# Show that $a^2 + b^2 =1$, $c^2 +d^2 = 1$, and $ac +bd = 0$ implies $a^2 + c^2 = 1$, $b^2+d^2=1$, and $ab+cd = 0$.

This question is really all about algebraic manipulation. However, it is motivated by trying to show that given a matrix $$A$$ with orthonormal rows $$A = \begin{pmatrix} a &b\\ c&d \end{pmatrix},$$

the columns must be orthonormal. I.e. we must prove that $$a^2 +b^2 = 1, \; c^2 +d^2 = 1, \; ac+bd = 0,$$ imply $$a^2 + c^2 = 1, \; b^2 + d^2 = 1, \; ab+cd = 0.$$

I have done a bunch of computation, but haven't been able to make any progress. This is my own question, I haven't seen this problem in a book or elsewhere.

NOTE: I know this is easy to solve using matrices, transposes, etc. This question is not to prove the statement. Rather show how we can do it with brute force algebra.

• en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity maybe will help?
– qwr
Commented Aug 1 at 17:31
• This is probably overkill, but in Sage I was able to verify with: R.<a,b,c,d> = QQ[]; I = ideal(a^2+b^2-1,c^2+d^2-1,a*c+b*d); (a^2+c^2-1) in I; (b^2+d^2-1) in I; (a*b+c*d) in I -- so presumably, a manual Groebner basis calculation could find the same thing. (Though the Groebner basis that's calculated, at least for the default monomial order, balloons to 6 generators.. And with the lexicographic order, it also results in 6 generators. So that would probably be a lot of work to do manually.) Commented Aug 1 at 17:38
• Generally see If $AB = I$ then $BA = I$. $\ \$ Commented Aug 1 at 18:51
• See this post for another elementary proof. Commented Aug 3 at 12:34

If you want a full answer, here is what I have: $$ac+bd=0\Rightarrow (ac)^2-(bd)^2=0$$ $$a^2c^2=b^2d^2$$ $$c^2(1-b^2)=b^2(1-c^2)$$ $$c^2-b^2c^2=b^2-b^2c^2$$ $$c=\pm b$$ Since $$ac+bd=0$$, you can also prove that, $$ab+cd=0$$. [Putting $$b=-c$$ or $$b=c$$ both works]

Moreover, $$b=\pm c$$ also proves easily that $$a^2+c^2=1$$ and $$b^2+d^2=1$$

• Oh nice, that's very direct. OP said they had "full pages of computation", so I went for something much more powerful. Good to see that brute force still works. Commented Aug 1 at 17:28
• Thanks. By the way, I learned today about the Brahmagupta–Fibonacci Two-Square identity from you. Warm regards from me to you too. Commented Aug 1 at 17:31

Hint: Show that $$ad - bc = \pm 1$$.

Further hint: Apply the Brahmagupta–Fibonacci Two-Square identity.

Then solve the system of linear equations $$ac+bd = 0, ad - bc = \pm 1$$ (with $$a, b$$ as constants) for $$c, d$$ to get that $$c = \mp \frac{b}{a^2+b^2 } = \mp b, d = \pm a$$.

Your desired result follows. We have proven something stronger, though you also knew that from orthonomal 2-matrices. (EG Rotation by $$\pm i$$ brings $$x+iy$$ to $$\mp y \pm ix$$)

• Thanks, I was thinking there had to be some cool identity with square that will help! I chose the other answer as it's quicker, but I would also choose this. Commented Aug 1 at 20:24

We have \begin{align}a^2+b^2&=1 \tag1\\ c^2+d^2&=1 \tag2 \\ ac+bd &=0 \tag3 \end{align}

Multiply $$(1)$$ and $$(2)$$ to give $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2 + (ad-bc)^2$$ where I use Brahmagupta's identity. Then one gets $$1=|ad-bc|$$. Then either $$2ad-2bc=-2 \tag4$$ or $$2bc-2ad=-2 \tag5$$ Based on this we consider two cases. Add $$(1)$$, $$(2)$$, and $$(4)$$ to give $$(a+d)^2+(b-c)^2=0$$ which gives $$a=-d$$ and $$b=c$$.

Or when in the second case we have $$(1)+(2)+(5)$$ gives $$(b+c)^2+(a-d)^2=0$$ so here $$b=-c$$ and $$a=d$$.

In conclusion we have $$a=\pm d$$ and $$b=\mp c$$, proving your claim.

For a unit vector $$(a,b)$$ in $$\mathbb{R}^2$$ there are exactly two unit vectors orthogonal to $$(a,b).$$ As the unit vectors $$(b,-a)$$ and $$-(b,-a)$$ are orthogonal to $$(a,b),$$ we get $$(c,d)=\pm (b,-a),$$ i.e. $$c=\pm b$$ and $$d=\mp a.$$ In both cases $$a^2+c^2=b^2+d^2=1$$ and $$ac+bd=a(\pm b)+b(\mp a)= 0$$

• OP said $7$ hours ago that they don't want proofs using matrices (which are dupes anyhow) Commented Aug 1 at 20:53
• @BillDubuque Now the conclusion is not formulated in terms of matrices. Commented Aug 2 at 0:35
• But it is really a matrix theoretic proof (with matrix language eliminated). Commented Aug 2 at 0:55
• @BillDubuque In my opinion it is a geometry theoretic proof. Commented Aug 2 at 3:31

I don't know if this is the solution you are looking for but this is the first thing that came into my mind:

Put parametrically $$a=cos(\theta),b=sin(\theta)$$ and $$c=cos(\phi),d=sin(\phi).$$

Then it it easy to see that the first condition is equivalent to $$cos(\phi)cos(\theta)+sin(\theta)sin(\phi)=0$$ Or $$cos(\phi-\theta)=0.$$ Thus we get $$\phi=\theta+\frac{(2n+1)\pi}{2}$$. Now I think it is easy to prove the claims. First 2 claims reduce to $$sin^2(\phi)+cos^2(\phi)$$=1.$Third claim can be seen as the two terms cancel out. In a monoid, from $$A \cdot A_1 = 1$$, and $$A_2 \cdot A = 1$$ we get $$A_1 \cdot A = 1$$. Indeed: $$A_1 A = (A_2 A) A_1 A = A_2 \cdot 1 \cdot A = 1$$ Now, back to matrices: $$A A^t = I$$ implies $$\Delta^2 = 1$$ ($$\Delta \colon = \det A$$). This can be done by hand, easy. Now use $$\operatorname{adj} A \cdot A = \Delta \cdot I$$ so $$\Delta \operatorname{adj}(A) \cdot A = I$$ From here we get $$A^t A = \Delta \operatorname{adj} (A) A \cdot A^t A = \Delta \operatorname{adj}( A ) A= I$$ Note: 1. the entries of $$A$$ are from a commutative ring with $$1$$. 2. The equalities above could be written component-wise $$\bf{Added:}$$ Here is how one could write the desired equalities as a consequence of the given ones. Consider the identity: $$A^t A - I = \Delta \operatorname{adj}(A) (A A^t-I) A - (\Delta^2-1)(A^t A - I)$$ Write the entries of $$A A^t-I$$ as $$p$$, $$q$$, $$r$$, $$s$$, use $$\Delta^2-1$$ as a consequence of $$p=q=r=s=0$$, fire your favorite CAS and get some equalities for the entries of $$A^t A-I$$. $$\bf{Added:}$$ From $$a^2+b^2=1$$ and $$c^2+d^2=1$$, and $$a c + b d = 0$$ we get $$1=(a^2+b^2)(c^2+d^2)-(a c + b d)^2 = a^2 d^2-2 a b c d + b^2 c^2 = (a d - b c)^2$$ Now consider the identity, where $$\Delta = (a d - b c)$$, $$p=a^2+b^2-1$$, $$q= a c + b d$$, $$s= c^2+d^2-1$$ $$\Delta( a d p + c d q - a b q - b c s, b d p + d^2 q-b^2 q - b d s, - b c p - c d q + a b q + a d s) - (\Delta^2-1)(a^2+c^2-1, a b + c d, b^2+d^2-1) = \\ = (a^2+c^2-1, a b + c d, b^2+d^2-1)$$ • OP said$7\$ hours ago that they don't want proofs using matrices (which are dupes anyhow). Commented Aug 1 at 20:34