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I was studying eigen values and suddenly they try to prove a property that states the only way eigenvalue of a matrix will be $0$ if the matrix itself was singular.

So, they went on to prove their statement by taking the determinant of $ A-\lambda I $

So, $ | A- \lambda I | = 0 $

Then they did this

$$ | A | - \lambda | I | =0 $$

Is this step valid? and if so what is the reason?

Another question: They tried to prove the characteristic equation for $ A-\lambda I $ and $ A^T -\lambda I $ is the same.

They took the determinant of the two matrices $ | A- \lambda I | $ and $ | A^T -\lambda I | $

Then they concluded that since $ | A | = | A^T | $ then the two aforementioned determinants should be the same. Were they using the same law?

Rodrigo de Azevedo
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Rifat
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    In general determinants are not additive,. but FYI, I think the most easiest (probably most common proof aswell) that a matrix with eigenvalue $\lambda = 0$ is singular is that if it has an eigenvalue of $0$ means $\exists v \ne 0 , Av = \lambda v = 0$ – Ziad Fakhoury Dec 23 '19 at 11:24
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    @ZiadFakhoury there exists nonzero $v$ so that $Av=0$. – YiFan Tey Dec 23 '19 at 11:25
  • That step is not valid. If a matrix has an eigenvalue $\lambda$, then $A-\lambda I$ is automatically singular. That does not mean $|A| = \lambda$. – Ninad Munshi Dec 23 '19 at 11:26
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    I general, that is wrong. If $A$ is an $n \times n$ matrix, then $|A - \lambda I|$ is a polynomial of degree $n$, but $|A| - \lambda |I|$ is a polynomial of degree $1$. – GEdgar Dec 23 '19 at 11:28
  • @NinadMunshi thank you. But, then how do they proof that characteristic equation for trasnposed matrices are same :/ – Rifat Dec 23 '19 at 11:28
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    The determinant for transposed matrices is the same. – GEdgar Dec 23 '19 at 11:29
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    $(A-\lambda I)^T = A^T - (\lambda I)^T = A^T - \lambda I$ because transposes are additive. Then they use the rule that transposes don't change determinants. – Ninad Munshi Dec 23 '19 at 11:29
  • @NinadMunshi wow! didn't see that one coming. Thank you so much ^_^ – Rifat Dec 23 '19 at 11:31
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    Determinates are the volume of the parallelopied defined by the vectors. When you multiply two matrices, the unit boxes of one are replaced by the boxes of the second, and the volume of the product is the product of the volume. – wendy.krieger Dec 23 '19 at 11:33
  • Where is this from? Some more context would be helpful in deciphering that step in the proof. If it’s a proof by contradiction, for instance, the authors might have made some assumption that justifies that step. Or, it could be a typo. – amd Dec 23 '19 at 20:59

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According to your explanation for the first question, it seems that the equation $$|A-\lambda I|=|A|-\lambda |I|$$ appaers in their attempt to ensure the following statement:

Statement 1: If $A$ is not singular, then $0$ cannot be an eigenvalue of $A$.

Statement 1 is equivalent to the following statement:

Statement 2: If $\lambda=0$ is an eigenvalue of $A$, then $A$ is singular.

Perhaps their proof is for Statement 2? If this is the case, then the equality holds trivially from the assumption $\lambda=0$.

As a matter of fact, the converse of Statement 2 is also correct. See Show that a matrix $A$ is singular if and only if $0$ is an eigenvalue.

modnar
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