# 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.

• 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

## 3 Answers

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.

• Thank you so much Arturo for your very clear explanation! I finally understand now. Thanks!! :) – xenon Jul 20 '11 at 19:02
• If the columns are pairwise orthogonal and have no zero entries, are the rows necessarily pairwise orthogonal? – Bohan Lu Dec 9 '14 at 19:22
• $Q^{T}Q=I$ (1). How do we know that $QQ^{T}=I$ as well? Because we can transpose both sides, and knowing that $I^T=I$ it follows that $QQ^{T}=I$? – user216094 Mar 22 '15 at 18:10
• Does it follow that all matrixes that are row orthogonal and column orthogonal are orthonormal sets? – Peter Oct 6 '17 at 9:29
• Columns are orthogonal implies $Q^{T}Q$ is a diagonal matrix. How do you write $QQ^{T} = I$ even after you add the additional assumption that they have 2-norm 1. I dont follow your logic. It looks like the OP made the a mistake and you continued the same error in your answer. – Hestenes May 19 '18 at 22:48

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 $12 \sqrt{70} \cong 100.4$).

• I edited this answer to make it clearer. Imo it's by far the best answer here, and probably deserves to be the accepted answer. – goblin Mar 29 '17 at 3:14

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.

• Having orthonormal basis as columns implies $Q^TQ = I$. However, $Q^TQ = I$ does not implies that $Q^T$ is the inverse. See pseudoinverse: en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse. – Mong H. Ng Jan 17 at 8:46
• @MongH.Ng The pseudoinverse is irrelevant here; the pseudoinverse is equal to the inverse when the latter exists, and if $A$ is a square matrix and there is no inverse at all, then it is not the case that $A^{\dagger}A=I$. You keep forgetting that $Q$ is a square matrix, and that has important implications. – Arturo Magidin Jan 17 at 20:53