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Consider the mapping $T:\mathbb{R}^n\mapsto\mathbb{R}^n$ defined by $T(\vec{x})=A\vec{x}$ where $A$ is a $n\times n$ matrix. Find the necessary and sufficient conditions on $A$ such that $\|T(\vec{x})\|=|\det A|\cdot\|\vec{x}\|$ for all $\vec{x}$. Here $\|\cdot\|$ denotes the norm.

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Hint: Consider the action on a standard basis of $\mathbb{R}^n$. – hardmath Dec 27 '12 at 18:44
Hi. Could you please elaborate a little more? Thanks. – Christmas Bunny Dec 31 '12 at 19:16
Do a computation that shows if $e_i,e_j$ are distinct standard basis vectors, $Ae_i$ and $Ae_j$ are orthogonal. – hardmath Jan 1 '13 at 4:08
up vote 1 down vote accepted

Restating the (Euclidean) norm identity:

$$ ||Ax||^2 = det(A)^2 ||x||^2 \;\;\; (*) $$

for all $x \in \mathbb{R}^n$. Since $det(A)^2$ is constant (independent of $x$), familiarity with Rayleigh quotients might lead one to conclude that $A^T A$ has eigenvalue $det(A)^2$ of (geometric) multiplicity $n$.

But we can prove $A^T A = det(A)^2 I$ with a couple of brief computations.

Let $e_i$ be the standard basis vector of $\mathbb{R}^n$ whose $i^{th}$ component is $1$. The diagonal entry $(A^T A)_{ii} = e_i^T (A^T A)e_i$ is then $det(A)^2$:

$$ ||Ae_i||^2 = det(A)^2 ||e_i||^2 = det(A)^2 $$

It remains to show any off-diagonal entries of $A^T A$ are zero. Suppose $i \neq j$. On one hand:

$$||A(e_i+e_j)||^2 = det(A)^2 ||e_i+e_j||^2 = 2 det(A)^2 $$

On the other hand expanding the "inner product" form:

$$ (A(e_i+e_j))^T A(e_i+e_j) = e_i^T(A^T A)e_i + 2 e_i^T(A^T A)e_j + e_j^T(A^T A)e_j $$

$$ ||A(e_i+e_j)||^2 = 2 det(A)^2 + 2 (A^T A)_{ij} $$

implying that $(A^T A)_{ij}$ is zero. $\; \therefore \; A^T A = det(A)^2 I$ .

Taking determinants of both sides:

$$ det(A)^2 = det(A)^{2n} $$

If dimension $n=1$ this doesn't place any restriction on $A$, and indeed every "linear transformation" $T: \mathbb{R} \to \mathbb{R}$ satisfies the norm identity $(*)$.

But if $n \gt 1$ this implies either $det(A)^2 = 0$ or $1$, resp. that $A$ is zero or orthogonal (since $A^T A = I$).

The converse is easy to see, that norm identity $(*)$ holds if $A$ is zero or orthogonal.

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Hint: If $\det A=0$, then everything is obvious. Otherwise consider $S=|\det A|^{-1}T$. It is norm preserving hence it is

unitary operator

So the following equalities holds


And you can rewrite them in terms of $T$.

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How do we show $S(\vec{x})=|\det A|^{-1}T(\vec{x})=|\det A|^{-1}A\vec{x}$ is norm preserving? When computing the norm, we don't necessarily have $\|Ax\|=|\det A|\cdot\|x\|$. Thank you! – Christmas Bunny Dec 27 '12 at 20:36
$\Vert S(x)\Vert=\Vert |\det A|^{-1}T(x)\Vert=|\det A|^{-1}\Vert T(x)\Vert=\Vert x\Vert$ – Norbert Dec 27 '12 at 20:45
So we get $AA^T=A^TA=(\det A)^2I$. Is there a way to show $|\det A|=1$ without using eigenvalues? Thank you! – Christmas Bunny Dec 31 '12 at 19:14
Take the determinant of both sides. For $n \gt 1$ this implies what you want. – hardmath Jan 1 '13 at 4:15
@YifengXu: No, we will not (except in dimension $n=1$). – hardmath Jan 1 '13 at 18:49

The condition is satisfied iff $T=0$ (ie, $A=0$) or $T$ is unitary (ie, $A$ is unitary).

If $T=0$ or $T$ is unitary, it is obvious that the condition is satisfied.

If $\det A =0 $, then clearly $T=0$, so suppose $\det A \neq 0$. Then $U = \frac{1}{\det A} A$ is a unitary operator. If $\lambda_1,...,\lambda_n$ are the eigenvalues of $A$, then the eigenvalues of $U$ are $\frac{\lambda_k}{\lambda_1 \cdots \lambda_n}$, and all have modulus $1$. It follows that $|\lambda_k|$ is a constant, and from this it follows that $|\lambda_k|=1$. Since $\det A = \lambda_1 \cdots \lambda_n$, we have $|\det A| =1$ and it follows that $A= (\det A) U$ is unitary.

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Why does $\frac{\lambda_k}{\lambda_1 \cdots \lambda_n}$ have modulus 1? Thanks so much! – Christmas Bunny Dec 27 '12 at 20:00
It is an eigenvalue of $U$ which is unitary. If $U$ is unitary, then $\|Ux\| = \|x\|$, hence if $Uv = \lambda v$, then you must have $|\lambda| = 1$. – copper.hat Dec 27 '12 at 20:01
I checked and I think you meant $U$ to be a unitary matrix as well, right? But why does $U$ necessarily satisfy $UU^T=U^TU=I$? – Christmas Bunny Dec 27 '12 at 20:30
You can express the inner product in terms of the norm (polarization identity), and use this along with $\|Ux\| = \|x\|$ to show that $\langle Uy, Ux \rangle = \langle y, x \rangle$ for all $x,y$. From this it follows that $U^*U = I$. – copper.hat Dec 27 '12 at 20:51
Well, I am assuming that you are using the Euclidean norm. Then $\langle y, x \rangle = \sum_n y_n x_n$. Look up the polarization identity. This expresses the (Euclidean) inner product in terms of the (Euclidean) norm, from which the desired result follows. – copper.hat Dec 27 '12 at 22:26

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