The sum of the rank and the nullity of a matrix is equal to the number of columns in the matrix. Therefore, the number of columns in  is equal to , and the number of columns in  - the number of rows in  - is  .

Furthermore, the rank of a matrix is equal to its transpose, so 

,

and

This makes  a matrix with the same number of rows as columns. It is square, and the statement is true.

The rank of a matrix can be found by performing row operations until it is reduced to row-echelon form. On matrix , we want to have leading 1's in each nonzero row, we want the leading 1's to go from upper left to lower right, and we want zero rows gathered at the bottom. Beginning with :

Perform the following row operations:

The matrix is now in row-echelon form. There are two nonzero rows, so the rank is 2. The sum of the rank and the nullity is equal to the number of columns, of which there are 5; this makes the nullity 3. 

The rank of a matrix can be found by performing row operations until it is reduced to row-echelon form. On matrix , we want to have leading 1's in each nonzero row, we want the leading 1's to go from upper left to lower right, and we want zero rows gathered at the bottom. Beginning with :

The following four operations can be performed simultaneously:

The next three can be performed simultaneously:

The matrix is now in reduced row-echelon form. There are two nonzero rows, so the rank is 2. The sum of the rank and the nullity is equal to the number of columns, of which there are 3, so the nullity is 1.

Two linear spaces are isomorphic if and only if they are of the same dimension, or, equivalently, the size of a basis of each includes the same number of elements.

 is the set of  matrices; each such matrix includes 4 elements, so it has four elements in one of its bases; for example, one such basis is

 is the set of all polynomials in  of degree 4 or less - that is, all polynomials of the form

Each polynomial is defined by five coefficients, so the dimension of  is 5. Each basis of  has five elements; for example, one such basis is

.

Since , the spaces are not isomorphic.

  is a vector space with dimension 2, so any basis for  must include exactly two vectors. This makes it possible that  forms a basis for ; also, this makes their linear independence necessary and sufficient for them to form a basis, since, if they are as such, they are a spanning set as well.

We can test for linear independence by forming the following matrix with their entries:

The rows (and the original vectors) are linearly independent if and only if 

,

so calculate the determinant:

Applying the rules of logarithms:

The rows of the matrix, and, consequently, the vectors  and , are linearly dependent. Furthermore,  and  do not span .