39
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ 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 $0$! $\endgroup$ Jul 20, 2011 at 18:38
  • $\begingroup$ 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? $\endgroup$
    – xenon
    Jul 20, 2011 at 18:43
  • 2
    $\begingroup$ 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. $\endgroup$ Jul 20, 2011 at 18:49

5 Answers 5

40
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ Thank you so much Arturo for your very clear explanation! I finally understand now. Thanks!! :) $\endgroup$
    – xenon
    Jul 20, 2011 at 19:02
  • $\begingroup$ If the columns are pairwise orthogonal and have no zero entries, are the rows necessarily pairwise orthogonal? $\endgroup$
    – Bohan Lu
    Dec 9, 2014 at 19:22
  • 3
    $\begingroup$ $Q^{T}Q=I$ (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 $QQ^{T}=I$? $\endgroup$
    – user216094
    Mar 22, 2015 at 18:10
  • 2
    $\begingroup$ 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. $\endgroup$ May 19, 2018 at 22:48
  • 3
    $\begingroup$ @Hestenes: For square matrices $A$ and $B$, $AB=I$ if and only if $BA=I$. Thus, for a square matrix $Q$, $Q^TQ=I$ if and only if $QQ^T=I$. $\endgroup$ May 20, 2018 at 0:01
10
$\begingroup$

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 $).

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$ Mar 29, 2017 at 3:14
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Mong H. Ng
    Jan 17, 2019 at 8:46
  • $\begingroup$ @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. $\endgroup$ Jan 17, 2019 at 20:53
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ use mathjax for eqns $\endgroup$
    – cineel
    Dec 17, 2021 at 16:57
  • $\begingroup$ There are six answers from ten years ago. Your answer adds nothing to them. $\endgroup$
    – amWhy
    Dec 17, 2021 at 18:28
  • 1
    $\begingroup$ Adds nothing, except brevity and simplicity $\endgroup$ Dec 17, 2021 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.