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Note: In the following, if \mathbb{F} is a set and both m and n are positive integers, then mxn matrices with entries from \mathbb{F}. The problem below has many applications. If T is a linear map from complex vector space V to itself, and \lambda is an eigenvalue of T, then \lambda is a simple eigenvalue of T if  ker((T-\lambda I)^2)\subset ker(T-\lambda I).

1. Suppose V is a vector space of dimension n over field \mathbb{F} where you may assume that \mathbb{F} is either \mathbb{R} or \mathbb{C}, and let T be a linear map from V to V. Show that V = ker(T) φ Range (T) if and only if ker(T)2 C ker (T) .

2. Use the previous result to prove the following. Assume that \mathbb{F} is \mathbb{R} or \mathbb{C}, and assume that dim(V)= n.

(a). If M\in \mathbb{F}^{n \times n} , and M is idempotent, then \mathbb{F}^n=ker(M)\oplus Range(M) .

(b). If T: V \to V is linear, and \lambda is an eigenvalue of T, then V=ker((T-\lambda I)^n)\oplus Range((T-\lambda I)^n) .

(c). If T: V \to V is linear, and \lambda is a simple eigenvalue of T, then V=ker((T-\lambda I))\oplus Range((T-\lambda I)) .

  




mxn

















V = ker(T) φ Range (T)
ker(T)2 C ker (T)














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Answer #1

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Note: In the following, if is a set and both and are positive integers, then matrices with entries from . The problem below has many applications. If is a linear map from complex vector space t...
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