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Let $T$ be an invertible linear operator on $\mathbb{R}^2$. Is it true that if $T$ has determinant $\pm 1$ , then $T$ and $T^{-1}$ have the same norm (the usual norm operator)?

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Note that this is not true for matrices larger than $2 \times 2$, e.g. the diagonal matrix with diagonal elements $[2,2,1/4]$ and its inverse don't have the same norm. –  Robert Israel Jul 20 '12 at 1:48
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2 Answers 2

If you use the usual norm on $\mathbb{R}^2$, then $\|A\|=\|A^T\|$, where $A^T$ denote the transpose of $A$.


Edit: Suppose $\|A\|\ne0$. For any $v\in\mathbb{R}^2$, $\|v\|=1$, we have $$ \|Av\|^2=\langle Av,Av\rangle = \langle v,A^TAv\rangle\le\|v\|\|A^TAv\|\le\|A^TA\|\le\|A^T\|\|A\| $$ so, $\|A\|^2\le\|A^T\|\|A\|$, and hence, $\|A\|\le\|A^T\|$. Analogously, $\|A^T\|\le\|A\|$, so $\|A\|=\|A^T\|$.


If $A=\pmatrix{a&b\\c&d}$ and $\det A=\pm1$, then $A^{-1}=\pm\pmatrix{d&-b\\-c&a}$. Let $B=(A^{-1})^T$. If $(x_1,x_2)\in\mathbb{R}^2$ then $\|A(x_1,x_2)\|=\|B(x_2,-x_1)\|$. From this, we have $\|A\|=\sup\limits_{\|x\|=1}\|Ax\|=\sup\limits_{\|x\|=1}\|Bx\|=\|B\|=\|A^{-1}\|$.

Edit: Let $(x_1,x_2)\in\mathbb{R}^2$. Then $\|A(x_1,x_2)\|=\|(ax_1+bx_2,cx_1+dx_2)\|=(dx_2+cx_1,-bx_2-ax_1)=\|B(x_2,-x_1)\|$, so for each $x\in\mathbb{R}^2$, $\|x\|=1$, we find $y\in\mathbb{R}^2,\|y\|=1,$ such that $\|Ax\|=\|By\|$, then $\|A\|\le\|B\|$. Analogously we have $\|B\|\le\|A\|$, and hence $\|A\|=\|B\|$.

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Yes, but the norm we are dealing with is the usual norm as linear operators not the Frobenius norm. –  david Jul 20 '12 at 3:14
    
Yuki, your last statement does not make any sense. You are using two different definitions of the norm. –  david Jul 20 '12 at 3:19
    
@david: I used the usual norm ($\|A\|=\sup\limits_{\|x\|=1}\|Ax\|$) –  Yuki Jul 20 '12 at 3:45
    
That is ok. Thanks. –  david Jul 20 '12 at 20:13
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If you define the matrix norm as spectral norm or 2-norm, which is the largest singular value $\mu$, then $T$ and $T^{-1}$ have the same norm if $|det(T)|=1$, because the singular values of the $T^{-1}$ is just the inverse of singular values of the $T$, and $|det(T)|=\mu_1 \mu_2 = 1$, so $T$ and $T^{-1}$ have the same norm $|\mu|$.

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That is not a norm. Consider $\pmatrix{0 & 1\cr 0 & 0\cr}$ whose only eigenvalue is $0$. –  Robert Israel Jul 20 '12 at 1:43
    
@RobertIsrael why 0 cannot be a norm? –  chaohuang Jul 20 '12 at 2:19
    
@chaohuang We usually require $\|x\| = 0 \Leftrightarrow x = 0$, no? –  Dylan Moreland Jul 20 '12 at 3:09
    
I changed the definition of norm via eigenvalue to singular value. I think this is OK. –  chaohuang Jul 20 '12 at 4:52
    
Yes, that's OK now. –  Robert Israel Jul 20 '12 at 5:57
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