# Criterion for existence of orthogonal operator with desired properties

Let $$V$$ be euclidean (hermitian) space. The set of vectors $$\{a_1,a_2,\dots,a_k\}$$ can be mapped under orthogonal operator to the set of vectors $$\{b_1,b_2,\dots,b_k\}$$ iff Gram matrices of each sets are equal, i.e. $$G(a_1,\dots,a_k)=G(b_1,\dots,b_k).$$

It seems to me quite a nice problem. Probably it should not be so difficult and let me show my attempts.

$$\Rightarrow$$ This is trivial because if the first set can be mapped to the second via orthogonal operator $$f$$ then $$f(a_i)=b_i$$ and $$(b_i,b_j)=(f(a_i),f(a_j))=(a_i,a_j)$$ the last inequality from definition of orthogonal operator $$f$$. It shows that corresponding elements of matrices $$G_a$$ and $$G_b$$ are equal which means that $$G(a_1,\dots,a_k)=G(b_1,\dots,b_k).$$

$$\Leftarrow$$ Suppose that $$G(a_1,\dots,a_k)=G(b_1,\dots,b_k)$$ and let $$(e)=\{e_1,\dots,e_n\}$$ be an orthonormal basis of $$V$$. Suppose $$a_i$$ has coordinates $$(a^1_i,\dots,a^n_i)$$ then $$[G_a]_{ij}=(a_i,a_j)=\sum \limits_{k=1}^{n}a^k_ia^k_j=[A^TA]_{ij}$$ which means that $$G_a=A^TA$$ and analogously $$G_b=B^TB$$ where $$A$$ and $$B$$ are $$n\times k$$ matrices, whose columns are coordinates of vectors $$\{a_1,\dots,a_k\}$$ and $$\{b_1,\dots,b_k\}$$, respectively. So we have that $$A^TA=B^TB$$.

My goal is to construct an operator $$f:V\to V$$ such that its matrix in the orthonormal basis $$(e)$$ is orthogonal matrix. Let's call this matrix $$M_f^{(e)}=M$$. As I said $$M$$ should be orthogonal and $$MA=B$$. I was trying to play with $$M=BA^T$$ or $$M=AB^T$$ but I failed.

So I would be very grateful if anyone can give useful idea or show the solution, please!

• It has occurred to me that this approach could be made easier if we use either the polar decomposition or the spectral theorem. Commented Apr 15, 2020 at 16:38
• @Omnomnomnom, Hmm could you show it, please?
– RFZ
Commented Apr 15, 2020 at 17:31
• See this post for the approach via the polar decomposition Commented Apr 15, 2020 at 17:38

A partial answer for the case that the sets $$\{a_1,\dots,a_k\}$$ and $$\{b_1,\dots,b_k\}$$ are linearly independent (or equivalently, the Grammian matrices are invertible).

Suppose that $$G(a_1,\dots,a_k) = G(b_1,\dots,b_k)$$. Let $$\{a_{k+1},\dots,a_n\}$$ and $$\{b_{k+1},\dots,b_n\}$$ be orthonormal bases for $$\{a_1,\dots,a_k\}^\perp$$ and $$\{b_1,\dots,b_k\}^\perp$$. Verify that $$G(a_1,\dots,a_n) = G(b_1,\dots,b_n)$$.

Note that a linear map $$f:V \to V$$ is orthogonal if and only if $$(f(x),f(y)) = (x,y)$$ for all $$x,y \in V$$. Show that if we take $$f$$ to be the unique linear map satisfying $$f(a_j) = b_j$$ for $$j=1,\dots,n$$, then $$f$$ satisfies this property and is therefore orthogonal.

An extension of this solution to the general case:

Because $$A^TA = B^TB$$, we have $$\ker A = \ker B$$. It follows that a set of vectors $$a_{j_1},\dots,a_{j_d}$$ will be linearly indepnendent if and only if the corresponding set $$b_{j_1},\dots,b_{j_d}$$ is linearly independent.

With that in mind, we can select a set $$a_{j_1},\dots,a_{j_d}$$ that forms a basis of $$\operatorname{span}(\{a_1,\dots,a_k\})$$ (which has dimension $$d$$). The corresponding set $$b_{j_1},\dots,b_{j_d}$$ forms a basis for $$\operatorname{span}(\{b_1,\dots,b_k\})$$. As before, we select vectors $$a_{d+1},\dots,a_{n}$$ and $$b_{d+1},\dots,b_n$$ that form bases for the respective orthogonal complements of the spans.

Now, it suffices to define $$f$$ to be the linear map satisfying $$f(a_{j_\ell}) = b_{j_\ell}$$ for $$\ell = 1,\dots,d$$ and $$f(a_\ell) = b_\ell$$ for $$\ell = d+1,\dots,n$$.

• Hmm. Thanks a lot! I will take a look at this but I guess that I got your idea.
– RFZ
Commented Apr 1, 2020 at 15:58
• @ZFR You're welcome! I've extended the proof to the general case now. An alternative approach here is to use something like Gram-Schmidt decomposition or the $QR$ factorizations of $A$ and $B$. Commented Apr 1, 2020 at 16:07
• Let me ask you a question: You are solving those problems very easily. OMG. How come? :)
– RFZ
Commented Apr 1, 2020 at 16:16
• @ZFR I recently finished my mathematics PhD and defended a thesis that focused on matrix analysis and linear algebra. I also answer questions like yours on this site in my spare time, and I tend to stick to the linear algebra questions when possible because those are the most appealing to me. All that is to say: I have a lot of practice. Commented Apr 1, 2020 at 16:20
• @ZFR The key point is that $A$ and $B$ have the same kernel. If $a_k$ can be produced as a linear combination of $a_{j_1},\dots,a_{j_d}$, then the corresponding linear combination of $b_{j_1},\dots,b_{j_d}$ produces $b_k$. Commented Apr 1, 2020 at 18:50