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.
Find the necessary and sufficient conditions on $A$ such that $\|T(\vec{x})\|=|\det A|\cdot\|\vec{x}\|$ for all $\vec{x}$.
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2Hint: Consider the action on a standard basis of $\mathbb{R}^n$. – 2012-12-27
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0Hi. Could you please elaborate a little more? Thanks. – 2012-12-31
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0Do a computation that shows if $e_i,e_j$ are distinct standard basis vectors, $Ae_i$ and $Ae_j$ are orthogonal. – 2013-01-01
3 Answers
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.
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
$SS^T=S^TS=I$
And you can rewrite them in terms of $T$.
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0How 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! – 2012-12-27
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1$\Vert S(x)\Vert=\Vert |\det A|^{-1}T(x)\Vert=|\det A|^{-1}\Vert T(x)\Vert=\Vert x\Vert$ – 2012-12-27
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0So we get $AA^T=A^TA=(\det A)^2I$. Is there a way to show $|\det A|=1$ without using eigenvalues? Thank you! – 2012-12-31
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0Take the determinant of both sides. For $n \gt 1$ this implies what you want. – 2013-01-01
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0If we take determinant of both sides, won't we just get $(\det A)^2=(\det A)^2$? – 2013-01-01
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1@YifengXu: No, we will not (except in dimension $n=1$). – 2013-01-01
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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0Why does $\frac{\lambda_k}{\lambda_1 \cdots \lambda_n}$ have modulus 1? Thanks so much! – 2012-12-27
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1It is an eigenvalue of $U$ which is unitary. If $U$ is unitary, then $\|Ux\| = \|x\|$, hence if $Uv = \lambda v$, then you must have $|\lambda| = 1$. – 2012-12-27
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0I checked http://en.wikipedia.org/wiki/Unitary_matrix and http://en.wikipedia.org/wiki/Unitary_operator I think you meant $U$ to be a unitary matrix as well, right? But why does $U$ necessarily satisfy $UU^T=U^TU=I$? – 2012-12-27
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0I guess I really want to ask how to show $U$ is unitary (since unitary implies $UU^T=I$)? Thanks for your time! – 2012-12-27
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1You 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 $U^*U = I$. – 2012-12-27
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0Since $\langle y, x \rangle = \|y\|\|x\|\cos\theta$, how do we show the angles are also preserved? – 2012-12-27
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1Well, 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. – 2012-12-27