Find a Basis for Nullspace, Row Space, and Range of a Matrix
Let $A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}$.
(a) Find a basis for the nullspace of $A$.
(b) Find a basis for the row space of $A$.
(c) Find a basis for the range of $A$ that consists of column vectors of $A$.
(d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.
Solution.
We first obtain the reduced row echelon form matrix corresponding to the matrix $A$.
We reduce the matrix $A$ as follows:
A=\begin{bmatrix}
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
\end{bmatrix}\\[6pt] \xrightarrow[R_3-R_1]{R_2-R_1}
0 &1 & -3 & 1 \\
0 & -1 & 3 & -1
\xrightarrow[R_3+R_2]{R_1-2R_2}
\end{bmatrix}.
The last matrix is in reduced row echelon form. That is,
\[\rref(A)=\begin{bmatrix}
\end{bmatrix}. \tag{*}\]
By the computation above, we see that the general solution of $A\mathbf{x}=\mathbf{0}$ is
x_1&=-9x_3-2x_4\\
x_2&=3x_3-x_4,
where $x_3$ and $x_4$ are free variables.
Thus, the vector form solution to $A\mathbf{x}=\mathbf{0}$ is
\mathbf{x}=\begin{bmatrix}
x_1 \\
=\begin{bmatrix}
-9x_3-2x_4 \\
3x_3-x_4 \\
=x_3\begin{bmatrix}
-9 \\
3 \\
\end{bmatrix}+x_4\begin{bmatrix}
It follows that the nullspace of the matrix $A$ is given by
\calN(A)&=\left\{ \mathbf{x}\in \R^4 \quad \middle | \quad \mathbf{x}= x_3\begin{bmatrix}
\end{bmatrix}, \text{ for all } x_3, x_4 \in \R^4 \right \}\\[6pt] &= \Span \left\{ \begin{bmatrix}
\end{bmatrix}, \begin{bmatrix}
\end{bmatrix} \right \}.
Thus, the set
\[\left\{ \begin{bmatrix}
\end{bmatrix} \right \}\] is a spanning set for the nullspace $\calN(A)$.
It is straightforward to see that this set is linearly independent, and hence it is a basis for $\calN(A)$.
Recall that the nonzero rows of $\rref(A)$ form a basis for the row space of $A$. (The row space method.)
Thus,
\[\left\{\begin{bmatrix}
\end{bmatrix}, \quad \begin{bmatrix}
\end{bmatrix} \right \}\] is a basis for the row space of $A$.
Recall that by the leading 1 method, the columns of $A$ corresponding to columns of $\rref(A)$ that contain leading 1 entries form a basis for the range $\calR(A)$ of $A$.
From (*), we see that the first and the second columns contain the leading 1 entries. Thus,
\end{bmatrix}\right \}\] is a basis for the range $\calR(A)$ of $A$.
Let us write $A_1, A_2, A_3$, and $A_4$ for the column vectors of the matrix $A$.
In part (c), we showed that $\{A_1, A_2\}$ is a basis for the range $\calR(A)$.
Thus, we need to express the vectors $A_3$ and $A_4$ as a linear combination of $A_1$ and $A_2$, respectively.
A shortcut is to note that the entries of third column vector of $\rref(A)$ give the coefficients of the linear combination for $A_3$. That is, we have
\[A_3=9A_1-3A_2.\] Similarly, the entries of the fourth column of $\rref(A)$ yield
\[A_4=2A_1+A_2.\]
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Tags: basis for a vector spaceleading 1 methodlinear algebralinear combinationnull spacerange of a matrixrow spacerow space methodvector space
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