Linear Algebra:Vector Space

Question

Can someone give a thorough explanation of this following fragmented text(source:Linear Algebra and its Applications)?

"The space of $3$ by $2$ matrices. In this case the “vectors” are matrices! We can add two matrices, and $A+B = B+A$, and there is a zero matrix, and so on. This space is almost the same as $\Bbb{R}^6$. (The six components are arranged in a rectangle instead of a column.) Any choice of $m$ and $n$ would give, as a similar example, the vector space of all $m$ by $n$ matrices."

I'm baffled by the dimension of a matrix and its coordinates.

Answer 1

Avoiding any technicalities and putting everything in a simple language: Note that the term vector simply means a point of a vector space. The fact that $\mathbb{R}^n$ is the simpler vector space and its points are of the form of a "vector" in the usual sense is what confuses you, I believe.

In the case of the real vector space $\mathbb{R}^3$, an arbitrary point $(x_1,x_2,x_3)$ is specified by 3 real numbers namely, $x_1,x_2$ and $x_3$. This explains why the dimension is 3. Now consider the $3\times 2$-matrices with real entries. An arbitrary point of this vector space is a matrix which has 6 entries. So in order to specify a random point you need 6 real numbers. This explains why the dimension of this vector space is 6. Analogously for the case of $m\times n$ matrices.

If the question is why the set of $m\times n$ matrices with real entries is a real vector space, then you need to recall the definition of the vector space and check (similarly with the case of $\mathbb{R}^n$) that it satisfies all required properties.

Answer 2

The set of all $m \times n$ matrices forms a vector space over $\mathbb{R}$ and its dimension is $mn$.

Answer 3

Just consider reordering the matrix by just writing the columns one below the other instead of one besides the other, so they form a big column. Because addition and scalar multiplication are element-wise, they continue to work as usual. Written in this case, you immediately see why as vector space this is just an $\mathbb R^6$, or in the more general case, $\mathbb R^{mn}$.

Of course things like matrix multiplication break with this reordering; but matrix multiplication is not a vector space property.

Answer 4

$3\times 2$ matrices looks like $$\begin{pmatrix}a_1&a_2\\a_3&a_4\\a_5&a_6\end{pmatrix}$$ and you add them as follows $$\begin{pmatrix}a_1&a_2\\a_3&a_4\\a_5&a_6\end{pmatrix}+\begin{pmatrix}b_1&b_2\\b_3&b_4\\b_5&b_6\end{pmatrix}=\begin{pmatrix}a_1+b_1&a_2+b_2\\a_3+b_3&a_4+b_4\\a_5+b_5&a_6+b_6\end{pmatrix}$$ also scalar multiplication: $$c\begin{pmatrix}a_1&a_2\\a_3&a_4\\a_5&a_6\end{pmatrix}=\begin{pmatrix}ca_1&ca_2\\ca_3&ca_4\\ca_5&ca_6\end{pmatrix}$$

Now compare all of this to vectors in $\mathbb{R}^6$, the vectors looks like

$$\begin{pmatrix}a_1\\a_2\\a_3\\a_4\\a_5\\a_6\end{pmatrix}$$ and similarly you add them as follows $$\begin{pmatrix}a_1\\a_2\\a_3\\a_4\\a_5\\a_6\end{pmatrix}+\begin{pmatrix}b_1\\b_2\\b_3\\b_4\\b_5\\b_6\end{pmatrix}=\begin{pmatrix}a_1+b_1\\a_2+b_2\\a_3+b_3\\a_4+b_4\\a_5+b_5\\a_6+a_6\end{pmatrix}$$

and finally scalar multiplication $$c\begin{pmatrix}a_1\\a_2\\a_3\\a_4\\a_5\\a_6\end{pmatrix}=\begin{pmatrix}ca_1\\ca_2\\ca_3\\ca_4\\ca_5\\ca_6\end{pmatrix}$$

Notice that these "vector spaces" behave exactly the same way, the only difference is how you draw the matrices. For vectors in $\mathbb{R}^6$ you have $6$ entries, and the same is true for the vector space of $3\times 2$-matrices, often denoted by $M_{3,2}(\mathbb{R})$.