How does one find the dual of a matrix?

[grzz]

TL;DR Summary

Does a matrix have a dual?

How does one find the dual of a matrix?
Thanks.

 

[martinbn]

Dual in what sense?

 

[grzz]

Dual in what sense?

In the sense that,
dual matrix times matrix = number, maybe complex times (unit matrix).

 

[renormalize]

In the sense that,
dual matrix times matrix = number, maybe complex times (unit matrix).

Can you specify precisely how this dual matrix differs from the more familiar inverse matrix?

 

[grzz]

Can you specify precisely how this dual matrix differs from the more familiar inverse matrix?

Thanks for your help.
But the trouble is that I never met with a dual matrix. All I know is that a dual matrix obeys the relation I gave in my 2nd post.

 

[Euge]

In the sense that,
dual matrix times matrix = number, maybe complex times (unit matrix).

The dual matrix in the first sense is ill-defined. For example, if $A$ is a $3\times 2$ matrix, there does not exist an $m\times n$ matrix $B$ such that $AB$ is a number.

In the second sense you mentioned, I suppose the dual of a complex square matrix $A$ is the adjugate of $A$, $\operatorname{adj} A$; indeed, $A(\operatorname{adj} A) = \det(A) I$.

 

[fresh_42]

Thanks for your help.
But the trouble is that I never met with a dual matrix. All I know is that a dual matrix obeys the relation I gave in my 2nd post.

There are many pairings that could be considered dual. A dual matrix to a matrix $A$ as a term isn't defined. You can transpose a matrix $A^\dagger$ , possibly invert a matrix $A^{-1}$, conjugate complex matrices $\overline{A},$ or write $A=\sum_{k=1}^r u_k\otimes v^*_k$ and consider $\sum_{k=1}^r u_k^*\otimes v_k$ as its dual.

 

[grzz]

Thanks Euge.
I forgot to mention that I was dealing with four vectors and square 4x4 matrices.
Hence

 

[grzz]

Hence your reply fits what I was asking for.

Another query of mine is whether there is a cross product of 4-vectors like there is with 3-vectors.

Thanks so.much.

 

[fresh_42]

Hence your reply fits what I was asking for.

Another query of mine is whether there is a cross product of 4-vectors like there is with 3-vectors.

Thanks so.much.

The cross product in $\mathbb{R}^3$ is a Lie product of a three-dimensional, real, simple Lie algebra. You can define several Lie products on $\mathbb{R}^4$ but none of them belongs to a simple Lie algebra. I'm not 100% sure, but I think that you don't even get all vectors as the result of such a product, i.e. some vectors cannot be retrieved as a result of the product.

I guess, no is the only answer that can be given without getting into details.

 

[mathwonk]

notice that the cross product of two vectors in 3 space is an oriented choice of vector orthogonal to the plane they span. Thus a cross product in 4 space should be a vector orthogonal to the 3-space spanned by the arguments. hence one should form cross product in 4 space using three argument vectors. I.e. given three vectors U,V,W in 4- space, yes there is a cross product vector UxVxW.

 

[Euge]

Another query of mine is whether there is a cross product of 4-vectors like there is with 3-vectors.

Thanks so.much.

The answer depends on what properties of the 3-dimensional cross product you want preserved in higher dimensions. For definiteness, let's suppose $n \ge 3$ and there is a "cross product" on $\mathbb{R}^n$, which we assume to be a continuous map $\times : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$ such that the cross product of any two independent vectors is nonzero, and $v\times w$ is always orthogonal to $v$ and $w$. Then $n$ must be either $3$ or $7$. Indeed, one can show that the existence of a cross product in this sense implies the existence of a continuous multiplication on the $n$-sphere $S^n$ that makes $S^n$ into an $H$-space. It was proven by J. F. Adams that $S^n$ is an $H$-space if and only if $n = 0, 1, 3$, or $7$.

 

[grzz]

The cross product in $\mathbb{R}^3$ is a Lie product of a three-dimensional, real, ...

I am not familiar with Lie Algebra. But thanks any way.

 

[fresh_42]

I am not familiar with Lie Algebra. But thanks any way.

That's easy in this case. Look at https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/#61-The-Pauli-Matrices. There are three matrices
$$
U=\dfrac{1}{2}\begin{pmatrix}0&i\\i&0\end{pmatrix}\ ,\
V=\dfrac{1}{2}\begin{pmatrix}0&1\\-1&0\end{pmatrix}\ ,\
W=\dfrac{1}{2}\begin{pmatrix}-i&0\\0&i\end{pmatrix}
$$
If we define the Lie algebra multiplication as $[X,Y]=X\cdot Y-Y\cdot X$ then we get
$$\begin{equation*} [U,V]=W\, , \,[V,W]=U\, , \,[W,U]=V \end{equation*}$$
which is another way to look at the cross-product. It has the same properties.

You cannot do this in four dimensions, i.e. with four matrices without losing some properties.

 

[WWGD]

Id believe duals are assigned to vector spaces * , not to matrices, so one may speak of the dual of the column/row/kernel. Or maybe, considering a matrix as a linear map $L : V \rightarrow W$, there is the canonical map $L^{*} : W^{*}\rightarrow V^{*}$.

* Or other Algebraic objects, though here vector spaces.