Question

linear algebra: inner product spaces?

suppose that {v1, v2, ... , vn} is an orthogonal set in Rn with the standard inner product. Let A be the matrix who jth column is vj, j = 1, 2, ..., n. prove or disprove: A is nonsingular.

answer: since some of the vectors vj can be zero, A can be singular.

I don't really understand the question and I don't get the answer.. how can A be singular just because some vectors in a column is zero?

Answer 1

if any vj is zero then the determinant of the matrix of those vectors is zero, so it is singular

Answer 2

but we didn't learn determinants yet so I think we're not suppose to know that yet :/

Answer 3

I personally am at a loss to explain that answer then :/
I can't see how you get to inner product space after determinants

Answer 4

before determinants*

Answer 5

@TuringTest the textbook has the determinants section first, and then inner product spaces, but our professor decided to leave the determinants for later..

Answer 6

well, remember that if Ax=0 has more than one solutions, then A will not be invertible, hence singular. now since the columns of A are already vj, j=1,2,...,n. then A could be written as A=[v1|v2|.....|vn]. now notice that Ax could be written as Ax=c1v1+c2v2+....cjvj.....+cnvn=0
now, without loss of generality, suppose if vj is the one which is the zero vector, then we have Ax=c1v1+c2v2+....cj(0)....+cnvn=0 now we have could infinite number of values of cj, and it will still be zero. Ofcourse for the other ci's for i=1,2,....,(j-1),(j+1),...,n, there will be a unique values since they are orthogonal to each other, hence linearly independent. but for cj, we could choose any value for it, and it will still give cjvj=0 since vj is a zero vector. therefore there will be more than one solution for Ax=0, hence A will not be invertible, therefore singular

Answer 7

Oh, I think I know now. Thanks! :)