Orthogonal matrix confusion I have a confusion about orthogonal matrix. 


*

*If columns of a square matrix are orthonormal to each other, is the matrix orthogonal?

*If yes, then are the rows of the matrix also orthonormal? Why?

*Why is it that QQ'=I? I get Q'Q=I but why QQ' is also I?


Thanks,
Tom
 A: *

*Yes, that's the definition.

*From $QQ^\top=I$.

*Since $Q^\top Q=I$, $Q^\top=Q^{-1}$. So $QQ^\top=QQ^{-1}=I$.

A: The key to your last question is that if a matrix $A$ has inverse $B$, then 
$$
AB = BA = I
$$
Where $I$ is the identity matrix.  That is, a finite square matrix always commutes with its inverse.  From there, it's clear that if $Q'Q=I$ (that is, $Q'$ is the inverse of $Q$), then $QQ'=I$
The easy answer to your second question is that $Q$ is an orthogonal matrix whenever $Q'Q=I$, which means that $QQ'=I$.  Now, if $Q$ is an orthogonal matrix, then 
$$(Q')'Q'=QQ'=I$$
which means that $Q'$ is orthogonal.  This in turn means that $Q'$ has orthonormal columns, which means that $Q$ has orthonormal rows.
A: If the columns of a square matrix are orthonormal, then the matrix is said to be orthogonal.

Note: The following are equivalent for a $n \times n$ matrix $P$.
$1)P$ is invertible and $P^{-1}=P^{T}$
$2)$ The rows of $P$ are orthonormal
$3)$ The columns of $P$ are orthonormal.
Hint: 
First recall that condition $(1)$ is equivalent to $PP^{T}=I$ because of the fact that a square matrix is invertible if and only if $A^{T}$ is invertible. Let $x_1,x_2,\ldots,x_n$ denote the rows of $P$. Then $x^T_j$ is the $j$th column of $P^{T}$, so the $(i,j)$-entry of $PP^{T}$ is $x_i \cdot x_j$. Thus $PP^{T}=I$ means that $x_i \cdot x_j=0$ if $i \neq j$ and $x_i \cdot x_j=1$ if $i=j$. Hence condition $(1)$ is equivalent to $(2)$. The proof of the equivalence of $(1)$ and $(3)$ is similar. 

If the columns of $Q$ are an orthonormal set, then that is equivalent to "the matrix is orthogonal", and since $I = Q^TQ = QQ^T$ and $(Q^T)^T = Q$, it follows that if $Q$ is orthogonal then so is $Q^T$, hence the columns of $Q^T$ (i.e., the rows of $Q$) form an orthonormal set as well. 
