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.

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The equation $QQ^T = I$ only holds if the columns of $Q$ form an orthonormal set of vectors, not merely an orthogonal one. After all, some columns might be equal to $0$! –  Arturo Magidin Jul 20 '11 at 18:38
Oh thanks for highlighting this! But still, if the columns are orthonormal and are unit vectors, why are its rows so coincidentally happen to be perpendicular too? If a column happens to be $0$, how it will no longer be an orthonormal set of vectors, will it? –  xenon Jul 20 '11 at 18:43
If the columns form an orthonormal set, then $QQ^T=I$, so $Q^T = Q^{-1}$, so $Q^TQ=(Q^T)(Q^T)^T=I$, so the columns of $Q^T$ are also an orthonormal set, and these are the rows of $Q$; it's not "coincidental", it's forced on by the very strong condition that the columns must form an orthonormal basis. If a column is $0$, then the set of columns is no longer orthonormal because that column is not a unit vector. But it can be an orthogonal set, if all nonzero columns are pairwise orthogonal. See my full answer below for an example. –  Arturo Magidin Jul 20 '11 at 18:49

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.

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Thank you so much Arturo for your very clear explanation! I finally understand now. Thanks!! :) –  xenon Jul 20 '11 at 19:02

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.

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