# Norm indueced by a matrix with eigenvalues bigger than 1

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1. Sep 23, 2015

### Diffie Heltrix

Suppose we pick a matrix M\in M_n(ℝ) s.t. all its eigenvalues are strictly bigger than 1.
In the question here the user said it induces some norm (|||⋅|||) which "expands" vector in sense that exists constant c∈ℝ s.t. ∀x∈ℝ^n |||Ax||| ≥ |||x||| .

I still cannot understand why it's correct. How can one pick this norm explicitly? The comments suggested dividing to diagonal case and normal Jordan form but I cannot see in both cases how to define this norm.

2. Sep 23, 2015

### BvU

Doesn't it say there that the smallest of the eigenvalues is a candidate for that norm ?

3. Sep 23, 2015

### Diffie Heltrix

So You map every $x=Id\cdot x$ to 1? That's not a norm.

4. Sep 23, 2015

### BvU

There's a c missing in your version:$$\forall A\in E_n(\mathbb{R}),\exists c>1: \forall x\in\mathbb{R}^n, |||Ax|||\ge c|||x|||$$the original at our colleagues site looks better.

I don't think I wanted to "map every $x=\mathbb I \cdot x$ to 1 ?" what gave you that impression ? (what is the ld in your post ?)

And I don't think you can do much better than the smallest eigenvalue $\lambda_i$. After all, if you pick that eigenvector $y_i$ as x then $||| Ax ||| = \lambda_i |||x|||$.

5. Sep 23, 2015

### Diffie Heltrix

So how the norm is defined precisely? Maybe $|||x|||_A=\lambda_i ||x||_1$? How lambda_i is connected here?

6. Sep 23, 2015

### BvU

Can't read that. What is $|||x|||=||Ax||_1$?

http://www.math.cuhk.edu.hk/course_builder/1415/math3230b/matrix%20norm.pdf [Broken] doesn't take the smallest , but the biggest K

Last edited by a moderator: May 7, 2017
7. Sep 23, 2015

### Diffie Heltrix

Again, given vector $x\in\mathbb{R}^n$, How can I define the norm associated with $A$ (denoted by $|||\cdot|||_A$) s.t. $|||Ax|||_A \ge c |||x|||_A$ (where $c>1$)? What is the connection to the minimal eigenvalue exactly?

8. Sep 24, 2015

### BvU

To me it seems they are one and the same thing...

9. Sep 25, 2015

### Hawkeye18

You want to prove the following statement: if $A$ is a matrix with eigenvalues $\lambda_k$, and $r=\min_k |\lambda_k|$, then for any $a<r$ you can find a norm $||| \,\cdot\,|||$ such that $||| Ax|||\ge a|||x|||$ for all vectors $x$.

We will be looking for a norm of form $|||x|||=\|Sx\|$, where $\|x\|$ is the standard norm in $\mathbb R^n$ or $\mathbb C^n$ and $S$ is an invertible transformation. Then the required estimate can be rewritten as $$\|SAx\|\ge a\|Sx\| \qquad \forall x,$$ or equivalently $$\|SAS^{-1} x\|\ge a\|x\| \qquad \forall x.$$
By the theorem about upper triangular respresentation there exists an orthonormal basis such that in this basis such that the matrix of $A$ in this basis is upper triangular, so we can fix this basis and assume without loss of generality that $A$ is upper triagular. We can write $A=D+T$, where $D$ is a diagonal matrix and $T$ is strictly upper triangular, i.e. all its non-zero entries are strictly above the main diagonal. Note that the diagonal entries are exactly eigenvalues of $A$.

Now, let $\epsilon>0$ be a small numer, and let our $S$ be a diagonal matrix with entries $1, \epsilon^{-1}, \epsilon^{-2}, \ldots, \epsilon^{-n-1}$ on the diagonal (exactly in this order). You that can see that the matrix $SAS^{-1}$ is obtained from $A$ as follows: the main diagonal remais the same as the main diagonal of $A$, the first diagonal above main is multiplyed by $\epsilon$, the second diagonal above main is multiplied by $\epsilon^2$, etc.

So, $SAS^{-1} = D+ T_\epsilon$, and by picking sufficiently small $\epsilon$ we can make the norm of $T_\epsilon$ as small as we want. For our purposes it is sufficient to get $\|T_\epsilon\|\le r-a$, can you see how to proceed from here?

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