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FYI - you can rearrange "alienr" to
make "linear" (and can rearrange "a real big
learn" to make "linear algebra"). Whoa!
Besides all that, this picture advertises the main equation in
Ch. 3:
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You'll see it in several definitions and proofs. This web page discusses several aspects of this equation and gives you some TF practice at the end.
| It's relation with Ax=0 |
Suppose the columns of A are called v1,v2, ... vn and the entries of x are called c1,c2 ,...cn. Then "the main equation" above is the same as Ax = 0. We can interpret the left hand side (LHS) as a linear combination of the column vectors v1,v2, ... vn and can interpret c1,c2 ,...cn as a solution of the homogeneous linear system. We know that it always has the trivial solution. But does it have nontrivial ones, too?? This depends on whether there are free variables - which depends on the size, shape and entries in A. If it does have nontrivial solutions, we say that the column vectors are linearly dependent and if not, that they are linearly independent.
If A is square, the situation is pretty simple - nonsingular means only trivial solutions, so the columns are linearly independent. Singular (and square) means nontrivial solutions, so the columns are linearly dependent. Likewise, if m<n the system has free variables, so the columns are always linearly dependent. If m>n, I don't know any shortcuts, but the GE process will tell whether there are free variables (and nontrivial solutions, and dependency).
| Other Situations |
In Chapter 3, we study other kinds of vectors (polynomials, etc) that cannot be combined into a matrix. Then, the main equation does not stand for Ax=0 anymore, and it's not really about linear systems. I might still refer to the c1,c2 ,...cn as a "solution" (though most people don't). The above vocabulary (in red) is still the same. You should find the definitions in the book, and check that they say the same things I have said here.
Example: 3(2x-1) + 6(4-x) -7(3) = 0x + 0 = 0. This matches the main equation, where the vectors are 2x-1 and 4-x and 3. The LHS is a linear combination of these vectors. The scalars are 3, 6 and -7. Since they aren't all zero, this is a nontrivial "solution" and we see that the set {2x-1 and 4-x and 3} is linearly dependent. I didn't even have to discuss which vector space this example is in, but it might be P2 or P3.
| When n=2 |
When n = 2, linear dependence has a simple meaning. The main equation is just c1v1 + c2 v2 = 0. And we know that at least one of the c's is nonzero, let's suppose that c1 ¹ 0. So, we can solve v1 for and get v1 = -(c2/c1)v2. So, v1 is a scalar multiple of v2. If they are vectors in R2 or R3 this means they are parallel.
Example: The vectors [2,3]T and [4,6]T are linearly dependent because v2 = 2v1. [maybe better - "because 2v1-v2 = 0"].
| When n>2 |
Then you should forget about scalar multiples and "parallel" arrows, and depend on the definitions and theorems. One of the main theorems says, for example, that a 3 dimensional space cannot contain a set of 4 linearly independent vectors. This was discussed above for column vectors (see the "if m>n" remarks), but is also true for spaces of polynomials, etc.
| When n=1 |
This case is a little silly. The equation c1v1 = 0 is only true if c1= 0 or v1 = 0. So, unless v1 = 0, a set S = {v1}, with only one vector, is independent.
| A quick note on proofs |
The main equation is related to several definitions, so it is the "link" you need in several Ch 3 proofs. Example: Prove that if v4 is a linear combination of v1,v2 and v3 then {v1,v2,v3,v4 } is linearly dependent. Proof: We are told that there are scalars such that c1v1+c2v2+c3v3=v4. So, c1v1+c2v2+c3v3-v4 = 0. This now matches the main equation with c4 = -1, which is nonzero. So, {v1,v2,v3,v4 } is linearly dependent.
| True-False Practice with | |