Vector spaces and span of linearly independent vectors

Question

I have a question. Are any n linearly independent vectors in a vector space V with dimV=n a bases for V? why or why not? How can I be sure that they span the vector space V? I'm asking this because I wish to prove that for any given Matrix of size n x n over a field F, if it has n distinct eigenvalues then it is diagonalisable. So if I consider the linear transformation associated with the the matrix A, T: $F^n--->F^n$ then dim$F^n$=n. But i'm not sure how to proceed after that, may someone please help? The question I asked, I think should help me because that would mean that the number of distinct eigenvalues is n and so is dim$F^n$, but even if that was the case, then i'm not sure how to proceed, may someone please help?

Answer 1

That depends upon the facts that you can use. If you can use the fact that, in a $n$-dimensional space, any set with more that $n$ vectors is linearly dependent, that's easy. If a set $S$ has exactly $n$ elements and $S$ is linearly independent, suppose that $\langle S\rangle$ is nor the whole $V$. Take $w\in V\setminus\langle S\rangle$. Then no element of $S\cup\{w\}$ can be written as a linear combination of the other elements of that set. Therefore, $S\cup\{w\}$ is linearly independent. So, a contradiction is reached.

Answer 2

If $V_n$ is a set of $n$ linearly independent vectors of $V$ and $V$ is a n-dimensional vector space, then these $n$ vectors will span the space. This follows as elementarily as it sounds, as $n$ linearly independent vectors belonging to a space with $n$ dimensions means you have one elemental for every dimension of that space that a combination of them could build any element of the space.

Answer 3

From the definition of a basis, a subset $B$ is a basis of space $V$ if its vector components are linearly independent and span $V$. To check the span requirement, any vector in $V$ should be a linear combination of all vectors in $B$.