# Linear dependence of two vectors is a necessary and sufficient condition for an equality

This is an example in the textbook.

To prove: for any real number** $$a_1, a_2, \cdots, a_n, b_1, b_2, \cdots b_n$$, linear dependence of the two vectors $$\left(a_1, a_2, \cdots, a_n\right)$$ and $$\left(b_1, b_2, \cdots, b_n\right)$$ is a necessary and sufficient condition for the equality:

$$\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(b_1^2+b_2^2+\cdots+b_n^2\right)=\left(a_1 b_1+a_2 b_2+\cdots+a_n b_n\right)^2$$

I take a three-dimensional vector as an example to prove its necessity:

Clear["Global*"];
listA = Table[a[i], {i, 1, 3}];
listB = m*listA;

Refine[Reduce[
Sum[a[i]^2, {i, 1, 3}]*Sum[(m*a[j])^2, {j, 1, 3}] ==
Sum[a[k]*m*a[k], {k, 1, 3}]^2],
m ∈ Reals && a[1] ∈ Reals &&
a[2] ∈ Reals && a[3] ∈ Reals]


True

But I cannot prove its sufficiency:

Clear["Global*"];
listA = Table[a[i], {i, 1, 3}];
listB = Table[b[i], {i, 1, 3}];
sol = Solve[
Sum[a[i]^2, {i, 1, 3}]*Sum[b[j]^2, {j, 1, 3}] ==
Sum[a[k]*b[k], {k, 1, 3}]^2, listA, Reals][[1]]
ResourceFunction["LinearlyIndependent"][{listA, listB}] /. sol


$$\left\{\mathrm{a}[2] \rightarrow \frac{\mathrm{a}[1] \times \mathrm{b}[2]}{\mathrm{b}[1]}, \mathrm{a}[3] \rightarrow \frac{\mathrm{a}[1] \times \mathrm{b}[1] \times \mathrm{b}[3]+\frac{a[1] b[2]^2 \mathrm{~b}[3]}{\mathrm{b}[1]}}{\mathrm{b}[1]^2+\mathrm{b}[2]^2}\right\}$$

True if condition...

ResourceFunction["LinearlyIndependent"][{listA, listB}] /. sol should give False to prove that listA and listB are linearly dependent.

How can I modify this code?

You can do some simplifications before computing matrix rank

u = {listA, listB} /. sol // Simplify
MatrixRank[u]
(*1*)


Matrix rank is smaller than the number of vectors, therefore, they are linearly dependent.

Create two linearly dependent symbolic vectors:

va = Array[Subscript[a, #] &, 3];
vb = lam  va;


Now, the left side of your equation is the product of the norm square of va and vb. The right side is the square of the scalar product:

(va . va) (vb . vb) == (va . vb)^2 // Simplify

(* True *)

• Thank you for your answer. But I have already solved this part of problem. Please take a look at the first part of my problem "I take a three-dimensional vector as an example to prove its necessity". What I want to know is that the second part: how to derive Linear dependence of va and vb from (va . va) (vb . vb) == (va . vb)^2? @Daniel Huber Sep 23, 2022 at 1:38
• THis is the "triangle inequality". va.va vb..vb = |va|^2 Cos[0] |vb|^2 Cos[0] and (va.vb)^2 = |va| ^2 |vb| ^2 Cos[va,vb] Therefore Cos[va,vb]==1 Sep 23, 2022 at 7:43
• I mean to solve this problem with the code of MMA. Now another answer has solved my problem. Anyway, thank you for your answer. Sep 23, 2022 at 14:44