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Intuitively I used to think that a $1\times 1$ matrix is simply a scalar number, I also saw this statement in books. However when I think about it now it doesn't make sense to me because of one problem.

Let $I$ be the $2\times 2 $ identity matrix, now $\lambda I$ is well defined if $\lambda $ is a scalar or if $\lambda $ is a $1 \times 2$ matrix. However if we want to make the statement that scalars and $1 \times 1$ matrices are the same then the $\lambda I$ would have to be defined for a matrix $\lambda$ which is not $1 \times 2$.

More generally, let $M \in \Bbb R^{n \times m}$, now linear algebra says that if $N$ is another matrix, the product $NM$ only exists if $N \in \Bbb R^{m \times n}$. However if $n$ and $m$ are not zero then $NM$ is not defined for a $1\times 1$ matrix $N$, while $NM$ is defined if $N$ is simply a scalar... So in conclusion we can simply state that $1\times 1$ matrices and scalars are not the same thing right? Is there any interesting theory behind this?

Slugger
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2 Answers2

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Your bottom line has it right, the way I see it:

Multiplication of a scalar by a vector or matrix is always defined for any vector space $V$ over some field $F$.

However, multiplication of a vector or matrix by another matrix or vector is not always defined.

Therefore, the $1 \times 1$ matrix is not a scalar.

Newb
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  • Isn't multiplication (cross and dot) of any two square matrices always defined? – JMCF125 Nov 15 '13 at 21:47
  • @JMCF125 When we are formally and generally considering vector spaces, that is not the case. It is only in certain cases that we can define matrix-matrix multiplication. – Newb Nov 15 '13 at 22:09
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I think we need to be careful with the word "is" here. Suppose that we're looking at matrices over the field $F$ (in your example $F= \mathbb{R}$). Then we call the elements $\lambda \in F$ as scalars and define the multiplication of a matrix $M$ by a scalar $\lambda$ in the standard way. Now, let's look at the space of $1 \times 1$ matrices with elements from $F$, let's denote this space $F^{1\times 1}$. Now, technically $F^{1 \times 1}$ is not the same space as $F$; however, there is a clear identification of $F$ with $F^{1\times 1}$ by $\lambda \leftrightarrow [\lambda]$. Moreover, $F^{1 \times 1}$ is again a field that behaves exactly like $F$. Now, let's ask your question: Are the elements of $F^{1 \times 1}$ scalars? The answer is, it depends! If you look at $F^{m \times n}$, the $m \times n$ matrices with elements in $F$, then technically no, since $[\lambda] \notin F$ but rather $[\lambda] \in F^{1 \times 1}$. However, $F^{1 \times 1}$ is again a field (which for all practical purposes is identical to $F$) and therefore it makes sense to define $\left(F^{1\times 1}\right)^{m \times n}$, the space of all $m \times n$ matrices with entries from $F^{1 \times 1}$. Now, with respect to $\left(F^{1\times 1}\right)^{m \times n}$, the elements in $F^{1 \times 1}$ are scalars, but the elements in $F$ are not (since the elements in these matrices come from $F^{1 \times 1}$ and not $F$). However, as before, $\left(F^{1\times 1}\right)^{m \times n}$ behaves identically as $F^{m \times n}$. So, when a book or other source claims that the scalars are the $1 \times 1$ matrices, they are making these identifications without explicitly mentioning them each time (because that would be a tedious task with no true benefit to the mathematical theories that result).

Tom
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  • This is actually a very good answer. Also - I don't know whether this was intentionally humorous - your first sentence had me chuckling. – Newb Nov 15 '13 at 22:11