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This is probably trivial, but a lot of time has passed since I studied these topics, so I'm a bit rusty.

So, let $\mathbf{A}$ be square matrix in $\mathbb{R}^{n\times n}$. We want to prove that there exists $c\in\mathbb{R}_{>0}$ such that the linear map $\mathbf{I}-c\mathbf{A}$ is a contraction. I need to prove that $\forall\ \mathbf{v},\mathbf{w}\in\mathbb{R}^{n}$, there exists $k\in[0,1[$ such that

$$||(\mathbf{I}-c\mathbf{A})(\mathbf{v}-\mathbf{w})||\leq k||\mathbf{v}-\mathbf{w}||$$

which reduces to proving that $\forall\ \mathbf{v}\in\mathbb{R}^{n}$,

$$||(\mathbf{I}-c\mathbf{A})\mathbf{v}||\leq k||\mathbf{v}||$$

Now, if I use an induced norm, I know that

$$||(\mathbf{I}-c\mathbf{A})\mathbf{v}||\leq ||\mathbf{I}-c\mathbf{A}||\cdot||\mathbf{v}||$$

so it all amounts to proving that $||\mathbf{I}-c\mathbf{A}||<1$. Now, from the triangle inequality and using the fact that $||\cdot||$ is an induced norm, I can get

$$1-c||\mathbf{A}||\leq||\mathbf{I}-c\mathbf{A}||\implies c||\mathbf{A}||\geq1-||\mathbf{I}-c\mathbf{A}||$$

but this doesn't seem to help me much. What am I missing?

DeltaIV
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    This looks false to me. Suppose $A= - I$ then you want $(1+c)I$ to be a contraction for some $c > 0$ which is clearly not true. Did you mean to include that $A$ is positive semidefinite for example? In that case your question has already been answered here. – Rammus Sep 27 '21 at 10:07
  • @Rammus nice! How did you come up with the counterexample? I had a look at the other question, but it doesn't seem a complete answer to me. It only proves that all eigenvalues of $\mathbf{I} - c\mathbf{A}$ have absolute value < 1, but how do you get from here to proving that the norm of $\mathbf{I} - c\mathbf{A}$ is < 1? – DeltaIV Sep 27 '21 at 21:20
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    If you're ever interested in whether a statement holds you can always try simple examples to gain confidence. The identity matrix is always one to check. Regarding the other question, is your matrix is symmetric then the norm is equal to the maximal absolute eigenvalue. – Rammus Sep 28 '21 at 08:29
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    Right, if $\mathbf{A}$ is symmetric then it's diagonalizable, so the result is trivial. Silly me! See https://math.stackexchange.com/a/603395/169160 – DeltaIV Sep 28 '21 at 08:49

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