Show that if the Gram–Schmidt process is applied to a linearly dependent vector system, then it outputs the 0 vector

Question

If the Gram–Schmidt process is applied to a linearly dependent vector system, it outputs the 0 vector

Can somebody help me with this problem

I thought taking the linear combination of the linearly dependent vectors ${\{a_1,\dots, a_n}\}$ $$\alpha_1a_1+\dots+\alpha_na_n=0$$ where at least one of $\alpha_i$ is nonzero. If we suppose the opposite that all the vectors ${\{b_1,\dots, b_n}\}$ that we get after the orthogonalization of ${\{a_1,\dots, a_n}\}$ are nonzero then we would have all $\alpha_i=0$ which would lead to a contradiction.

But I don't know how to proceed.

Answer 1

Note that if $a_i$'s are linearly dependent then at some point we have $a_j \in span(a_1, a_2, ... a_{j-1})$.

Indeed, consider a zero linear combination and let $a_j$ be the last element with non-zero coefficient - then bringing $a_j$ to the other side and dividing by $\alpha_j$ we obtain $a_j$ as a linear combination of $a_1, a_2, ..., a_{j-1}$.

Showing that it will result in $b_j$ being zero is left as an exercise for the reader.:)