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If Q is an orthogonal matrix that has orthonormal vectors as columns, I can total understand the following result $$Q^TQ=I$$ but $QQ^T=I$ only if Q is a square matrix and not if Q is a rectangular matrix. I cannot understand why the second result is true. When I'm doing $QQ^T=I$ I'm not taking dot product of orthonormal vectors and yet it yields as identity matrix for a square matrix. How is this possible?

Edit: Orthonormal changed to Orthogonal

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    $\begingroup$ What is your definition of orthonormal matrix? It usually goes by specifying an orthonormal matrix is a square matrix. $\endgroup$
    – Clement C.
    Feb 23 at 20:45
  • $\begingroup$ Orthonormal matrices are square so $Q$ square. $\endgroup$ Feb 23 at 20:53
  • $\begingroup$ Using this, it follows that a square matrix with orthonormal columns also has orthonormal rows. $\endgroup$
    – azif00
    Feb 23 at 21:08
  • $\begingroup$ So there goes my question....I create a square orthogonal matrix with orthonormal vectors as columns....why do the rows have to be orthogonal? $\endgroup$
    – Orpheus
    Feb 23 at 21:10
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    $\begingroup$ If the columns of a rectangular-but-not-square $Q$ are orthogonal, that does tell you that $Q^T Q$ is the identity matrix of the right size (i.e. $I_n$ if $Q$ is $m \times n$). This is because the entries of $Q^T Q$ are the pairwise dot products of the columns of $Q$. If $Q$ is not square ($m \neq n$) you can deduce that $Q$ has fewer columns than rows (because its columns are $n$ orthornormal vectors in a space of dimension $m$), i.e., $m > n$. No hope of $QQ^T$ being $I_m$ e.g. because $QQ^T$ has the same rank as $Q^T Q = I_n$, i.e. $n$, and this is smaller than $m$, the rank of $I_m$. $\endgroup$ Feb 23 at 21:13
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Whenever $Q$ is not a square matrix , the result may not be true.

For example, take $$Q= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix} $$

Clearly then , $$Q^{T} Q= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} $$ But, $$Q Q^{T}= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$

If $Q$ is a square matrix such that $Q^{T}Q=I$

Clearly then $Q$ is an invertible matrix i.e. $Q^{-1}$ exists.

So, now $Q=QI $ $\implies Q=Q(Q^{T}Q)$ $\implies Q=(QQ^{T})Q$ $\implies (I-QQ^{T})Q=0$ $\implies QQ^{T}=I (\text{as, $Q$ is invertible })$

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  • $\begingroup$ So you mean to say that....$Q^TQ=I=QQ^T$ works for square matrices because I of the same order and for the rectangular matrices they yield I of different orders? $\endgroup$
    – Orpheus
    Feb 24 at 6:18
  • $\begingroup$ I have already said in example (in my answer) for rectangular matrices,, $QQ^{T} $ may not even identity matrix $I$, please see it carefully! $\endgroup$
    – A learner
    Feb 24 at 8:14
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Suppose that $Q \in \mathbb{R}^{n \times n}$ has orthonormal columns $q_1,\dots,q_n \in \mathbb{R}^{n \times 1}$. Then the map $L_Q : \mathbb{R}^{n \times 1} \to \mathbb{R}^{n \times 1}$ given by $L_Q(x) = Qx$ for all $x \in \mathbb{R}^{n \times 1}$ is bijective since it takes the standard basis of $\mathbb{R}^{n \times 1}$ to the basis $q_1,\dots,q_n$ (why $n$ orthogonal vectors in $\mathbb{R}^{n \times 1}$ form a basis?). Then there exists $P \in \mathbb{R}^{n \times n}$ such that $(L_Q)^{-1} = L_P$, which means that $QP=I=PQ$. Moreover, since $Q^\textsf T Q = I$, it follows that $$P = IP = (Q^\textsf T Q)P = Q^\textsf T (QP) = Q^\textsf TI = Q^\textsf T.$$

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