Question

Good morning. :) I just finished a problem where I was asked to determine if a set of three vectors Rn form a basis for Rn. I got the answer right (by algorithm, mostly). Could someone please help me understand *why* it's right? Work attached.

Answer 1

Is it that any value of a could satisfy the conditions for (x,y,z) ?

Answer 2

you established that the 3 initial vectors are independent when you found a non-zero determinant. that means that none of those vectors is some combination of the other 2. that means that you have 3 vectors that in some combination or other cover each and every point in R3 space, not as efficiently as <1,0,0>, <0,1,0>, <0,0,1>, but suffcient nevertheless to identify every single spot in R3.

to see this, note that the first 2 vectors will establish a plane. if the 3rd is some combination of these 2, the all 3 lie on the same plane, can only ever identify points in that plane, and can probably d so in a number of different ways.

hence whilst these vectors are not orthogonal, as i, j, k are, they cover all of R3 in a unique way.

does that help?

Answer 3

Ah, yes, that totally helps. So when I take the determinant of the three vectors, I'm essentially taking the cross-product, which when equal to zero means that I have dependent, or overlapping vectors, right?

Answer 4

good analogy.

the matrix determinant -- provided the vectors A, B and C are put in as rows or columns and the det calculated in the right way -- is actually A • (B x C) (!!!), ie the scalar triple product.

A x B gives a vector (n) that is perp to the plane formed by B & C.

if the overall triple product A • (B x C) is zero, A lies in the same plane as B and C as it too is perp to normal vector n.

if however A • (B x C) is non zero, then A is not totally in the same plane. this means that you can access all of R3, albeit in a pretty inefficient way compared to orthogonal vectors.

Answer 5

Ah, ok! Thanks so much for your help, and for the clear explanation. It was easy enough from the book to see how, but not as easy to understand why.

Answer 6

@eighthourlunch: thank you, this analogy has really helped me too. vectors are awesome!!!