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Let $A$ be a real $2\times 2$ matrix such that $\det A=1$, show that $\|A\|=\left\|A^{-1}\right\|$.

Any hint would be appreciated, thanks.

EDIT: $\|\cdot\|$ is the operator norm $\|A\|=\max_{\|x\|=1}\|Ax\|$, all vector norms are Euclidean norms.

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2  
What is $\|\cdot \|$? –  Patrick Li Dec 15 '12 at 17:07
    
op norm, added an edit –  Shai Deshe Dec 15 '12 at 17:12
    
Uses the facts: (1) $det(I_{n\times n})=1$, (2) $A^{-1}A=AA^{-1}=I_{n\times n}$, (3) $\|A B\|\leq \|A\|\cdot\| B\|\leq |trace(A^TB)|$ (4), (5) $det(I_{n\times n})=\|I_{n\times n}\|=1$ and (6) $trace(I_{n\times\n})=1$. –  Elias Dec 15 '12 at 17:13
    
Don't you mean $tr(I_n)=n$? –  Shai Deshe Dec 15 '12 at 17:34

2 Answers 2

Hint: Use the definition of the inverse matrix

If A is invertible, then $$ \det(\mathbf{A}) \mathbf{A}^{-1}=\mathrm{adj}(\mathbf{A}) , $$ where $\mathrm{adj}(\mathbf{A})$ is the adjugate matrix.

The adjugate of the 2 × 2 matrix

$$\mathbf{A} = \begin{pmatrix} {{a}} & {{b}}\\ {{c}} & {{d}} \end{pmatrix}$$

is $$\operatorname{adj}(\mathbf{A}) = \begin{pmatrix} \,\,\,{{d}} & \!\!{{-b}}\\ {{-c}} & {{a}} \end{pmatrix}.$$

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Recall the singular value decomposition of $A$ as $A_{2 \times 2} = U \begin{bmatrix}\Sigma_{11} & 0\\ 0 & \Sigma_{22} \end{bmatrix}V^*$, then $A^{-1} = V^{*^{-1}} \begin{bmatrix}\dfrac1{\Sigma_{11}} & 0\\ 0 & \dfrac1{\Sigma_{22}} \end{bmatrix} U^{-1}$. Since $\det(A) = 1$, we have that $\Sigma_{11} \Sigma_{22} = 1$. Also, recall that $$\Vert B \Vert_2 = \max \{\sigma_i: \sigma_i \text{ is a singular value of }B\}$$

Hence, $\Vert A \Vert_2 = \Sigma_{11}$ and $\Vert A^{-1} \Vert_2 = \dfrac1{\Sigma_{22}} = \Sigma_{11}$. Hence, we get that $$\Vert A \Vert_2 = \Vert A^{-1} \Vert_2$$

The equality also hold good for the Frobenius norm since $$\left \Vert A \right \Vert_{F} = \sqrt{\Sigma_{11}^2 + \dfrac1{\Sigma_{11}^2}} = \left \Vert A^{-1} \right \Vert_F$$

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