# Eigenvalue products

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

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 $\text{det} A > 0,$ 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 $\text{det} A > 0$. 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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But it still will be true that the detA will be less than 1 if all the eigenvalues are less than 1. –  Q.matin Nov 24 '12 at 2:26
You are asked to prove that IF det A > 1, then at least one eigenvalue must be > 1. If at least one eigenvalue is not greater than one, then det A $\leq$ 1. –  amWhy Nov 24 '12 at 2:29
@Q.matin : Your problem does not seem to be with the equations but with the logic behind the argument. Marvis tried to explain this as well as he could in his answer, so I suggest you give some deep thinking on it and ask a question afterwards if you still have problems understanding his explanation. –  Patrick Da Silva Nov 24 '12 at 2:31
Yes I know that. But how does that proof I gave a contradiction? Maybe, I am overthinking because I really do not know. –  Q.matin Nov 24 '12 at 2:33
My pleasure! You're welcome. I consolidated the comments in my answer. –  amWhy Nov 24 '12 at 2:45
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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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But I dont understand how that proof I gave saitisfies the question being asked. Because the proof I gave isnt a contradiction, isnt the proof true? That if every product that is less than 1 will be equal to an answer that is less than 1. –  Q.matin Nov 24 '12 at 2:30
@Q.matin You have proved that $$\text{If all the eigenvalues \lambda satisfy \vert \lambda \vert \leq 1, then \det(A) \leq 1.}$$ The statement you have proved is the same statement as $$\text{If \det(A) > 1, then there is at-least one eigenvalue \lambda that satisfies \vert \lambda \vert > 1.}$$ –  user17762 Nov 24 '12 at 2:34
That makes more sense. The contradiction at the end of the proof threw me off. Thanks! –  Q.matin Nov 24 '12 at 2:36
Also: The product of the eigenvalues of a matrix is $\pm$ its determinant.
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