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In an exercise, a teacher asked me to show that a given linear transformation from $\mathbb R^2$ to itself was a contraction. The matrix of the transformation with respect to the canonical basis is the following: $$\left( \begin{matrix} \frac{1}{2}\sin(x_0) & \frac{1}{2}\\ \frac{1}{2}& 0 \end{matrix} \right)$$

for some constant $x_0$. I noticed that the determinant of this matrix is $-1/4$ and this means that after the transformation all areas are scaled down by a factor of $1/4$ but this is not enough to prove that $f$ is a contraction.

I tried setting $x=(x_1,x_2)$ and $y=(y_1,y_2)$ and directly proving that $||f(x)-f(y)||\leq L||x-y||$ for some $L\in (0,1)$ but I wasn't able to do so. How can this be solved?

kabenyuk
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656475
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1 Answers1

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Note that a linear map satisfies $f(x) - f(y) = f(x-y)$. Next you can use the fact that $\|f(v)\|\leq \sigma_\text{max}\|v\|$ where $\sigma_\text{max}$ is the largest singular value of $f$. (See this post for a proof of this fact.) In your case $f$ is Hermitian, so that would just be the largest absolute value of its eigenvalues. This means that you need to show that the eigenvalues of $f$ have absolute values less than $1$.

Elliot Yu
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