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Prove that if $detA > 1$ then $A$ has at least one eigenvalue with $|\lambda |> 1$.

The answer says:

If all $|\lambda_j | \le 1$ then so is their product $1 \ge |\lambda_1 ...\lambda_n| = |detA|$, which is a contradiction.

How is that a contradiction? If you have all eigenvalues less then 1 then it must follow that all their products must be less than one, so I can't see how it is a contradiction?

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Perhaps it makes more sense if you rewrite the inequality and state the proof in steps:

Premise: $\text{det}A>1.$

Suppose, for the sake of contradiction, that there is NOT an eigenvalue $\lambda_i > 1$. In other words, assume that for all $\lambda_i,\;1\leq i \leq n,\;\lambda_i \leq 1$.

Then if all $|\lambda_j | \leq 1$, their product: $|\lambda_1 ...\lambda_n| = |\text{det}A| \leq 1$.

This is a contradiction to the premise: $\text{det} A > 1$.

Hence, the supposition is false, and we conclude that if it's the case that $\text{det} A >1$, then it must be the case that at least one eigenvalue must be greater than 1.

EDIT:

It is a contradiction because if we take as true that $\text{det} A > 1,$ but at the same time suppose that all eigenvalues $|\lambda_i| \leq 1$, then their product must be less than or equal to one, but their product is equal to $\text{det} (A)$, so $\text{det} (A)$ would then be $\leq 1$ This contradicts what we took to be true: $\text{det} A > 1.$

The assumption leading to the contradiction was that all eigenvalues were less than or equal to 1, so that assumption cannot be true, if the $\text{det} A > 1$. Hence, it follows, that if $\text{det} A >1$, at least one eigenvalue must be greater than 1. Which is exactly what you are to prove.

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    My pleasure! You're welcome. I consolidated the comments in my answer.2012-11-24
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Your proof is not a proof by contradiction. Rather you have essentially proved the contrapositive of the statement.

The contraposition of $A \implies B$ is $\lnot B \implies \lnot A$.

In your case, you want to prove that $\{\det(A) > 1\} \implies \{A \text{ has at-least one eigenvalues $>1$}\}$ The contraposition of the statement is $\lnot \{A \text{ has at-least one eigenvalues $>1$}\} \implies \lnot \{\det(A) > 1\}$ $\lnot \{A \text{ has at-least one eigenvalues $>1$}\} = \{A \text{ has all eigenvalues }\leq1\}$ $\lnot \{\det(A) > 1\} = \{\det(A) \leq 1\}$ Hence, the contraposition of the statement is $\{A \text{ has all eigenvalues }\leq1\} \implies \{\det(A) \leq 1\}$

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    That makes more sense. The contradiction at the end o$f$ the proof threw me off. Thanks!2012-11-24
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The open unit disk in the complex plane is closed under multiplication. Think about it.

Also: The product of the eigenvalues of a matrix is $\pm$ its determinant.

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    All complex numbers whose distance to the origin is less than 1.2012-11-24