How to prove that a diference between the same component of two vectors is less than or equal to the norm of the vector diference?

Question

how can i prove that a diference between the same component of two vectors is less than or equal to the norm of the vector diference? i mean supose that

$A=(a_1,...,a_m)$ and $B=(b_1,...,b_m)$ both in R$^{m}$ then:

$a_k-b_k$ les than or equal to $||A-B||$.

Thank You.

Answer 1

For an orthonormal basis in any inner product space this follows from the Cauchy-Schwarz inequality, i.e. $|C.D| \le ||C||.||D||$ . Put $D$ = one of the (orthonormal) basis vectors $e_i $ and then $|c_i| = |C.e_i| \le ||C||.||e_i|| = ||C||$. Now put $C = A - B$, so that $a_i - b_i \le |a_i - b_i| \le ||A - B||$.

If the basis is not orthogonal then this may not be true. Consider a basis in 2D of two almost parallel unit vectors: the components for a short vector which is not parallel to either of them will be large whereas the vector length is small.