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I have 2 questions -

  1. If $B = \{x(i) : i \in I \}$ be a basis of a subspace $S$ ( finite or infinite dimensional ) of a vector space $V$ and $C = \{ y(i) + S : i = 1 \text{ to } n \}$ is a basis of the finite dimensional space $V/S$ , then is it true that $B \cup \{ y(i) : i = 1 \text{ to } n \}$ is a basis to $V$?. If it is true what do I have to show to prove the statement ?

  2. Suppose $f : X \to R$ and $g : X \to R^n$ be 2 linear maps where kernel $g$ is a subset of kernel $f$ . I have shown that there exists a linear map $h : g(X) \to R$ such that $f = h(g)$. Can I extend h to R ^n without using Hann - Banach Theorems ?

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up vote 2 down vote accepted
  1. Yes. Take $x\in X$, then $\hat{x}=\sum a_i\hat{y_i}$, which says $x-\sum a_i y_i\in S$. From here should be easy to show that your set is spanning. Linear independence is also immediate.

  2. Yes. $h(t):=f(x)$, where $x$ is such that $g(x)=t$. Linearity and the condition on the kernels ensure $h$ is well defined.

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For 1, I had shown exactly that . Independence is OK . But how do you define a basis in infinite dimensional vector space? – Ester Aug 27 '12 at 21:33
Ricky, whether $V$ is finite-dimensional or not, $B$ is a basis for $V$ means every element of $V$ is a finite linear combination of elements of $B$, and if a finite linear combination of elements of $B$ is zero then all the coefficients are zero. – Gerry Myerson Aug 28 '12 at 0:26
@Gerry If we take R over Q then how can you express e^x as a finite linear combination of elements of Q – Ester Aug 28 '12 at 9:08
Ricky, I don't understand your question. The reals are a vector space over the rationals. There is a basis $B$, consisting of real numbers, such that every real number is a finite linear combination of elements in $B$, the coefficients being rational. But $e^x$ is not a real number, so how does it enter the discussion? – Gerry Myerson Aug 28 '12 at 13:47
I mean e is a real number but how can we express it as a linear combination of rationals – Ester Aug 29 '12 at 19:44

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