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$\begin{pmatrix}a\\b\\c\end{pmatrix}\begin{pmatrix}x&y&z\end{pmatrix}=\begin{pmatrix}ax&ay&az\\bx&by&bz\\cx&cy&cz\end{pmatrix}$ by regular matrix multiplication. But if

$\begin{pmatrix}a\\b\\c\end{pmatrix}$ is an element of the double dual space I believe this product would be a number:

$\begin{pmatrix}a\\b\\c\end{pmatrix}\begin{pmatrix}x&y&z\end{pmatrix}=(cx+by+az)$

Is one of these correct? If I stumble across it what should I do?

Also since I'm teaching this to myself, I tried my best but if I have any other wrong intuition, notation, formulas, or the question is unclear please offer a suggestion so I can fix it.

Edit: I expanded the second product.

Pineapple Fish
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  • Yes you're right. By definition, the dual space $V^*$ of a vector space $V$ is a collection of all linear map $\omega : V \to \Bbb{F}$. – Kelvin Lois Jun 11 '18 at 16:46
  • $$\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}a\b\c\end{pmatrix}=xa+yb+zc$$ – AccidentalFourierTransform Jun 11 '18 at 16:48
  • @Sou but this is a vector acting on a covector, not the other way around, right? Is any arbitrary vector in a vector space or double dual vector space? – Pineapple Fish Jun 11 '18 at 16:49
  • @BenjaminThoburn, this product correspond to $v\otimes f$. – janmarqz Jun 11 '18 at 17:39
  • Your matrix allows you to pair a couple of vector via the universal formula $$(v,w)\mapsto vAw^{\top}$$ which upon $v=w$ we get a map $V\to\Bbb R$ called quadratic form. – janmarqz Jun 11 '18 at 17:58
  • @janmarqz I edited the question slightly. I understand $v^T Aw$ but how is this the same as $vAw^T$. In your response, I'm assuming that f is a function? Also how is the matrix pairing vectors relevant? Thank you! – Pineapple Fish Jun 13 '18 at 15:31
  • First $$\begin{pmatrix}a\b\c\end{pmatrix}\begin{pmatrix}x&y&z\end{pmatrix}=cx+by+az$$ makes no sense for an usual product of matrices. – janmarqz Jun 13 '18 at 19:34
  • Second $v^{\top}Aw$ is like $$(x,y)\begin{pmatrix}a_{11}&a_{12}\a_{21}&a_{22}\end{pmatrix}\begin{pmatrix}z\t\end{pmatrix}$$ which gonna give you a scalar – janmarqz Jun 13 '18 at 19:36

1 Answers1

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Matrix algebra consists of a set of rules for adding and multiplying rectangular arrays of appropriate sizes. According to these rules your first formula is correct.

Matrix algebra is the proper computational machine when working with vectors $x\in V$, where $V$ is a finite dimensional vector space, linear maps $A:\>V\to W$, and functionals $\phi\in V^*$. Common usage treats vectors as "column vectors", functionals (e.g., gradients) as "row vectors", and matrices of linear maps as "acting on the left". In particular the evaluation of a $\phi\in \bigl({\mathbb R}^3\bigr)^*$ on a vector $x\in{\mathbb R}^3$ appears as $$\bigl[\phi_1 \ \phi_2 \ \phi_3\bigr]\>\left[\matrix{x_1\cr x_2\cr x_3\cr}\right]=\phi_1x_1+\phi_2x_2+\phi_3 x_3\ .$$

But matrix algebra with its restriction to arrays of dimension $\leq2$ is not the proper tool when it comes to multilinear algebra where tensors of higher rank, etc., appear. In the same vein, if you consider a vector $x\in V$ as an element in the double dual space, i.e., as a functional on $V^*$, you can no longer expect that the conventions set up in linear algebra 101 can handle this without ado. In this sense your third formula, interpreted as a matrix product, is false even if you correct the typos on the RHS.

Christian Blatter
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  • So the matrix multiplication introduced in introductory linear algebra courses isn't really multiplication? – Pineapple Fish Jun 14 '18 at 14:21
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    What makes you think it is not? What I'm saying is that it cannot handle in a whim every idea linear algebraists cook up. – Christian Blatter Jun 14 '18 at 14:44
  • Btw, what is RHS (guessing right hand side?), and which equation is the third? About the response, is it just that you have to define what each type each vector is and if and how it's a functional? Could you ever figure out if it's a vector, a dual, or double dual without clarification. This is confusing to me since I thought there was only one way to multiply two vectors, matrices, etc.. If one was a special case of the other I'd understand, but they aren't and in your answer it sounded like everything is treated as a linear/multilinear functional. Thanks for responding! – Pineapple Fish Jun 23 '18 at 01:59