Alright, let's work through this proof clearly and step by step. Here's how we can confirm that any eigenvalue of matrix ( A ) (where ( A^3 = A )) must be -1, 0, or 1:

Step 1: Start with the core definition of eigenvalues #

First, let's recap the basics to make sure we're aligned. If ( \lambda ) is an eigenvalue of ( A ), there exists a non-zero eigenvector ( \mathbf{v} ) such that:
[ A\mathbf{v} = \lambda\mathbf{v} ]
This relationship is the foundation of our proof—we'll build everything from here.

Step 2: Compute ( A^3\mathbf{v} ) in two ways #

We know ( A^3 = A ), so let's calculate ( A^3\mathbf{v} ) using two different approaches:

1. Using the eigenvalue property repeatedly:
   First, ( A^2\mathbf{v} = A(A\mathbf{v}) = A(\lambda\mathbf{v}) = \lambda(A\mathbf{v}) = \lambda^2\mathbf{v} ).
   Then, ( A^3\mathbf{v} = A(A^2\mathbf{v}) = A(\lambda^2\mathbf{v}) = \lambda^2(A\mathbf{v}) = \lambda^3\mathbf{v} ).
2. Using the given condition ( A^3 = A ):
   Directly substitute to get ( A^3\mathbf{v} = A\mathbf{v} = \lambda\mathbf{v} ).

Step 3: Equate the two expressions and solve for ( \lambda ) #

Since both results equal ( A^3\mathbf{v} ), we can set them equal to each other:
[ \lambda^3\mathbf{v} = \lambda\mathbf{v} ]
Rearrange terms to isolate the scalar factor:
[ (\lambda^3 - \lambda)\mathbf{v} = 0 ]

Here's the key detail: ( \mathbf{v} ) is a non-zero eigenvector. A scalar multiplied by a non-zero vector can only equal the zero vector if the scalar itself is zero. So we solve the polynomial equation:
[ \lambda^3 - \lambda = 0 ]
Factor the left-hand side to find roots:
[ \lambda(\lambda^2 - 1) = 0 ]
[ \lambda(\lambda - 1)(\lambda + 1) = 0 ]

The solutions are exactly ( \lambda = 0 ), ( \lambda = 1 ), and ( \lambda = -1 ).

Conclusion #

That's all there is to it! We've proven that any eigenvalue of ( A ) must be one of these three values—no other eigenvalues are possible for a matrix satisfying ( A^3 = A ).

内容的提问来源于stack exchange，提问作者HAC