3
$\begingroup$

Given that:

$x_1 x_2 + y_1 y_2 + z_1 z_2=0$,

$x_1 x_3 + y_1 y_3 + z_1 z_3=0$,

$x_2 x_3 + y_2 y_3 + z_2 z_3=0$,

$x_1^2+y_1^2+z_1^2=1$,

$x_2^2+y_2^2+z_2^2=1$,

$x_3^2+y_3^2+z_3^2=1$

How to prove the following:

  1. $x_1^2+x_2^2+x_3^2=1$,

  2. $x_1 y_1 + x_2 y_2 + x_3 y_3=0$

$\endgroup$
10
  • $\begingroup$ What have you tried? What Linear Algebra are you allowed to use? What’s the context for the question? $\endgroup$
    – Eric
    Commented Apr 26, 2023 at 0:44
  • $\begingroup$ The two statements are required by definition of an orthonormal basis. Are you sure that the title fits with your question? $\endgroup$
    – hff1
    Commented Apr 26, 2023 at 0:46
  • $\begingroup$ Do you know that for square matrix $M = \pmatrix{\mathbf v_1 & \mathbf v_2 & \mathbf v_3}$ ($3\times3$ in your case), if there is a left inverse $L$ such that $LM=I$, then $ML=I$ too? $\endgroup$
    – peterwhy
    Commented Apr 26, 2023 at 0:56
  • $\begingroup$ @Eric, Actually, by leveraging the property of a rotation matrix (i.e., any rotation matrix is a orthogonal matrix with det=1), the above two statements could be directly proved. However, I want to just use the 6 equations I gave to prove this. I know they suffice, but I don't know how to manipulate them to come up with an algebraic prroof. $\endgroup$
    – Xiren Zhou
    Commented Apr 26, 2023 at 1:10
  • $\begingroup$ Please delete all the $\cdot$s. We are multiplying real numbers, not taking dot products of vectors. Write these as vector statements. $\endgroup$ Commented Apr 26, 2023 at 1:15

1 Answer 1

0
$\begingroup$

With Linear Algebra, this is straightforward.

Let $U$ be the matrix whose rows are the 3 vectors $(x_i,y_i,z_i)$. Then, you’re given that $U U^T=I$. Therefore $U$ is invertible with inverse $U^T$. Since right inverses are left inverses, we also have $U^T U=I$ which simplifies to your desired equations.

-

How can we do this without directly relying on the concept of invertibility in Linear algebra?

I’m still going to use Linear algebra to simplify the computations, but you could imagine writing this all out as linear equations of the underlying 9 variables.

We have that $|U||U^T|=1$, so $|U| \neq 0$.

Let $\mathbb{adj}(A)$ be the adjugate matrix defined as the transpose of the matrix of cofactors. If you expanded this out, then by the cofactor definition of determinant, it satisfies $A \mspace4mu\mathbb{adj}(A) = \mathbb{adj}(A) A=|A| I$. This is just a simple equality if your write it out in terms of the underlying 9 variables.

Let’s try simplifying $\mathbb{adj}(U) U U^T U$.

Collapsing from the left side, we have: $\mathbb{adj}(U) U U^T U=|U|I U^T U= |U| U^T U$

Collapsing the middle first, we have: $$\mathbb{adj}(U) U U^T U=\mathbb{adj}(U) I U = \mathbb{adj}(U) U=|U| I$$

Thus, $$|U| U^T U =\mathbb{adj}(U) U U^T U= |U| I$$

Since $|U|\neq 0$, we can divide by it:

$$U^T U=I$$

Expanding this gives the desired result.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .