Difference between revisions of "TDSM 8.15"
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2) If for <math>k = n</math> the statement is true, then: | 2) If for <math>k = n</math> the statement is true, then: | ||
| − | <math>A^n = \lambda^n v</math> | + | <math>A^n v = \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> | <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> | ||
Latest revision as of 07:40, 29 November 2017
Matrix [math]A[/math] has an eigenvector [math]v[/math] with eigenvalue [math]\lambda[/math].
[math]\Rightarrow A v = \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 v = \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.
