The rank is equal to the dimension of the row space and the column space (both spaces always have the same dimension). This matrix has three rows and five columns, which means the largest possible number of vectors in a basis for the row space of a $\displaystyle 3 \times 5$ matrix is $\displaystyle 3$, so this is the largest possible rank.

According to the Rank + Nullity Theorem, 

$\displaystyle Rank(A)+Nullity(A)= columns$

Since the matrix has $\displaystyle 5$ columns, we can rearrange the equation to get

$\displaystyle Nullity(A) = 5- Rank(A)$

So to make the nullity as small as possible, we need to make the rank as large as possible.

The rank is equal to the dimension of the row space and the column space (both spaces always have the same dimension). This matrix has three rows and five columns, which means the largest possible number of vectors in a basis for the row space of a $\displaystyle 3 \times 5$ matrix is $\displaystyle 3$, so this is the largest possible rank.

Hence the smallest possible nullity is $\displaystyle 5-(3) =2$.

The rank is equal to the dimension of the row space and the column space (both spaces always have the same dimension). This matrix has seven rows and two columns, which means the largest possible number of vectors in a basis for the column space of a $\displaystyle 7 \times 2$ matrix is $\displaystyle 2$, so this is the largest possible rank.

According to the Rank + Nullity Theorem, 

$\displaystyle Rank(A)+Nullity(A)= columns$

Since the matrix has $\displaystyle 2$ columns, we can rearrange the equation to get

$\displaystyle Nullity(A) = 2- Rank(A)$

So to make the nullity as small as possible, we need to make the rank as large as possible.

The rank is equal to the dimension of the row space and the column space (both spaces always have the same dimension). This matrix has three rows and five columns, which means the largest possible number of vectors in a basis for the row space of a $\displaystyle 7 \times 2$ matrix is $\displaystyle 2$, so this is the largest possible rank.

Hence the smallest possible nullity is $\displaystyle 2-(2) =0$.

The sum of the rank and the nullity of any matrix is always equal to to the number of columns in the matrix. Therefore, a matrix with four columns and rank 3, such as $\displaystyle B$, must have as its nullity $\displaystyle 4- 3= 1$.

A subset $\displaystyle S$ of a vector space is a subspace of that vector space if and only if it meets two criteria. Both will be given and tested, letting 

$\displaystyle S = \left \{ f(x)| f(x) = 2n x + n , n \in \mathbb{R}\right \}$.

This can be rewritten as

$\displaystyle S = \left \{ f(x)| f(x) = n( 2 x +1) , n \in \mathbb{R}\right \}$

One criterion for $\displaystyle S$ to be a subspace is closure under addition; that is:

If $\displaystyle a, b \in S$, then $\displaystyle a+b \in S$.

Let $\displaystyle f, g \in S$ as defined. Then for some real $\displaystyle n, m$:

$\displaystyle f(x) = n( 2 x +1)$

$\displaystyle g(x) = m( 2 x +1)$

$\displaystyle (f+g)(x) = f(x) + g(x)$

$\displaystyle = n( 2 x +1) + m( 2 x +1)$

$\displaystyle = (n + m)( 2 x +1)$

It follows that $\displaystyle f+g \in S$.

The second criterion for $\displaystyle S$ to be a subspace is closure under scalar multiplication; that is:

If $\displaystyle a \in S , c \in \mathbb{R}$, then $\displaystyle ca \in S$

Let $\displaystyle h \in S$ as defined. Then for some real $\displaystyle p$:

$\displaystyle h(x) =p (2 x +1)$

$\displaystyle (ch)(x) = c (h(x))$

$\displaystyle = c \cdot p (2 x +1)$

$\displaystyle =( c p )(2 x +1)$

It follows that $\displaystyle ch \in S$

$\displaystyle S$, as defined, is a subspace of $\displaystyle C(-\infty , \infty)$.

A subset $\displaystyle S$ of a vector space can be proved to not be a subspace of the space by showing that the zero of the space is not in $\displaystyle S$. 

Let $\displaystyle S$ be the subset in question, and let $\displaystyle f(x) = 0$, the zero function, which is in $\displaystyle C(-\infty, \infty)$. This cannot be expressed as $\displaystyle f(x) = 3nx + n + 1$ for any $\displaystyle n$. It if could then

$\displaystyle 3n = 0$, in which case $\displaystyle n = 0$,

and

$\displaystyle n+ 1 = 0$, in which case $\displaystyle n = -1$.

By contradiction, $\displaystyle 0 \notin S$. $\displaystyle S$ is not a subspace of $\displaystyle C(-\infty, \infty)$.

Let $\displaystyle S = \left \{ f(x)| f(x) = |nx | , n \in \mathbb{R}\right \}$.

We can show through counterexample that this is not a subspace of $\displaystyle C(-\infty , \infty)$.

Let $\displaystyle f(x) = |3x|$. This is an element of $\displaystyle S$.

One condition for $\displaystyle S$ to be a subspace of a vector space is closure under scalar multiplication. Multiply $\displaystyle f(x)$ by scalar $\displaystyle -1$. The product is the function 

$\displaystyle g(x) = -1 f(x) = -|3x|$.

$\displaystyle g(x) \ne S$.

This violates a criterion for a subspace, so $\displaystyle S$ is not a subspace of $\displaystyle C(-\infty , \infty)$.

Let $\displaystyle S$ be the set of all functions $\displaystyle f$ on $\displaystyle (-\infty, \infty)$ such that $\displaystyle f^{-1}$ is defined. 

A sufficient condition for $\displaystyle S$ to not be a subspace of $\displaystyle C (-\infty, \infty)$ is that the zero of the set - which here is the zero function $\displaystyle f(x) = 0$ - is not in $\displaystyle S$. $\displaystyle f \notin S$ because the zero function - a constant function - does not have an inverse ( $\displaystyle f(0) = f(1) = 0$, violating a condition of an invertible function). It follows that $\displaystyle S$ is not a subspace of  $\displaystyle C (-\infty, \infty)$.

A set $\displaystyle S$ is a subspace of a vector space if and only if two conditions hold, both of which are tested here.

The first condition is closure under addition - that is:

If $\displaystyle a , b \in S$, then $\displaystyle a+ b \in S$

Let $\displaystyle f , g \in S$ as defined. Then for some $\displaystyle n, m \in \mathbb{R}$,

$\displaystyle f(x) = \left\{\begin{matrix} 0,& x< 0\\ nx,& x \ge 0 \end{matrix}\right.$ 

and

$\displaystyle g(x) = \left\{\begin{matrix} 0,& x< 0\\mx,& x \ge 0 \end{matrix}\right.$.

$\displaystyle (f+ g )(x) = \left\{\begin{matrix} 0+ 0,& x< 0\\nx + mx,& x \ge 0 \end{matrix}\right.$

or

$\displaystyle (f+ g )(x) = \left\{\begin{matrix} 0 ,& x< 0\\(n + m)x,& x \ge 0 \end{matrix}\right.$,

$\displaystyle f +g \in S$. The first condition is met.

The second condition is closure under scalar multiplication - that is:

If $\displaystyle a \in S$ and $\displaystyle c$ is a scalar, then $\displaystyle ca \in S$

Let $\displaystyle h \in S$ as defined. Then for some $\displaystyle p \in \mathbb{R}$, 

$\displaystyle h(x) = \left\{\begin{matrix} 0,& x< 0\\ px,& x \ge 0 \end{matrix}\right.$

For any scalar $\displaystyle c$,

$\displaystyle c \cdot h(x) = \left\{\begin{matrix} c \cdot 0,& x< 0\\ c \cdot px,& x \ge 0 \end{matrix}\right.$,

and

$\displaystyle (c h)(x) = \left\{\begin{matrix} 0,& x< 0\\ (c p)x,& x \ge 0 \end{matrix}\right.$

$\displaystyle c h \in S$. The second condition is met. 

$\displaystyle S$, as defined, is a subspace.