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## Homework Statement

Prove true or false.

If A^2+A=0 then λ=1 may not be an eigenvalue.

## Homework Equations

To find the eigenvalues of A I find the solutions to det(λ-A).

The definition of an eigenvalue from my understanding, AX = λX.

A(A+I) = 0

## The Attempt at a Solution

I'm unable to find the connection between the restriction of A^2+A=0 and its effects on the eigenvalues.

Since A^2 + A =0 , I've thought of some matrices where this is true.

Scenarios:

A = 0

A =\begin{pmatrix}

-1\end{pmatrix}

A = \begin{pmatrix}

-1 & 0\\

0 & -1\end{pmatrix}

A = \begin{pmatrix}

-1 & -1\\

0 & 0\end{pmatrix}

It seems that when you have -1 on the diagonal the first equation holds true and the eigenvalues are -1 or 0. With that being said, this statement

*appears true*, but I have no idea how to find scenarios which could be counterexamples nor do I have a formal proof if the statement is indeed true.

**Clarification 1:**Since I've only been able to find matrices with -1 (after reduction) on the diagonal to satisfy A^2 + A = 0, it appears that the eigenvalue will be always negative because when I find the characteristic polynomial its always in the form of (λ+1)^n which gives me a negative eigenvalue.

Any help would be appreciated.

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