Column Vectors orthogonal implies Row Vectors also orthogonal? If the column vectors of a matrix $A$ are all orthogonal and $A$ is a square matrix, can I say that the row vectors of matrix $A$ are also orthogonal to each other?
From the equation $Q \cdot Q^{T}=I$ if $Q$ is orthogonal and square matrix, it seems that this is true but I still find it hard to believe. I have a feeling that I may still be wrong because those column vectors that are perpendicular are vectors within the column space. Taking the rows vectors give a totally different direction from the column vectors in the row space and so how could they always happen to be perpendicular?
Thanks for any help.
 A: Recall that two vectors are orthogonal if and only if their inner product is zero. You are incorrect in asserting that if the columns of $Q$ are orthogonal to each other then $QQ^T = I$; this follows if the columns of $Q$ form an orthonormal set (basis for $\mathbb{R}^n$); orthogonality is not sufficient. Note that "$Q$ is an orthogonal matrix" is not equivalent to "the columns of $Q$ are pairwise orthogonal".
With that clarification, the answer is that if you only ask that the columns be pairwise orthogonal, then the rows need not be pairwise orthogonal. For example, take
$$A = \left(\begin{array}{ccc}1& 0 & 0\\0& 0 & 1\\1 & 0 & 0\end{array}\right).$$
The columns are orthogonal to each other: the middle column is orthogonal to everything (being the zero vector), and the first and third columns are orthogonal. However, the rows are not orthogonal, since the first and third rows are equal and nonzero.
On the other hand, if you require that the columns of $Q$ be an orthonormal set (pairwise orthogonal, and the inner product of each column with itself equals $1$), then it does follow: precisely as you argue. That condition 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. 
A: This condition says that $Q^{-1} = Q^t$.  That means that you have
$$Q^tQ = Q Q^t = I.$$
Yes, if the rows are orthonormal (basis -- oops my omission), so are the columns.
A: None of the existing answers quite worked for me, so...
If $A$ has orthonormal columns, $A^T A=I$. $A^T A$ is full of dot products of pairs of columns of $A$. By orthogonality, these are all zero off of the diagonal. By orthonormality, these are all one on the diagonal.
If $A$ is rectangular, $A^T$ cannot be a true inverse of $A$. If $A$ is square with orthonormal (or even independent non-zero) columns, we have $A^{-1} A=I$, suggesting that $A^T$ could be $A^{-1}$. But how to rule out that there may be more than one $B$ such that $BA=I$?
Well, if $XY=Z$, every column of $Z$ is a linear combination of the columns of $X$, and every row of $Z$ is a linear combination of the rows of $Y$. But if $X$ has independent columns, there's precisely one unique linear combination of those columns that gives any particular vector result (including each column of $Z$), so $Y$ is uniquely determined by $X$ and $Z$. Similarly if $Y$ has independent rows, each row of $X$ is uniquely determined by $Y$ and $Z$.
The columns of $A$ are orthonormal so clearly independent. If (and only if) $A$ is square, its rows are independent. And by those uniqueness properties, if $A^T A=I$, we can conclude that $A^T=A^{-1}$.
Since $A^T=A^{-1}$, $A A^T=I$ also - the rows of $A$ are orthonormal because $A A^T$ is made of dot products of pairs of rows which are all one on the diagonal, zero elsewhere (as earlier with $A^T A$, but for rows not columns).
I'm still assuming $B A=I$ implies $A B=I$ with the same $B=A^{-1}$ both ways, so...
If $B$ is the left inverse of $A$, and $C$ is the left inverse of $B$, then $BA=I$ and $CB=I$ by definition. Multiplying both sides of $BA=I$ gives $CBA=C$, so $A=C$ -- the left inverse of the left inverse is the original matrix. We knew that anyway because the left inverse is the transpose (good job - otherwise I'd have to prove the existence of the left inverse of the left inverse). But substituting $A=C$, $BA=I$ becomes $BC=I$ and $CB=I$ becomes $AB=I$, so left inverses are right inverses too.
This answer still isn't a full self-contained proof because I haven't proved the uniqueness properties for linear combinations of independent vectors, or the properties of dot products, or that matrix multiplication is associative.
A: Even if $A$ is non-singular, orthogonality of columns by itself does not guarantee orthogonality of rows. Here is a 3x3 example:
$$ A = \left( \begin{matrix}
1 & \;2 & \;\;5 \\ 2 & \;2 & -4 \\ 3 & -2 &  \;\;1
\end{matrix} \right) $$
Column vectors are orthogonal, but row vectors are not orthogonal.
On the other hand, orthonormality of columns guarantees orthonormality of rows, and vice versa.
As a footnote, one of the forms of Hadamard's inequality concerns the absolute value of the determinant of a matrix given the norms of the column vectors.  That absolute value will be maximum when those vectors are orthogonal. The determinant, in absolute value, will be equal to the product of the norms. In the case of the above matrix, as the columns are orthogonal, 84 is the maximum possible absolute value of the determinant $-$  det(A) is -84 $-$ for column vectors with the given norms ($\sqrt {14}, 2\sqrt 3$ and $ \sqrt {42}$ respectively).
Although $det(A)=det(A^T)$, Hadamard's inequality does not imply neither orthogonality of the rows of A nor that the absolute value of the determinant is maximum for the given norms of the row vectors ($ \sqrt{30}, 2\sqrt 6$ and $ \sqrt{14}$ respectively; their product is $ 12 \sqrt{70} \cong 100.4 $).
A: Proof that a square matrix with orthonormal columns must have orthonormal rows
Let a square matrix Q have Orthonormal columns qi, then QT Q = I  (since qi qj= 1 when i=j else 0)
For Orthonormal rows we need Q QT = I which follows from Q QT = Q (I) QT = Q (QT Q) QT = (Q QT)(Q QT) which requires Q QT = I
