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 $ab+cd = 0$.

Question

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.

Answer 1

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 $b^2+d^2=1$

Answer 2

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 $\mp y \pm ix$)

Answer 3

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.

Answer 4

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 $$ac+bd=a(\pm b)+b(\mp a)= 0$$

Answer 5

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.

Answer 6

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 $\Delta = (a d - b c)$, $p=a^2+b^2-1$, $q= a c + b d$, $s= c^2+d^2-1$

$$\Delta( a d p + c d q - a b q - b c s, b d p + d^2 q-b^2 q - b d s, - b c p - c d q + a b q + a d s) - (\Delta^2-1)(a^2+c^2-1, a b + c d, b^2+d^2-1) = \\ = (a^2+c^2-1, a b + c d, b^2+d^2-1)$$