ROW AND COLUMN SPACE OF A MATRIX

ROW & COLUMN SPACES OF A MATRIX:

Theorem: The non-zero rows in any row-echelon form of a matrix A form a basis for the
row space of A.

Theorem: The columns with the leading 1's in any row-echelon form of a matrix A form a basis for the column space of A.

Thm: The row and column space of A have the same dimension called the rank of A. (the dimension is the number of vectors in a basis).

Theorem: The vectors that define the solution set of Ax = 0 are a basis for the nullspace of A. The dimension of the nullspace is called the nullity of A.

Theorem: If a matrix A has n columns, then rank(A) + nullity(A) = n.

Theorem: The zero vector space has dimension 0.

Example 1: Find a basis for the row space, column space and nullity of matrix A.

here's how we reduce it to row-echelon form:

with 2 more steps, (r1 – 3r2) and (r1 – 3r3) we can reduce it to I 3× 3,
so the set { i, j, k }, the standard basis for R³,
forms a basis for both the row and column spaces,
and the 3-dimensional zero vector is the basis for the nullity.

This matrix has Rank = 3 and nullity = 0.

Since A is a square matrix, we could have determined all this by showing that
the determinant of A is not equal to zero. (see "Mini-Series Theorem" file).

Example 2: Find a basis for the row space, column space and nullity of matrix A.

,

We put it in row echelon form to get: , and solve for the nullspace basis.

Now there are 3 non-zero rows which is 3 equations in 4 unknowns, ( w, x, y, z ),
so we need 1 parameter. When we set y = s, we get

w = s x = – s y = s z = 0 s R

The vector: forms a basis for the nullspace so the nullity = 1.

A basis for the row space is the set of the 3 non-zero rows in the matrix:

, , so Rank(A) = 3.

A basis for the column space is the set of the 3 columns with the leading 1's.

, ,

These vectors are not unique. Had we reduced the matrix further, to this:

,

these vectors would form the row-space basis: , ,,

and these would be a basis for the column space: , ,

Both the row and column space of this matrix are 3-dimensional subspaces of R 4.

Finding a Basis for The Space Spanned by a Given Set of Vectors

A common question on this topic is to find the specific vectors from a given set that form a basis for the space spanned by the set of vectors. In such a case, we write the transpose of the matrix of row vectors and reduce it to row-echelon form. We do this because columns don't move like rows do (or can) during elimination, so once we find the columns with the leading 1's, we'll know exactly which of the original vectors correspond to those columns. These will be our basis vectors.

Example 3: a) Find a subset of these vectors that form a basis for the space spanned by them.

, ,, ,

Using each vector as a column, we get the matrix:

which reduces to .

The leading 1's are in columns 1, 2 and 4, so those 3 vectors form a basis for the space spanned by this set.

The basis vectors are: , , , and therefore the other 2 vectors must be linear combinations of these 3.

b) Find the linear combination of the basis vectors for the 2 vectors remaining in the set.

In the matrix, the 3rd column vector , , therefore the original column vector is the same linear combination of the original vectors.

,.

In the matrix, the 5th column vector , , so the original

5th vector is

Note: if asked to find a basis for the row space of matrix A consisting entirely of row vectors from A we do the same thing. We reduce At to row echelon form, then choose the columns with the leading 1's, which will indicate the original vectors in the basis.

Practice:

1. Find a subset of these vectors that forms a basis for the space spanned by the vectors then express each vector not in the basis as a linear combination of the basis vectors.

a) v1 = (1, 0, 1, 1), v2 = ( – 3 , 3, 7, 1), v3 = (–1, 3, 9, 3), v4 = (–5, 3, 5, –1).

b) v1 = (1, – 2, 0, 3), v2 = ( 2 , – 4, 0, 6), v3 = (–1, 1, 2, 0), v4 = (0, –1, 2, 3).

2. Find a basis for the subspace of R 4 spanned by these vectors.

.

Solutions:

1. a) We set up the matrix with the vectors as columns and row reduce:

which reduces to .

The 1st and 2nd columns have leading 1's, so v1 and v2 form the basis.

Since the 3rd column in the row-reduced matrix = 2 (col1) + (col2), v3 = 2v1 + v2.

Since the 4th column in the row-reduced matrix = – 2 (col1) + (col2), v4 = – 2v1 + v2.

1. b) We set up the matrix with the vectors as columns and row reduce:

which reduces to .

The 1st and 3rd columns have leading 1's, so v1 and v3 form the basis.

Since the 2nd column in the row-reduced matrix = 2 (col1), v2 = 2v1 .

Since the 4th column in the row-reduced matrix = (col1) + (col3), v4 = v1 + v3.

2. Since we're just looking for a basis rather than selecting vectors from the set to form a basis, we'll set up the vectors as rows of a matrix and reduce it to reduced row-echelon form. The non-zero rows will form the basis for the row space spanned by these vectors.

which reduces to .

All 3 rows are non-zero, therefore they form a basis for the row space which is a 3-dimensional subspace of R 4.

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