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  matrix is , so this is the largest possible rank.

According to the Rank + Nullity Theorem, 

Since the matrix has  columns, we can rearrange the equation to get

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  matrix is , so this is the largest possible rank.

Hence the smallest possible nullity is .

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  matrix is , so this is the largest possible rank.

According to the Rank + Nullity Theorem, 

Since the matrix has  columns, we can rearrange the equation to get

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  matrix is , so this is the largest possible rank.

Hence the smallest possible nullity is .

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 , must have as its nullity .

A subset  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 

.

This can be rewritten as

One criterion for  to be a subspace is closure under addition; that is:

If , then .

Let  as defined. Then for some real :

It follows that .

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

If , then 

Let  as defined. Then for some real :

It follows that 

, as defined, is a subspace of .

A subset  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 . 

Let  be the subset in question, and let , the zero function, which is in . This cannot be expressed as  for any . It if could then

, in which case ,

and

, in which case .

By contradiction, .  is not a subspace of .

Let .

We can show through counterexample that this is not a subspace of .

Let . This is an element of .

One condition for  to be a subspace of a vector space is closure under scalar multiplication. Multiply  by scalar . The product is the function 

.

.

This violates a criterion for a subspace, so  is not a subspace of .

Let  be the set of all functions  on  such that  is defined. 

A sufficient condition for  to not be a subspace of  is that the zero of the set - which here is the zero function  - is not in .  because the zero function - a constant function - does not have an inverse ( , violating a condition of an invertible function). It follows that  is not a subspace of  .

A set  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 , then 

Let  as defined. Then for some ,

and

.

or

,

. The first condition is met.

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

If  and  is a scalar, then 

Let  as defined. Then for some , 

For any scalar ,

,

and

. The second condition is met. 

, as defined, is a subspace.