I have been wondering recently if a $1 \times 1$ matrix reprsents a scalar, and after doing some reading I'm still not satisfied. I've decided to ask this question from a different perspective: does a scalar represent a $1 \times 1$ map?

Because any matrix is a linear map between two bases, that would imply that a $1 \times 1$ matrix is an isomorphism between two 1 dimensional spaces.

So in the function/map $ f(x) = rx $ where r is some number, isn't $r$ just a scalar? This is a map and so can't this map be represented by the matrix $(r)$? Does that not, then, imply that a $1 \times 1$ and a scalar are the same?

Obviously, multiplication between a $1 \times 1$ matrix and a $n\times m$ matrix where $n \neq 1$ is not defined, but $1 \times 1$ matrices seem to behave like scalars in all other cases, so why isn't an exception made? And what is the fundamental difference between a $1 \times 1$ map and a scalar, if there is one?

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    $\begingroup$ I wouldn't say a matrix is a linear map. $\endgroup$ – user137731 Jun 10 '16 at 22:42
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    $\begingroup$ Why? Any matrix represents a linear map. $\endgroup$ – user3210986 Jun 10 '16 at 22:45
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    $\begingroup$ Once you've chosen a basis for both the domain and codomain sure -- a matrix can represent a linear map. $\endgroup$ – user137731 Jun 10 '16 at 22:47
  • $\begingroup$ I don't see how the answer you accepted answers your question. It only talks about scalars for a one-dimensional vector space, which is so trivial as to be a field itself. $\endgroup$ – user21820 Jun 11 '16 at 6:28
  • $\begingroup$ @user21820 I believe (but could be entirely wrong) that's what OP meant by "$1\times 1$ map": a map between 1D vector spaces. $\endgroup$ – user137731 Jun 11 '16 at 13:46

A vector space is a two-sorted structure, one sort being vectors and the other sort being scalars. Scalars act linearly on vectors via scalar multiplication, and a scalar multiplication is nothing more than a scaling operation, so it is no surprise why "scalar" has that name. Of course, scaling operations are (meant to be) linear operators on vectors, and for finite-dimensional vector spaces, linear operators on vectors are isomorphic to matrices (with operator composition corresponding to matrix multiplication).

Specifically, given any finite-dimensional vector space $V$:

$c \cdot v = \pmatrix{ c & 0 & \cdots & 0 \\ 0 & c & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & c } \pmatrix{ x \\ y \\ \vdots \\ z }$ for any scalar $c$ and vector $v$ in $V$, where $\pmatrix{ x \\ y \\ \vdots \\ z }$ represents $v$.

Note that the above is valid regardless of which basis we use for $V$. If you choose a different basis then $v$ will have a different representation, but the result after the matrix multiplication will be the representation of the same scaled $v$, namely $c \cdot v$ here.

So it is not correct to think of a scalar as a $1 \times 1$ matrix, even though it is true that the field of scalars alone by itself is isomorphic to the field of $1 \times 1$ matrices. That is just a natural coincidence, but whenever we think of scalars it is with respect to a larger vector space structure, in which case scalars are not at all $1 \times 1$ matrices. For an $n$-dimensional vector space, scalars correspond to a certain kind of $n \times n$ matrices as shown above.

Note that the correspondence is not at all equality. That is why the same collection of scalars can be used in many different vector spaces with different (or even non-existent) dimension. It is only with respect to each vector space structure that scalars are isomorphic to a particular collection of matrices.

Since not all vector spaces have matrix representations, scalars are in general just (isomorphic to) a special collection of linear maps that are also a field under pointwise addition and composition.

Exactly the same thing applies to scalar multiplication of matrices, which is nothing more than composition of a scaling operation after the linear map represented by the matrix.

Specifically, given any finite-dimensional vector space $V$:

$c M = \pmatrix{ c & 0 & \cdots & 0 \\ 0 & c & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & c } M$ for any scalar $c$ in $V$ and matrix $M$ over a basis of $V$.

  • $\begingroup$ There is another approach of considering $c \cdot v$ to be $v \pmatrix{c}$ where $\pmatrix{c}$ would be a $1 \times 1$ matrix, but this is not the intended meaning of matrices, which represent linear maps, unless you first show that vectors themselves can be treated as linear maps, which depending on your point of view may not be valid even though it works out for finite-dimensional vector spaces. $\endgroup$ – user21820 Jun 11 '16 at 6:26
  • $\begingroup$ "in many different vector spaces with different (or even non-existent) dimension." — what would be a vector space without dimension? $\endgroup$ – celtschk Jun 11 '16 at 7:41
  • $\begingroup$ @celtschk: It depends heavily on your set-theoretic axioms. If you only have ZF, some vector spaces may not even have a basis, and even if it does it may have bases with different cardinality (mathoverflow.net/q/93242). $\endgroup$ – user21820 Jun 11 '16 at 8:43
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    $\begingroup$ Thank you for this long answer. As far as I can tell, you are essentially saying the same thing as snulty and Bye_World, but this definitely clarifies the entire thing and the foundations more. $\endgroup$ – user3210986 Jun 12 '16 at 7:43
  • $\begingroup$ @user3210986: You're welcome and I'm glad that my answer has cleared it up. If you have any further queries I'll be glad to help. $\endgroup$ – user21820 Jun 12 '16 at 8:03

I'll expand on snulty's answer by giving an example.

Let's take a linear map $f:\Bbb R\to \Bbb R$ defined by $f(x) = 3x$. First I choose the standard basis, $\{1\}$, for both the domain and codomain of this map. Then clearly the matrix which represents this map is $(3)$.

But let's try a different pair of bases. I'll choose the basis $\{2\}$ for the domain and $\{3\}$ for the codomain of this function. Then the matrix which represents this map is $\left(2\right)$. How do I know? Well let's check what happens when we try to transform the "vector" $6$. The component of $6$ wrt the basis $\{2\}$ is $(3)$ so we multiply out the matrices $$\left(2\right)(3) = \left(6\right)$$ which is the component of the vector $18$ wrt the basis $\{3\}$.

So as you can see the $3$ in the definition of the map is a member of the field $\Bbb R$ and is independent of the bases chosen for the domain and codomain of the linear function. However, the sole entry of the matrix which represents the function depends on the bases we choose for the domain and codomain of our mapping.

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    $\begingroup$ It's probably worth noting that if you demand that the bases of domain and codomain be the same, then the matrix is definitely just $(3)$. One can certainly find situations where this interpretation is useful (e.g. $f$ is a map from a vector space to itself) and situations where the interpretation elaborated in your post is useful (e.g. $f$ is a map between unrelated copies of a vector space). $\endgroup$ – Milo Brandt Jun 11 '16 at 1:34

Any linear map from a one dimensional vector space to another is given by multiplication by a constant, or a $1\times 1$ matrix, once you've chosen basis (bases).

The $1\times 1$ matrix however changes depending on the choice of basis. A scalar in linear algebra is usually and element of the field the vector space is over, which is independent of the choice of basis for the vector space itself.


If $A$ is an $n \times m$ matrix representing some linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$, then it is helpful to note that the $j$th column of $A$ is the image of the $j$th basis vector (e.g., if you are using the standard basis, then it is the vector with a $1$ in the $j$th component, and all other components zero). Sticking with the standard basis, the columns of $A$ are elements of the image space $\mathbb{R}^n$.

If $A=[r]$ is $1 \times 1$, then yes it is a linear map $\mathbb{R} \to \mathbb{R}$. Under the standard basis, this does take the form $x \mapsto rx$ as you noted. From the previous paragraph, you can also think of the single entry $r$ as the image of the element $1$ (since $r \cdot 1 = r$), and clearly $r$ is an element of the image space $\mathbb{R}$.

I am not sure what your confusion is concerning scalars. The entries of vectors and matrices are scalars. In particular, a $1$-dimensional vector and a $1\times 1$ matrix can be thought of as a scalar, but can also still be interpreted as a vector or a matrix / linear map respectively.


Yes, a scalar $s \in \mathbb{F}$ does represent the map $$ T : \mathbb{F} \to \mathbb{F} : T(x) = s x $$

Because of $$ T(\alpha x + \beta y) = s(\alpha x + \beta y) = \alpha s x + \beta s y = \alpha T(x) + \beta T(y) \quad (\alpha, \beta, x, y \in \mathbb{F}) $$ it is a linear map.


In terms of matrix algebra, there are two senses in which a scalar $r$ determines a map:

  • It acts on vector spaces, sending $v \mapsto rv$
  • It acts on linear transformations, sending $T \mapsto rT$.

The former should be thought of as the linear transformation $r I_V$, where $I_V$ is the identity transformation on the vector space $V$.

For the the second form, there is a way to equate a scalar with a $1 \times 1$ matrix by using the tensor product: specifically, the linear transformation

$$ V \xrightarrow{rT} W $$

that sends $v \mapsto r T(v)$ can be rewritten as a linear transformation

$$ V \to \mathbb{R} \otimes V \xrightarrow{[r] \otimes T} \mathbb{R} \otimes W \to W $$

where the maps send:

  • $v \mapsto 1 \otimes v$
  • $[s] \otimes v \mapsto ([r][s]) \otimes T(v)$
  • $[s] \otimes w \mapsto sw$

where I've identified vectors in $\mathbb{R}^n$ with $n \times 1$ matrices.

So, if we always identify $V$ and $\mathbb{R} \otimes V$ with the standard maps used above, then

$$ rT = [r] \otimes T$$

For example, we can see this when $V$ is a standard vector space and $T$ a matrix, by the Kronecker product of matrices.


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