Homework Help: Linear Algebra proofs

1. Dec 12, 2009

rhyno89

1. The problem statement, all variables and given/known data
Ok so I am stick on three proofs for my linear algebra final adn help on any of all of them would really help with my studying

For the first 2 assume that A is an nxn matrix

1.If the collumns of A span Rn then the homogenous system Ax = 0 has only the trivial solution
2. If the collumns of A are linearly independent, then the columns of A span Rn

These 2 have to be proved without referencing other parts of the invertible matrix theorem

And then,

3. Be able to prove: If A is an nxn matrix then lambda is an eigencalue of A if and only if det(A-lamda*In) = 0

2. Relevant equations

3. The attempt at a solution

the first two i have a better idea at than the third, since its an nxn i know that if it spans Rn then it has a pivot in every row and coincidently in every row and therefore every column. This same fact can be used to explain number 2 with them being linearly independent. My problem with these two is that Im having trouble since I cant reference them being part of the Invertible matrix theorem.

For the third one I think I am on the right track but not sure
The determinat of A-IL must equal zero because the determinant of A is simply the product of the eigenvalues. If you replace each eigenvalue into the determinat one at a time and multiply it by I, one of the entries will be replaced by zero and any other vlues multiplied by zero will result in zero. As a result, the determinat must equal zero.

Any help would be great

2. Dec 12, 2009

rasmhop

A somewhat different take on 1 and 2 than yours (I suspect yours could be formalized as well, but I find the following approach more natural):
For 1: In other words you are told that the column space of A equals R^n. Then what is the rank and nullity of A? Knowing the nullity what can you tell about the space of solutions?

For 2: In other words the n-columns are linearly independent in the column space whose dimension is at most n. Can n-linearly independent vectors possibly fail to span a n-dimensional space?

For 3: For $\lambda$ to be an eigenvalue means that there exists a non-zero vector v such that,
$$Av = \lambda v = \lambda I_n v$$
which is equivalent to:
$$(A-\lambda I_n)v = 0$$
If $A-\lambda I_n$ has an inverse B, try to left-multiply the above equality by B.

For the other direction assume $\det(A-\lambda I_n)=0$ which is equivalent to $A-\lambda I_n$ not having an inverse, or having nullity >0. Since the nullity is >0 the null space contains a non-zero vector v which you can show is a eigenvector with eigenvalue $\lambda$.

3. Dec 12, 2009

rhyno89

Ok I think i can take it from here, thanks a lot you really saved me with this

4. Dec 13, 2009

HallsofIvy

In (1) you have n vectors that span an n dimensional space.

In (2) you have n vectors in an n dimensional space.

Do you recall that a basis for a space has three properties- but that any two are enough to prove the third?