A test to determine whether these matrices form a basis is to set up a matrix with each row comprising the coefficients of one polynomial, and performing row reductions until the matrix is in row-echelon form.   is a vector space of dimension 4, so these four polynomials will form a basis if and only if the resulting   matrix has rank 4. The initial matrix is:

Perform the following row operations:

The matrix is now in row-echelon form. Each row has a leading 1, so the matrix has rank 4, the dimension of . It follows that the four matrices , , , and  comprise a basis of .

Linearly dependent sets have no limit to the number of vectors they can have.

A set of elements in a vector space forms a basis of the space if and only if satisfies two conditions:

1) It must be a spanning set, so that every other vector in the space can be written as a linear combination of those elements; and,

2) It must be a linearly independent set, so each vector can be so written uniquely.

We can test for both conditions by forming a matrix with the vectors and determining the rank of the matrix by way of row operations. The matrix is 

Perform the following row operations:

We now have a matrix in row-echelon form. All three rows have nonzero entries, so the rank of the matrix is equal to 3, the dimension of .  does form a basis of .

The sum of the rank and the nullity of a matrix is equal to the number of columns in the matrix. Since the matrix in question has six columns, for the rank and the nullity to be equal, they must each be 3.

A set of elements in a vector space forms a basis of the space if and only if satisfies two conditions:

1) It must be a spanning set, so that every other vector in the space can be written as a linear combination of those elements; and,

2) It must be a linearly independent set, so each vector can be so written uniquely.

We can test for both conditions by forming a  matrix with the entries of the four matrices and determining the rank of the matrix by way of row operations. The matrix is 

Perform the following row operations in order to get the matrix to a row-equivalent row-echelon form, and to therefore determine its rank:

We now have a matrix in row-echelon form. All four rows have nonzero entries, so the rank of the matrix is equal to 4, the dimension of . The set  forms a basis of .

A  matrix is nonsingular if and only if its rank is . This is not the case with , so  does not exist. 

The sum of the rank and the nullity of any matrix is the number of columns in the matrix, so , whose rank and nullity are 4 and 2, respectively, must have 6 columns.  Also, the rank of the transpose  is always equal to that of  itself.  therefore has rank 4 and nullity 3, so it has 7 columns - the number of rows in . This makes  a matrix with seven rows and six columns - a  matrix.

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.