More Examples on Matrix Space

Example 1

Consider all

m\times n

matrices, here

\(n\)

and

\(m\)

are constant (say

\(m=5\)

and

\(n=17\)

).
Let's denote space of all

m\times n

matrices by

\mathcal{M}

.
Now consider a subset of all

m\times n

matrices,

Consider all $m\times n$ matrices with rank $$=1$$ .
Let's denote space of all $m\times n$ matrices with rank $$=1$$ by $\mathcal{Q}$ .

Is space of all

m\times n

matrices with rank $$=1$$

(\mathcal{Q})

is a subspace of space of all

m\times n

matrices

(\mathcal{M})?

Or say Is

(\mathcal{Q})

a subspace of

(\mathcal{M})?

For $(\mathcal{Q})$ to be a subspace, first it had to be a vector space.

So Is

(\mathcal{Q})

a vector space?

Space of all $m\times n$ matrices with rank $$=1$$ $(\mathcal{Q})$ do not form a vector space of all $m\times n$ matrices.

But Why?
Consider two rank $$1$$ matrices $A\in\mathcal{Q}$ and $B\in\mathcal{Q}$ .
For $\mathcal{Q}$ to be a vector space, linear combination of $$A$$ and $$B$$ must $\in\mathcal{Q}$ .
Let's consider an example (say $$m=2, n=3$$ )
$A= \begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$ and $B= \begin{bmatrix} 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix}$

Here $$A$$ and $$B$$ both are rank $$1$$ matrices so $A\in\mathcal{Q}$ and $B\in\mathcal{Q}$ .

$A+B= \begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix}$
$$A$$ and $$B$$ are rank $$1$$ matrices so they $\in\mathcal{Q}$ .
But $$A+B$$ has rank = $$2$$ .
$$A+B$$ is a linear combination of $$A$$ and $$B$$ but because $$A+B$$ has rank = $2\Rightarrow$ $A+B\notin \mathcal{Q}$ .
So $\mathcal{Q}$ is not a vector space.

Example 2

Consider all vectors in

\mathbb{R}^4

Let's denote space of all vectors in

\mathbb{R}^4

\mathcal{S}

.
Now consider a subset of all vectors in

\mathbb{R}^4

Consider all vectors in $\mathbb{R}^4$ where there elements sum up to $$0$$ .
for example assume a vector $\vec{v}\in\mathbb{R}^4$ .
$\vec{v}=\begin{bmatrix} \color{red}{v_1\in\mathbb{R}}\\ \color{blue}{v_2\in\mathbb{R}}\\ \color{green}{v_3\in\mathbb{R}}\\ \color{brown}{v_4\in\mathbb{R}}\\ \end{bmatrix}$ And we want $\color{red}{v_1} + \color{blue}{v_2} + \color{green}{v_3} + \color{brown}{v_4} =0$
Let's denote space of all vectors in $\mathbb{R}^4$ where there elements sum up to $$0$$ by $\mathcal{Q}$ .
(So $\vec{v}\in\mathcal{Q}$ )

Is space of all vectors in

\mathbb{R}^4

where there elements sum up to

\(0\)

(\mathcal{Q})

is a subspace of space of all vectors in

\mathbb{R}^4

(\mathcal{S})?

Or say Is

\mathcal{Q}

a subspace of

\mathcal{S}?

Yes, $\mathcal{Q}$ a subspace of $\mathcal{S}$ .

But How? Consider vectors in $\mathbb{R}^4$ where there elements sum up to $$0$$ , $\vec{v}\in\mathcal{Q}$ and $\vec{w}\in\mathcal{Q}$ .
For $\mathcal{Q}$ to be a subspace of $\mathcal{S}$ linear combination of $\vec{v}$ and $\vec{w}$ must $\in\mathcal{Q}$ .
$\vec{v} = \begin{bmatrix} v_1\in\mathbb{R}\\ v_2\in\mathbb{R}\\ v_3\in\mathbb{R}\\ v_4\in\mathbb{R}\\ \end{bmatrix}$ , $\vec{w} = \begin{bmatrix} w_1\in\mathbb{R}\\ w_2\in\mathbb{R}\\ w_3\in\mathbb{R}\\ w_4\in\mathbb{R}\\ \end{bmatrix}$
There is a constraint on $\vec{v}$ and $\vec{w}$ , we want elements of $\vec{v}$ and $\vec{w}$ sums to $$0$$ , $\sum_{v_i\in\vec{v}}v_i=0 \\ \sum_{w_i\in\vec{w}}w_i=0 \\$ Let's take linear combination of $\vec{v}$ and $\vec{w}$ and if every linear combination of $\vec{v}$ and $\vec{w}$ $\in\mathcal{Q}$ then it's a subspace of all vectors in $\mathbb{R}^4$ .
$\alpha\cdot\vec{v} +\beta\cdot\vec{w} = \alpha\cdot \begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4\\ \end{bmatrix} + \beta\cdot \begin{bmatrix} w_1\\ w_2\\ w_3\\ w_4\\ \end{bmatrix} ;\quad$ $\alpha\in\mathbb{R}, \beta\in\mathbb{R}$
Say $\alpha\cdot\vec{v} +\beta\cdot\vec{w} = \vec{r}$
$\vec{r} = \begin{bmatrix} \alpha\cdot v_1 + \beta\cdot w_1 \\ \alpha\cdot v_2 + \beta\cdot w_2 \\ \alpha\cdot v_3 + \beta\cdot w_3 \\ \alpha\cdot v_4 + \beta\cdot w_4 \\ \end{bmatrix}$
For $\mathcal{Q}$ to be a subspace of $\mathcal{S}$ , $$r$$ must $\in\mathcal{Q}$ .
And if $r\in\mathcal{Q}$ then $\sum_{r_i\in\vec{r}}r_i=0$ Proof:
$\sum_{r_i\in\vec{r}}r_i = \alpha\cdot (v_1 + v_2 + v_3 + v_4) + \beta\cdot (w_1 + w_2 + w_3 + w_4)$
$\sum_{r_i\in\vec{r}}r_i = \alpha\cdot (0) + \beta\cdot (0)$ $\sum_{r_i\in\vec{r}}r_i = 0$ So space of all vectors in $\mathbb{R}^4$ where there elements sum up to $$0$$ $(\mathcal{Q})$ is a subspace of space of all vectors in $\mathbb{R}^4$ $(\mathcal{S})$