Question #161871

The function f1 and f2 are linearly independent if there is a real number X such that k1f1(x) + k2f2(x)=0

(2) Prove that if V is a subspace of a vector space W and if V is infinite demensial then so in W.

(3) Prove that if A is digonalizble matrix then the rank of A is the number of non-zero eigen values of A.


1
Expert's answer
2021-02-24T06:41:38-0500

2) By Theorem:

A vector space VV is infinite dimensional if and only if there is a sequence of vectors v1,v2,v3,v4,...v_1,v_2,v_3,v_4,... such that for all nZ+n\isin Z^+ the subsequence v1,v2,...,vnv_1,v_2,...,v_n

is linearly independent.

Since V is a subspace of a vector space W, then the given sequence of vectors exists in W also. So, W is infinite dimensional.


3) Let's say D is the diagonalized form of A. Then the diagonal elements of D are precisely the eigenvalues of A. The rank of D is the number of linearly independent columns. Obviously this equals the number of non-zero eigenvalues. Since the rank of A and the rank of D are the same, the conclusion follows.


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