 Feb 27 '13 at 20:30
  • $\begingroup$ The $k$-th column of the matrix $A$ is $A e_k$ where $e_k$ is the $k$-th element of the canonical basis. $\endgroup$ – Emanuele Paolini Feb 27 '13 at 20:39

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