4

Assume $A,B \in \mathbb{R}^{m\times n}$, how can you prove the following:

$A^\mathrm{T}A=B^\mathrm{T}B \Leftrightarrow \exists$ orthogonal $Q$ such that $A=QB$

or is there a counterexample? Intuitively it makes sense to me, but I haven't found a nice proof yet. I have tried it through using SVD, but the non-uniqueness of the decomposition makes problems. I would be happy for some suggestions!

Sungjin Kim
  • 20,102
Dimitar Ho
  • 1,559
  • Possibly helpful: Your statement is wrong if we replace $\mathbb{R}$ by $\mathbb{C}$. Thus, we need something specific to the fact that $\mathbb{R}$ is formally real. That could be the fact that $\operatorname{Ker}\left(C^T C\right) = \operatorname{Ker} C$ for any matrix $C$ over the reals. – darij grinberg Jul 09 '14 at 19:41
  • Yes, it does help. This fact shows that $\operatorname{Ker} A = \operatorname{Ker} B$. Now, working with linear maps instead of matrices, we can let $P$ be the canonical projection from $\mathbb{R}^{n}$ to the quotient space $\mathbb{R}^{n} / \operatorname{Ker} A = \mathbb{R}^{n} / \operatorname{Ker} B$. Then, each of $A$ and $B$ factor through $P$, that is, we have $A = A'P$ and $B = B'P$ for some linear maps $A'$ and $B'$ from that quotient space. These linear maps $A'$ and $B'$ are injective. Now, $A^T A = B^T B$ becomes $P^T A'^T A' A = P^T B'^T B' P$. – darij grinberg Jul 09 '14 at 19:45
  • This easily yields $A'^T A' = B'^T B'$ thanks to the surjectivity of $P$. The nice thing about the two maps $A'$ and $B'$ is that they are injective (as opposed to $A$ and $B$, which may and may not be). Hence, your question is reduced to the case of both $A$ and $B$ being injective (because if $A' = QB'$, then clearly $A = A'P = QB'P = QB$). – darij grinberg Jul 09 '14 at 19:47
  • 1
    Now, assuming that $A$ and $B$ are injective, we can look at the map $B\left(\mathbb{R}^n\right) \to A\left(\mathbb{R}^n\right)$ which sends every $Bv$ to $Av$ (this is well-defined, because $B$ is injective and so every element of $B\left(\mathbb{R}^n\right)$ can be written as $Bv$ for exactly one $v$). This map is an isometry (because check that $\left(Bv, Bv'\right) = \left(Av, Av'\right)$ for any $v$ and $v'$). By Witt's theorem ( http://en.wikipedia.org/wiki/Witt%27s_theorem ), this isometry can be extended to an isometry of $\mathbb{R}^m$ to itself. This extended isometry is your $Q$ ! – darij grinberg Jul 09 '14 at 19:50
  • I hope someone with more time at hand can expend some of it to write this up as an answer, possibly avoiding the overkill(?) of using Witt's theorem. Nice question! – darij grinberg Jul 09 '14 at 19:51

2 Answers2

1

As a preparation for my exam, I tackled the problem of solving this old (and not quite resolved) question and came up with a nice solution. I hope the people who stared the problem will like the answer :).

Let's prove a more general theorem: Let $A: V \rightarrow W$ and $B: V \rightarrow W$ be linear operators on finite-dimensional inner-product spaces over $\mathbb{R}$ or $\mathbb{C}$. Then the following holds:

$A^{*}A = B^{*}B \Leftrightarrow \exists U: W \rightarrow W$ s.t. $U^*U = I_{W}$ and $UA = B$

Proof:

$\Leftarrow$: is trivial.

$\Rightarrow$: Using the Polar-Decomposition theorem, we know that $A = U_A (A^*A)^\frac{1}{2}$ and $B = U_B(B^*B)^\frac{1}{2}$, where ${U_A}^{*}U_A = {U_B}^{*}U_B = I_W$. Now, since $B^*B = A^*A$, we have $U_B{U_A}^*A = U_B{U_A}^*U_A (A^*A)^\frac{1}{2} = U_B (A^*A)^\frac{1}{2} = U_B (B^*B)^\frac{1}{2} = B$. Thus, per construction we found our desired $U$ to be $U = U_B{U_A}^*$.

Dimitar Ho
  • 1,559
1

$\Leftarrow$ is trivial.

for $\implies$: consider $A = (A_1, \dots A_d)$ as a family of columns. Then the hypothesis writes $A_i\cdot A_j = B_i\cdot B_j$.

Consider a subfamily that generates $\text{span }\{A_k\}$. You can find a set $I$ such as $\text{span }\{A_k\} = \text{span }\{A_i, i\in I\} $.

As for every matrix $C$, $\text{rank }C = \text{rank }C^TC$, you get that $ \{B_i, i\in I\} $ is a basis of $\text{span }\{B_k\}$.

Eventually, consider the linear map $$ f: \text{span }\{B_k\}\to \text{span }\{A_k\} \\ i\in I \implies f(B_i) = A_i $$

This is an orthogonal transformation and you can define it in any way to $R^d$, keeping the orthogonality. Its matrix $Q$ is such as $QB = A$.

mookid
  • 28,236