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I was wondering. Clearly, we cannot multiply a (1x1)-matrix with a (4x3)-matrix; However, we can multiply a scalar with a matrix. This suggests a difference.

On the other hand, I was, for example, in an econometrics lecture today, where we had for a (Tx1)-vector $\underline{û}=\left( \begin{array}{c} û_1\\ \vdots\\ û_T\end{array}\right)$:

$S_{ûû}:= \sum_{i=1}^T û_i^2$ shall be minimized. We see see that $S_{ûû}=\underline{û}^T\underline{û}$.

Well, formally, shouldn't it be $(S_{ûû})=\underline{û}^T\underline{û}$ or $S_{ûû}=\det(\underline{û}^T\underline{û})$, to ensure that we stay in the space of matrices and not suddenly go to the space of scalars? So here, the professor (physicist) not only treats $\underline{û}^T\underline{û}$ like a scalar, but also calls it a scalar. Is this formally legit or a wrong simplification (though it does not seem to have any impact, and surely makes life easier)?

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  • $\begingroup$ I think you can get away with it if you define the product of a $1 \times 1$ matrix with a larger matrix as the tensor product of the matrices, possibly after/combined with a natural transformation back to the original space. $\endgroup$
    – Lord_Farin
    Oct 23, 2012 at 14:51
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    $\begingroup$ A minor note: Some people write $|A|$ for the determinant of a matrix $A$. For those that do, conflating $1\times 1$ matrices with scalars poses a notational conundrum, since $|[-2]|\neq |-2|$. $\endgroup$
    – Cam McLeman
    Oct 24, 2012 at 13:26

3 Answers 3

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It's just a scalar in the sense that the ring of $1\times 1$ matrices over a field $K$ is isomorphic to $K$ (by the map $[x]\mapsto x$), but, as you observed, when you're considering the interaction of matrices of different sizes, then you have to treat them differently.

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Any matrix $A$ carries with it a type $(m,n)$ with $m$, $n\in{\mathbb N}_{\geq1}$. In fact such an $A$ is nothing else but a map $$A:\quad[m]\times[n]\to K\ ,\qquad (i,k)\mapsto a_{ik}\ .$$ When ${\rm type}(A)={\rm type}(B)$ then the sum $A+B$ is defined, and if ${\rm type}(A)=(m,n)$, ${\rm type}(B)=(n,p)$ then the product $AB$ is defined and has type $(m,p)$.

When $m=n=1$ then $A=[a]$ for a single number $a$ in the ground field $K$, e.g., $a\in{\mathbb R}$. Unfortunately there is no established notation to extract this $a$ out of the matrix $A$, just the same as there is no notation to extract the element $a$ out of the one-element set $\{a\}$. At any rate the map $[a]\mapsto a$ is well defined.

In the reverse direction things are more worrying. With any $c\in K$ we can form the $(1,1)$-matrix $[c]$ in a unique way. But note that the product $[c] \,A$ is only defined if $A$ has just one row (i.e., is of type $(1,n)$), and the product $A\, [c]$ is only defined if $A$ has just one column (i.e., is of type $(m,1)$).

Contrasting with this is the fact that the scalar multiple $c\,A$ is defined for all $c\in K$ and any matrix $A$, whatever its type. The effect of left-multiplying $A$ by the scalar $c$ is, that all elements of $A$ are multiplied by $c$. If you want to realize that by means of a matrix product you have to replace the scalar $c$ by a square diagonal matrix ${\rm diag}(c,c,\ldots, c)$ of the appropriate size.

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  • $\begingroup$ I know this is a little late, but you can extract the entry in $[a]$ using $[a]_{11}$. It's ugly, but it works. In a pure set theory (no atoms) $a$ can be extracted from $\{a\}$ using $\bigcup$. If there are atoms, I suppose you could extend the definition of $\bigcup$ conventionally so that $\bigcup\{a\}=a$ whether or not $a$ is a set, but that could be confusing. $\endgroup$
    – dfeuer
    Nov 15, 2013 at 22:10
  • $\begingroup$ "There is not established notation to extract this $a$". What about $\text{trace}$? Or $\text{det}$? $\endgroup$
    – Mike Earnest
    Mar 10, 2021 at 7:04
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No, a one-by-one matrix is not a scalar.

Let's assume you want to multiply a one-by-one matrix $[c]$ with a 3x3 matrix $A$. To do this realize, that you can always extend a matrix to fit the dimensions by adding rows and columns of zeros. So you get $$ \begin{bmatrix} c \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} c & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} ca_{11} & ca_{12} & ca_{13} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ But multiplying a scalar $c$ with $A$ is different. The scalar acts like a square diagonal matrix. So you get $$ c \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} c & 0 & 0 \\ 0 & c & 0 \\ 0 & 0 & c \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} ca_{11} & ca_{12} & ca_{13} \\ ca_{21} & ca_{22} & ca_{23} \\ ca_{31} & ca_{32} & ca_{33} \end{bmatrix} $$

From this we see: A one-by-one matrix $[c]$ always acts like a matrix with a single eigenvalue $c$. A scalar $c$ on the other hand is a shortcut for a n-dimensional matrix with $n$ equal eigenvalues $c$.

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  • $\begingroup$ You can't multiply matrices of incompatible sizes just by adding zeros! Mathematically it is not at all valid. $\endgroup$
    – user856
    Feb 25, 2017 at 21:34
  • $\begingroup$ You replied to a four year old question to explicitly show what was mentioned in the question $\endgroup$
    – Hugh
    Feb 25, 2017 at 21:37
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    $\begingroup$ The problem is that $\begin{bmatrix} c \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \ne \begin{bmatrix} c & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, because $\begin{bmatrix} c \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ is not defined in the first place. $\endgroup$
    – user856
    Feb 26, 2017 at 19:41
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    $\begingroup$ In your WolframAlpha examples, the outcome of the matrix multiplication does change: the result of the first is a $2\times2$ matrix and the result of the second is $3\times3$. It even says so right on the WolframAlpha page. $\endgroup$
    – user856
    Feb 26, 2017 at 19:43
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    $\begingroup$ Boss, you're genius. The bunch of commentators here missed the moot point, "why do you want to multiply". In geometric language, it will be either a transformation or a scaling. Ultimately a m-by-n matrix has to live in a k-dimensional space where k=max(m,n). If the 1x1 matrix lives on any one axis it will converge the whole dimensional plane on to its own axis (which is your first example). If the 1x1 matrix has to be interpreted as a scalar, it is nothing but scaled unit vectors on each dimension of that plane (which is your second example) $\endgroup$
    – manikawnth
    Aug 19, 2018 at 15:41

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