# Orthonormal Matrix and Transpose.

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

• What is your definition of orthonormal matrix? It usually goes by specifying an orthonormal matrix is a square matrix. Feb 23 at 20:45
• Orthonormal matrices are square so $Q$ square. Feb 23 at 20:53
• Using this, it follows that a square matrix with orthonormal columns also has orthonormal rows. Feb 23 at 21:08
• So there goes my question....I create a square orthogonal matrix with orthonormal vectors as columns....why do the rows have to be orthogonal? Feb 23 at 21:10
• 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 $I_m$. Feb 23 at 21:13

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

• 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? Feb 24 at 6:18
• I have already said in example (in my answer) for rectangular matrices,, $QQ^{T}$ may not even identity matrix $I$, please see it carefully! Feb 24 at 8:14

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 $$P = IP = (Q^\textsf T Q)P = Q^\textsf T (QP) = Q^\textsf TI = Q^\textsf T.$$