Linearly Independent Sets and Basis
Gram-Schmidt Process

Rank and Nullity

Example 1.

 

We want to find a basis for the row space of

(1)   \begin{equation*}A =\left (\begin{array}{cccc}4 & 0 & 2 & 3 \\ 2 & -1 & 2 & 0 \\ 5 & 2 & 2 & 1\end{array}\right ) \end{equation*}

We know the row space of A is the subspace of \mathbb{R}^{4} spanned by the row vectors of A. We also know that row-equivalent matrices have the same row space. Of course, this leads us to the fact that when we reduce A to row-echelon form, say matrix B, the non-zero row vectors of B form a basis for the row space of A . Therefore,

(2)   \begin{equation*}\left (\begin{array}{cccc}4 & 0 & 2 & 3 \\ 2 & -1 & 2 & 0 \\ 5 & 2 & 2 & 1\end{array}\right ) \rightarrow \left (\begin{array}{cccc}4 & 0 & 2 & 3 \\ 0 & -1 & 1 & -\frac{3}{2} \\ 0 & 0 & \frac{3}{2} & -\frac{23}{4}\end{array}\right ) \end{equation*}

implies the set \{\left (4 ,0 ,2 ,3\right ) ,\left (0 ,1 , -1 , -\frac{3}{2}\right ) ,\left (0 ,0 ,\frac{3}{2} , -\frac{23}{4}\right )\} is a basis for the row space of A . Note that this basis has 3 elements, the rank of A is equal to 3.

 

Example 2.

 

Suppose we are given three vectors \{\left ( -2 , -4 ,4\right ) ,\left (3 ,6 , -6\right ) ,\left ( -1 , -2 ,2\right )\}. We are interested to find a basis for the subspace of \mathbb{R}^{3} which is spanned by this set. Use the members of the given set to define a matrix as followed

(3)   \begin{equation*}A =\left (\begin{array}{ccc} -2 & -4 & 4 \\ 3 & 6 & -6 \\ -1 & -2 & 2\end{array}\right ) \end{equation*}

 

Now reducing A and collecting the nonzero rows will result in a basis for the row space of A (which is a subspace of \mathbb{R}^{3}). We write

(4)   \begin{equation*}\left (\begin{array}{ccc} -2 & -4 & 4 \\ 3 & 6 & -6 \\ -1 & -2 & 2\end{array}\right ) \rightarrow \left (\begin{array}{ccc} -2 & -4 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right ) \end{equation*}

 

The set with one member \{\left ( -2 , -4 ,4\right )\} is a basis for the row space of A and so it is a basis for the subspace of \mathbb{R}^{3} which is spanned by the original given set. Please note we also know that rank of A is equal to 1 since the row space of A has dimension 1.

 

Example 3.

 

Suppose we are given the matrix

(5)   \begin{equation*}\left (\begin{array}{cc}2 & -1 \\ -8 & 4\end{array}\right ) \end{equation*}

We know the set of all solutions of the homogeneous system of linear equations

(6)   \begin{equation*}\left (\begin{array}{cc}2 & -1 \\ -8 & 4\end{array}\right )\left (\begin{array}{c}x \\ y\end{array}\right ) =\left (\begin{array}{c}0 \\ 0\end{array}\right ) \end{equation*}

is a subspace of \mathbb{R}^{2} . In fact, this set os solutions is called the nullspace of A . Reducing the matrix A results in

(7)   \begin{equation*}\left (\begin{array}{cc}2 & -1 \\ -8 & 4\end{array}\right ) \rightarrow \left (\begin{array}{cc}2 & -1 \\ 0 & 0\end{array}\right ) \end{equation*}

and therefore we get

(8)   \begin{equation*}2x -y =0 \end{equation*}

Set y =t and get

(9)   \begin{equation*}x =\frac{t}{2} \end{equation*}

It is clear that any solution space of the given homogeneous system looks like

(10)   \begin{equation*}\left (\begin{array}{c}\frac{t}{2} \\ t\end{array}\right ) =t\left (\begin{array}{c}\frac{1}{2} \\ 1\end{array}\right ) \end{equation*}

for any real t . The vector \left (\begin{array}{c}\frac{1}{2} \\ 1\end{array}\right ) also provides a basis for the nullspace of A . Note that dimension of the nullspace, that is nullity, is equal to 1.

Leave a Reply