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,

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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 $(\mathcal{M})?$

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

So Is $(\mathcal{Q})$ a vector space?

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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 $m=2, n=3$)

$$A= \begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \text{ and } B= \begin{bmatrix} 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix}$$

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 $\mathbb{R}^4$ by $\mathcal{S}$.
Now consider a subset of all vectors in $\mathbb{R}^4$

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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 $\mathcal{S}?$

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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 $\mathbb{R}^4$ $(\mathcal{S})$