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Math people:

This is a pretty simple question, but I am having trouble finding an answer. I did not find an answer in the Similar Questions, and I apologize ahead of time if this is a duplicate.

Here is my question: if $A$ is a real square matrix, and $\|A\mathbf{x}\| = \|\mathbf{x}\|$ for all real column vectors $\mathbf{x}$ of the right length, is $A^T A = I$, that is, is $A$ an orthogonal matrix? (the converse is true, of course). I am pretty sure the answer is "yes", but it is surprisingly hard to verify this, and I would like at least a link to a proof.

EDIT: This is in fact true, and it appears as Proposition 5.9 in the document at . This is in the thirty-third page of a forty-two page document, and it is not apparent whether there is an elementary proof.

Stefan (STack Exchange FAN)

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up vote 1 down vote accepted

Hint: $||Ax||^2=(Ax)^T(Ax)=x^T(A^TA)x$. (EDIT: I no longer think this hint is actually useful)

Alternative hint: $A$ is orthogonal iff the columns of $A$ are orthogonal. The norm being preserved means that the inner product is as well. (i.e. $\langle x,y\rangle = \langle Ax,Ay\rangle.)$ The columns of $A$ are the images of the usual basis under the transformation that $A$ represents, so since inner products are preserved, the columns are orthonormal also.

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I can see how to use your first hint to prove the converse of what I want, but not what I want (you're missing a ^2, but it's obvious what you mean). Your alternative hint definitely works. I am teaching this material this Thursday. Very few of my students are math majors. While not agonizing, this proof is sufficiently difficult that I'll probably just ask my students to take my word for the fact (as opposed to the converse, which is easy). I don't think the text I am using includes $A$ inducing an isometry in the list of equivalent conditions to being orthogonal. – Stefan Smith Mar 26 '13 at 23:45
just so there is no misunderstanding, I am not criticizing your proof. I doubt there is an easier one. – Stefan Smith Mar 26 '13 at 23:45
@Stefan Smith Good point, I think you're right! Very sorry about that, I don't think there's a way to use the first hint to show what you want, sorry again! – Tom Oldfield Mar 27 '13 at 0:07
Since $\langle \mathbf{x},\mathbf{y} \rangle = \langle A\mathbf{x},A\mathbf{y} \rangle$ , $\mathbf{x}^T(A^T A - I)\mathbf{y} = 0$ for all $\mathbf{x},\mathbf{y}$. Letting $B = A^T A - I$, $\mathbf{e}_i^T B \mathbf{e}_j = b_{i,j} = 0$ for all $i, j$, and $A^T A - I = 0$. – Stefan Smith Mar 27 '13 at 0:09
@StefanSmith Certainly, but that doesn't follow immediately from the original hint, only the alternative one. (I think). Although I feel that the introduction of $B$ is unnecessary, simply using that $e_i^TA^TAe_j=\delta_{ij}$ shows that $A^TA = I$. – Tom Oldfield Mar 27 '13 at 0:12

All you need to do to show that $A$ is orthogonal is to show that each column of $A$ has length 1, and any that two columns of $A$ are orthogonal.

Let $e_i$ be the vector with $1$ in the $i$th component, and $0$ in the other components. Then the $i$th column of $A$ is $A e_i$, and $||A e_i || = ||e_i|| = 1$.

Now, consider the vector $v = e_i + e_j$. Since $A$ preserves lengths, $Av$ must have length $2$. But if the columns of $A$ are $c_i$, the length of this is $2 + 2c_i\cdot c_j$, which shows that $c_i \cdot c_j = 0$.

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Prof. Shor: Once again, you have answered one of my basic linear algebra questions, and your answer is remarkably elegant. However, I have already accepted another good answer, and it would be unsporting to unaccept it and accept yours. Your answer is simple enough that I might present it to my students. Thanks. – Stefan Smith Mar 27 '13 at 1:15
My answer is essentially the same proof as the one you accepted, only made much more concrete. – Peter Shor Mar 27 '13 at 1:17
I didn't recognize the connection. I will compare the two answers until I see it. – Stefan Smith Mar 27 '13 at 1:18
$||c_i + c_j||^2 = \sum_k (c_{ik} + c_{jk})^2 = \sum c_{ik}^2 + c_{jk}^2 + 2c_{ik}c_{jk} = 2+2\sum_k c_{ik}c_{jk} = 2 +2c_i \cdot c_j$ for those that do not see it immediately (as I did not). – adam W Mar 27 '13 at 2:12

You could start with $<A^TAx,x>=<Ax,Ax>=||Ax||^2=||x||^2=<x,x>$. So $<Ix,x>=<Ax,x>$ for all x. ($<>$ is the inner/dot product.) The by observe that $0=<A^TAx,x>-<Ix,x> = <(A^TA-I)x,x>$ for all x and conclude that $A^TA-I$ must be the zero matrix.

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You say (in a comment) that you would like to give a proof to "non math majors". Speaking as quite the opposite of a non math major, I find that the SVD (singular value decomposition) is very important. Maybe you can use it: $$ A = USV^*$$ $$ ||Ax|| = ||USV^*x|| = ||x|| \quad \forall x \Rightarrow \exists X \quad \text{s.t.}\quad X^*X=I \wedge ||AX|| = ||USV^*X|| = ||X|| $$ and by the property of unitary $X$ (and unitary $U$ and $V$) that may be converted to $$||USV^*|| = ||I||=||S||$$ Which gives $S$ with only unitary elements on the diagonal (otherwise if some element on the diagonal of $S$ is not unitary then some $x$ gives $||Ax|| \ne ||x||$), so $S^*S = I$ and $$A^*A = (VS^*U^*)(USV^*) = VS^*SV^* = VV^* = I$$

and of course as you seem to be using reals, use $\star^\top $ instead of $\star^*$.

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