TDSM 8.15

Matrix [math]A[/math] has an eigenvector [math]v[/math] with eigenvalue [math]\lambda[/math].

[math]\Rightarrow A = \lambda v[/math]

[math]\Rightarrow A^2 v = (A \cdot A) \cdot v = A \cdot (A \cdot v) = A \cdot (\lambda v) = \lambda (A \cdot v) = \lambda \cdot (\lambda v) = \lambda^2 v[/math]

So [math]v[/math] is also an eigenvector for [math]A^2[/math], and its coresponding eigenvalue is [math]\lambda ^2[/math]

Generalization: for [math]A^k[/math], for [math]2 \leq k \leq n[/math], it has an eigenvector [math]v[/math] and its coresponding eigenvalue is [math]\lambda ^k[/math]

Proof: using inductive proof:

1) For [math]k = 2[/math]: the statement is true.

2) If for [math]k = n[/math] the statement is true, then:

[math]A^n = \lambda^n v[/math]

[math]\Rightarrow A^{n+1} v = (A^n \cdot A) \cdot v = A^n \cdot (A \cdot v) = A^n \cdot (\lambda v) = \lambda (A^n \cdot v) = \lambda \cdot (\lambda^n v) = \lambda^{n+1} v[/math]

So for [math]k = n+1[/math] the statement is also true.

[math]\Rightarrow[/math] The statement is proved to be true.