# Prove that the following application between $M_{n×n}(\Bbb R)$ and $M_{n^2×n^2}(\Bbb R)$ is an isometry such that $\det φ(X)≠ 0$ whether $\det X≠ 0$

Let be $$\varphi:M_{n\times n}(\Bbb R)\rightarrow M_{n^2\times n^2}(\Bbb R)$$ the application defined through the equation $$\varphi(X):=\sum_{i,j,h=1}^{n}x^{i,j}\hat e_{h+(i-1)n}\otimes\hat e_{j+(h-1)n}$$ for any $$X\in M_{n\times n}(\Bbb R)$$. So I ask me if $$\varphi(X)$$ is invertible when $$X$$ is and moreover observing that $$\varphi$$ is a linear application I ask also to me if $$\varphi$$ is an isometry. So first of all I suppose $$n=3$$ so that $$\varphi(X)=\begin{pmatrix}x_{1,1} && x_{1,2} && x_{1,3} && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && x_{1,1} && x_{1,2} && x_{1,3} && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && x_{1,1} && x_{1,2} && x_{1,3} \\x_{2,1} && x_{2,2} && x_{2,3} && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && x_{2,1} && x_{2,2} && x_{2,3} && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && x_{2,1} && x_{2,2} && x_{2,3} \\x_{3,1} && x_{3,2} && x_{3,3} && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && x_{3,1} && x_{3,2} && x_{3,3} && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && x_{3,1} && x_{3,2} && x_{3,3} \end{pmatrix}$$ for any $$X\in M_{3\times 3}(\Bbb R)$$ and thus I observed that $$\varphi(X)\cdot\big(\varphi(Y^T)\big)^T=\begin{pmatrix}x_{1,1} && x_{1,2} && x_{1,3} && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && x_{1,1} && x_{1,2} && x_{1,3} && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && x_{1,1} && x_{1,2} && x_{1,3} \\x_{2,1} && x_{2,2} && x_{2,3} && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && x_{2,1} && x_{2,2} && x_{2,3} && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && x_{2,1} && x_{2,2} && x_{2,3} \\x_{3,1} && x_{3,2} && x_{3,3} && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && x_{3,1} && x_{3,2} && x_{3,3} && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && x_{3,1} && x_{3,2} && x_{3,3} \end{pmatrix}\cdot\begin{pmatrix}y_{1,1} && 0 && 0 && y_{1,2} && 0 && 0 && y_{1,3} && 0 && 0 \\y_{2,1} && 0 && 0 && y_{2,2} && 0 && 0 && y_{2,3} && 0 && 0 \\y_{3,1} && 0 && 0 && y_{3,2} && 0 && 0 && y_{3,3} && 0 && 0 \\0 && y_{1,1} && 0 && 0 && y_{1,2} && 0 && 0 && y_{1,3} && 0 \\0 && y_{1,2} && 0 && 0 && y_{2,2} && 0 && 0 && y_{2,3} && 0 \\0 && y_{3,1} && 0 && 0 && y_{3,2} && 0 && 0 && y_{3,3} && 0 \\0 && 0 && y_{1,1} && 0 && 0 && y_{1,2} && 0 && 0 && y_{1,3} \\0 && 0 && y_{2,1} && 0 && 0 && y_{2,2} && 0 && 0 && y_{2,3} \\0 && 0 && y_{3,1} && 0 && 0 && y_{3,2} && 0 && 0 && y_{3,3} \end{pmatrix}=\dots=\varphi(X\cdot Y^T)$$ from which the statement follows observing that $$\text{rank}\,\varphi(I)=\text{rank}\,\begin{pmatrix}1 && 0 && 0 && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && 1 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && 1 && 0 && 0 \\0 && 1 && 0 && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 1 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && 0 && 1 && 0 \\0 && 0 && 1 && 0 && 0 && 0 && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 1 && 0 && 0 && 0 \\0 && 0 && 0 && 0 && 0 && 0 && 0 && 0 && 1 \end{pmatrix}=9$$ because this implies that $$\det\varphi(X)\cdot\det\varphi(X^{-T})=\det\big(\varphi(X)\cdot\varphi(X^{-T})\big)=\det\big(\varphi(X\cdot X^{-1})\big)=\det\big(\varphi(I)\big)\neq 0$$ from which by the zero-product property I conclude that the quantity $$\det\big(\varphi(X)\big)$$ is not zero. Moreover observing that $$\text{tr}\big(\varphi(X)\big)=\text{tr}(X)$$ for any $$X\in M_{3\times 3}(\Bbb R)$$ I conclude that $$\big\langle\varphi(X),\varphi(Y)\big\rangle=\text{tr}\big(\varphi(X)\cdot\varphi(Y)^T\big)=\\\text{tr}\Big(\varphi(X)\cdot\varphi\big((Y^T)^T\big)^T\Big)=\text{tr}\big(\varphi(X\cdot Y^T)\big)=\text{tr}\big(X\cdot Y^T\big)=\langle X,Y\rangle$$ so that $$\varphi$$ is an isometry. So the statement is srely true for $$n=3$$, provided that the argumetations I gave are correct. Anyway I think that the statement, at least the first part, can be proved using the following conjecture which I did not able to prove

CONJECTURE

Any isometry $$\phi$$ between $$M_{n\times n}(\Bbb R)$$ and $$M_{n^2\times n^2}(\Bbb R)$$ is such that $$\det\big(\phi(X)\big)\neq 0$$ provided that $$X$$ is not singular.

Anyway I think that it is better to pose a dedicated question to this.

So I ask if the argumetation I gave are correct and if can be generalised to an arbitrary $$n\in\Bbb N$$. So could anyone help me, please?

• Your conjecture is not correct. For example, consider the isometry for $n=2$ given by $$\phi \pmatrix{x_{11} & x_{12}\\ x_{21} & x_{22}} = \pmatrix{x_{11} & x_{12}&0&0\\ x_{21} & x_{22} &0&0\\0&0&0&0\\0&0&0&0}.$$ Commented Dec 26, 2021 at 23:21
• @BenGrossmann Okay, thanks for the counterexample. Anyway I have to point out that the expression of $\varphi$ for any arbitrary $n\in\Bbb N$ is a bad generalization of the expression of $\varphi$ for $n=3$. So can you say how write $\varphi$ for an arbitrary $n$? Then are my argumentation correct? Finally how implement it to the general case? Could you help me, please? Commented Dec 26, 2021 at 23:26
• Forgive for the bother. Commented Dec 26, 2021 at 23:26
• If we use $\otimes$ to denote the Kronecker product, we could write $$\varphi(X) = \pmatrix{X \otimes e_1 & X \otimes e_2 & \cdots & X \otimes e_n}$$ Commented Dec 26, 2021 at 23:30
• The two interpretations of $\otimes$ correspond to competing interpretations of the tensor product. Your version (where the tensor product of two vectors is a matrix) corresponds to "dyadics" and the classical treatment of a tensor product. The Kronecker product corresponds to the "modern treatment" of tensor products. Commented Dec 26, 2021 at 23:45

At the risk of being a bit confusing, I will use $$\otimes$$ to denote the Kronecker product. We have $$\varphi(X) = \pmatrix{X \otimes e_1 & X \otimes e_2 & \cdots & X \otimes e_n}.$$ From there, we find that $$\langle \varphi(X), \varphi(Y) \rangle = \sum_{i=1}^n \langle X \otimes e_i, Y \otimes e_i\rangle = \sum_{i=1}^n \langle X,Y \rangle = n \langle X, Y \rangle.$$ So, it is not true that $$\varphi$$ is an isometry. However, $$\psi(X) := \varphi(X)/n$$ is an isometry.

Regarding the comment: note that $$\varphi(X) \varphi(Y^T)^T = \\ \pmatrix{X \otimes e_1 & \cdots & X \otimes e_n} \pmatrix{Y \otimes e_1^T \\ \vdots \\ Y \otimes e_n^T} = \\ (X \otimes e_1)(Y \otimes e_1^T) + \cdots + (X \otimes e_n)(Y \otimes e_n^T) =\\ (XY) \otimes (e_1e_1^T) + \cdots + (XY) \otimes (e_ne_n^T) = \\ (XY) \otimes (e_1e_1^T + \cdots + e_ne_n^T) = \\ (XY) \otimes I.$$ Note that in general, if $$A$$ is $$m \times n$$ with entries $$a_{ij}$$, then $$A \otimes I = \pmatrix{ a_{11} I & \cdots & a_{1n} I \\ \vdots & \ddots & \vdots \\ a_{m1} I & \cdots & a_{mn} I}.$$

• Okay, so $\varphi$ is not an isometry: however at least is the equality $$\varphi(X)\cdot\big(\varphi(Y^T)\big)^T=\varphi(X\cdot Y^T)$$ true? Commented Dec 26, 2021 at 23:42
• Anyway I think that it is false because if it was true then $\varphi$, at least for $n=3$, would be an isometry, right? Commented Dec 26, 2021 at 23:49
• @AntonioMariaDiMauro See my latest edit Commented Dec 26, 2021 at 23:52
• Okay, so $$\det\big(\varphi(X)\varphi(Y^T)^T\big)=\det\big((XY)\otimes I\big)=\Big(\det(XY)\Big)^n\cdot\big(\det I\big)^n=\big(\det(XY)\big)^n$$, right? Commented Dec 26, 2021 at 23:55
• @AntonioMariaDiMauro That's correct Commented Dec 26, 2021 at 23:55