# Vector space of $m\times n$ matrix $\Bbb R^{m\times n}$ vs vector space of a vector of size $mn$ $\Bbb R^{mn}$?

I have a very basic question in linear algebra. Every vector of size $$n$$ with each entry from $$\Bbb R$$ lies in the space $$\Bbb R^n$$, where $$\Bbb R^n$$ is the cartesian product of $$n$$ copies of the set $$\Bbb R$$.

But how is it different from the vector space of matrices? It is written in my books that the vector space of matrices is $$\Bbb R^{m\times n}$$, but what does that mean? What is $$\Bbb R^{m\times n}$$? Because if it is the same as $$\Bbb R^{mn}$$, then how is an $$m\times n$$ matrix different from an $$mn$$ dimensional vector?

• Matrices have multiplication, whereas vectors in $\mathbb{R}^{mn}$ may not. Commented Oct 29, 2016 at 20:09
• A vector space does not require multiplication Commented Oct 29, 2016 at 20:11

Both spaces are isomorphic. For instance take $$\phi:M_{m,n}(\Bbb R) \rightarrow \Bbb R^{mn}$$ the morphism such that for $$i_1\in\{1, ..., n\}$$ and $$j_1\in\{1, ..., m\}$$ we have $$\phi([\delta_{i, i_1}\delta_{j, j_1}]_{i, j})$$ is the vector where all coordinates are 0 except the $$i_1m + j_1$$ -eth which is 1. Such a morphism exists since $$[\delta_{i, i_1}\delta_{j, j_1}]_{i, j}$$ is a basis of $$M_{m,n}(\Bbb R)$$.

You'll find that $$\ker\phi = \{0\}$$ and since $$\dim M_{m,n}(\Bbb R) = \dim \Bbb R^{mn} = mn$$, $$\phi$$ is bijective.

Thus a matrix in $$M_{m,n}(\Bbb R)$$ may be viewed as a vector in $$\Bbb R^{mn}$$.

Simply put, a matrix of $$M_{m,n}(\Bbb R)$$ is never anything more than $$nm$$ numbers, and neither is a vector in $$\Bbb R^{mn}$$.

• Hi, I'm really sorry but that was a little verbose for me to understand. My mathematics background is a little shaky, if possible, could you please elaborate your answer/break it down in simpler words? Thanks in advance :) Also, the way I was thinking about it up until now was that $\mathbb{R}^{m\times n}$ is a space of $\mathbb{R}^n$ vector spaces. Is that incorrect? Commented Oct 29, 2016 at 22:32
• Formally two spaces are isomorphic when there exists a certain kind of bijection between them. Informally, two spaces are isomorphic when they behave in the same manner, they are so similar that any property concerning the first one will apply to the second one. You can also consider each of the $m$ columns of your matrix as a vector of length $n$ (this space is also isomorphic) or the lines as vectors of length m. What you get is that $M_{n, m}(\Bbb R) \sim \Bbb R^{nm} \sim {R^n}^m \sim {R^m}^n$ ($\sim$ means that both sets are isomorphic, ie they have the same properties) Commented Oct 29, 2016 at 22:38
• So then $\mathbb{R}^{m\times n}$ is NOT a space of vector spaces, but instead a simple vector space? Commented Oct 29, 2016 at 23:11
• If I am not mistaken, that is correct Commented Oct 30, 2016 at 8:07
• What's more : all real vector space of dimension $n$ are isomorphic to $\Bbb R^n$. (Informally dimension is the number of real values you need to have to caracterise any vector) Commented Oct 30, 2016 at 8:21

Both spaces are isomorphic as vector spaces. In fact, often we will "flatten" matrices by converting them to $$1\times(NM)$$ vectors in just the way you are talking about. If the only thing you care about is the vector space addition and scalar multiplication, then there is no difference.

Where there is a difference is that the space of matrices has with it some extra structure. For instance, if you have $$N = M$$, meaning that you have square matrices, then you can multiply the matrices together to get a new matrix. This turns the space of $$N \times N$$ matrices into an algebra, which is a vector space with an additional way to "multiply" vectors together. Of course, the space of "flattened" vectors can also be turned into an algebra as well, so there is no real material difference, but it's important to note that when we talk about matrices we are implicitly including this extra structure as well.