I have this "matrix" and it is 1-dimensional. That is,

$${\bf B} = \mu_1 + \mu_2 - \lambda_A$$

where $\mu_1$, $\mu_2$, and $\lambda_A$ are scalars.

From a technical perspective in mathematics, am I allowed to call this a "matrix"?

  • $\begingroup$ The meaning of $\mathbf{B}$ is not at all clear. $\endgroup$ – AJY Apr 12 '17 at 2:31
  • $\begingroup$ It has no particular meaning - It's a 1x1 "matrix" (as an example) $\endgroup$ – PiE Apr 12 '17 at 2:39
  • $\begingroup$ So $\mathbf{B}$ is just a scalar? $\endgroup$ – AJY Apr 12 '17 at 2:39
  • $\begingroup$ Yes. I just wanted to give the scalar matrix notation $\endgroup$ – PiE Apr 12 '17 at 2:40
  • 1
    $\begingroup$ Well one of the big important things about matrices is that they encode linear transformations. For $1$-dimensional vector spaces, the only linear maps are just scalar multiplication, so the $1 \times 1$ matrix that encodes it is just that scalar. So yes, you can treat $\mathbf{B}$ as a $1 \times 1$ matrix that acts on a $1$-dimensional vector space, but that's just because it's a scalar. $\endgroup$ – AJY Apr 12 '17 at 2:44

(I'm assuming you're referring to real numbers here.) Any scalar $k\in \mathbf{R}$ can be thought of as a $1\times 1$ matrix. This is because it encodes a linear transformation $k:\mathbf{R}\to \mathbf{R}$ given by $k(x)=kx$. In particular, $k(x+y)=kx+ky=k(x)+k(y)$ and for any scalar $\lambda \in \mathbf{R}$, $k(\lambda x)=k\lambda x=\lambda k x=\lambda k(x)$.

So, in short: yes.

As an afterthought, I suppose this should work in any field $\mathbf{F}$ where we think of $\mathbf{F}$ as a $1-$dimensional vector space over itself.


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