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

Xiren Zhou
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  • What have you tried? What Linear Algebra are you allowed to use? What’s the context for the question? – Eric Apr 26 '23 at 00:44
  • The two statements are required by definition of an orthonormal basis. Are you sure that the title fits with your question? – hff1 Apr 26 '23 at 00:46
  • 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? – peterwhy Apr 26 '23 at 00:56
  • @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. – Xiren Zhou Apr 26 '23 at 01:10
  • Please delete all the $\cdot$s. We are multiplying real numbers, not taking dot products of vectors. Write these as vector statements. – Ted Shifrin Apr 26 '23 at 01:15
  • @hff1, The six conditions already forms an orthogonal basis by definition. Of course, the two statements always hold for any orthogonal basis. I use such title cuz it's in essence the property of orthogonal basis; but since I'm looking for a pure algebraic proof which is expected to only use the six equations as conditions, you can just ignore the title and deem it as irrelevant if u want. – Xiren Zhou Apr 26 '23 at 01:17
  • @TedShifrin, modified. – Xiren Zhou Apr 26 '23 at 01:20
  • @peterwhy, Yes I know. – Xiren Zhou Apr 26 '23 at 01:22
  • Are you saying that you want brute force high school algebra but will not allow basic linear algebra? Why? – Ted Shifrin Apr 26 '23 at 01:46
  • Do you know that from your six givens and let $M = \pmatrix{\mathbf v_1 & \mathbf v_2 & \mathbf v_3}$, then $M^T M = I$, and so $MM^T = I$? Two entries of $MM^T$ gives your required results. – peterwhy Apr 26 '23 at 01:48

1 Answers1

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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.

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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.

Eric
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