Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[
\mathbf{v}_{1}=
\begin{bmatrix}
1 \\ 2 \\ 2 \\ -1
\end{bmatrix}
,\;\mathbf{v}_{2}=
\begin{bmatrix}
1 \\ 3 \\ 1 \\ 1
\end{bmatrix}
,\;\mathbf{v}_{3}=
\begin{bmatrix}
1 \\ 5 \\ -1 \\ 5
\end{bmatrix}
,\;\mathbf{v}_{4}=
\begin{bmatrix}
1 \\ 1 \\ 4 \\ -1
\end{bmatrix}
,\;\mathbf{v}_{5}=
\begin{bmatrix}
2 \\ 7 \\ 0 \\ 2
\end{bmatrix}
.\] Find a basis for the span $\Span(S)$.

 

We will give two solutions.

Solution 1.

We apply the leading 1 method.
Let $A$ be the matrix whose column vectors are vectors in the set $S$:
\[
A=
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
.\] Applying the elementary row operations to $A$, we obtain
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
\xrightarrow[R_4+R_1]{\substack{R_2-2R_1 \\ R_3-2R_1}}
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
0 & 1 & 3 & -1 & 3 \\
0 & -1 & -3 & 2 & -4 \\
0 & 2 & 6 & 0 & 4
\end{bmatrix}\\[6pt] \xrightarrow[R_4-2R_2]{\substack{R_1-R_2 \\ R_3+R_2}}
\begin{bmatrix}
1 & 0 & -2 & 2 & -1 \\
0 & 1 & 3 & -1 & 3 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 2 & -2
\end{bmatrix}
\xrightarrow[R_4-2R_3]{\substack{R_1-2R_3 \\ R_2+R_3}}
\begin{bmatrix}
1 & 0 & -2 & 0 & 1 \\
0 & 1 & 3 & 0 & 2 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}=\rref(A).
\end{align*}
Observe that the first, second, and fourth column vectors of $\rref(A)$ contain the leading 1 entries.
Hence, the first, second, and fourth column vectors of $A$ form a basis of $\Span(S)$.
Namely,
\[\left\{ \begin{bmatrix}
1 \\
2 \\
2 \\
-1
\end{bmatrix}, \begin{bmatrix}
1 \\
3 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
4 \\
-1
\end{bmatrix}\right \}\] is a basis for $\Span(S)$.

Solution 2.

Let
\[
A=
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
.\] Then $\Span(S)$ is the column space of $A$, which is the row space of $A^{T}$. Using row operations, we have
\[
A^{T}=
\begin{bmatrix}
1 & 2 & 2 & -1 \\
1 & 3 & 1 & 1 \\
1 & 5 & -1 & 5 \\
1 & 1 & 4 & -1 \\
2 & 7 & 0 & 2
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & 2 & -1 \\
0 & 1 & -1 & 2 \\
0 & 3 & -3 & 6 \\
0 & -1 & 2 & 0 \\
0 & 3 & -4 & 4
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 4 & 5 \\
0 & 1 & -1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 0 & -1 & -2
\end{bmatrix}
\] \[
\to
\begin{bmatrix}
1 & 0 & 0 & -13 \\
0 & 1 & 0 & 4 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
.
\] Therefore, the set of nonzero rows
\[
\left\{
\begin{bmatrix}
1 \\ 0 \\ 0 \\ -13
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 1 \\ 0 \\ 4
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 0 \\ 1 \\ 2
\end{bmatrix}
\right\}
\] is a basis for the row space of $A^{T}$, which equals $\Span(S)$.

The post Find a Basis for the Subspace spanned by Five Vectors first appeared on Problems in Mathematics.

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:
\begin{align*}
A=\begin{bmatrix}
2 & 4 & 6 & 8 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_1}
\begin{bmatrix}
1 & 2 & 3 & 4 \\
1 &3 & 0 & 5 \\
1 & 1 & 6 & 3
\end{bmatrix}\\[6pt] \xrightarrow[R_3-R_1]{R_2-R_1}
\begin{bmatrix}
1 & 2 & 3 & 4 \\
0 &1 & -3 & 1 \\
0 & -1 & 3 & -1
\end{bmatrix}
\xrightarrow[R_3+R_2]{R_1-2R_2}
\begin{bmatrix}
1 & 0 & 9 & 2 \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form. That is,
\[\rref(A)=\begin{bmatrix}
1 & 0 & 9 & 2 \\
0 &1 & -3 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}. \tag{*}\]

(a) Find a basis for the nullspace of $A$.

By the computation above, we see that the general solution of $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
x_1&=-9x_3-2x_4\\
x_2&=3x_3-x_4,
\end{align*}
where $x_3$ and $x_4$ are free variables.
Thus, the vector form solution to $A\mathbf{x}=\mathbf{0}$ is
\begin{align*}
\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=\begin{bmatrix}
-9x_3-2x_4 \\
3x_3-x_4 \\
x_3 \\
x_4
\end{bmatrix}
=x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}.
\end{align*}
It follows that the nullspace of the matrix $A$ is given by
\begin{align*}
\calN(A)&=\left\{ \mathbf{x}\in \R^4 \quad \middle | \quad \mathbf{x}= x_3\begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix}, \text{ for all } x_3, x_4 \in \R^4 \right \}\\[6pt] &= \Span \left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}.
\end{align*}
Thus, the set
\[\left\{ \begin{bmatrix}
-9 \\
3 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
-1 \\
0 \\
1
\end{bmatrix} \right \}\] 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 $\calN(A)$.

(b) Find a basis for the row space of $A$.

Recall that the nonzero rows of $\rref(A)$ form a basis for the row space of $A$. (The row space method.)

Thus,
\[\left\{\begin{bmatrix}
1 \\
0 \\
9 \\
2
\end{bmatrix}, \quad \begin{bmatrix}
0 \\
1 \\
-3 \\
1
\end{bmatrix} \right \}\] is a basis for the row space of $A$.

(c) Find a basis for the range of $A$ that consists of column vectors 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,
\[\left\{\begin{bmatrix}
2 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
4 \\
3 \\
1
\end{bmatrix}\right \}\] is a basis for the range $\calR(A)$ 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 $A$.

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
\[A_4=2A_1+A_2.\] Click here if solved 94

The post Find a Basis for Nullspace, Row Space, and Range of a Matrix first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
1 & -1 & 0 & 0 \\
0 &1 & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 2 & 2 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}.\]

(a) Find a basis for the null space $\calN(A)$.

(b) Find a basis of the range $\calR(A)$.

(c) Find a basis of the row space for $A$.

(The Ohio State University, Linear Algebra Midterm)
 

Solution.

(a) Find a basis for the null space $\calN(A)$.

To find a basis for the null space $\calN(A)$, we first find an algebraic description of $\calN(A)$.
Recall that $\calN(A)$ consists of the solutions of the homogeneous system $A\mathbf{x}=\mathbf{0}$. Let us find the solution of this system by applying elementary row operations to the augmented matrix $[A\mid \mathbf{0}]$ as follows.
\[ \left[\begin{array}{rrrr|r}
1 & -1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 \\
1 & -1 & 0 & 0 & 0 \\
0 & 2 & 2 & 2 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array} \right] \xrightarrow[R_4-2R_2]{R_3-R_1}
\left[\begin{array}{rrrr|r}
1 & -1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array} \right] \xrightarrow{R_1+R_2}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 1 & 0 \\
0 & 1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array} \right].\] Hence the general solution satisfies
\begin{align*}
x_1&=-x_3-x_4\\
x_2&=-x_3-x_4
\end{align*}
and hence the elements in $\calN(A)$ are of the form
\[\mathbf{x}=\begin{bmatrix}
x_1\\ x_2\\ x_3 \\x_4
\end{bmatrix}=\begin{bmatrix}
-x_3-x_4\\ -x_3-x_4\\ x_3 \\x_4
\end{bmatrix}
=x_3 \begin{bmatrix}
-1\\ -1\\ 1 \\0
\end{bmatrix}
+x_4 \begin{bmatrix}
-1\\ -1\\ 0 \\1
\end{bmatrix}
.\]

It follows that the set
\[\left\{ \,\begin{bmatrix}
-1\\ -1\\ 1 \\0
\end{bmatrix}, \,
\begin{bmatrix}
-1\\ -1\\ 0 \\1
\end{bmatrix} \, \right\}.\] is a spanning set for $\calN(A)$.
It is straightforward to see that these vectors are linearly independent.
Hence it is a basis for $\calN(A)$.

(b) Find a basis of the range $\calR(A)$.

Note that we have already computed the reduced row echelon form matrix for $A$ in part (a) (just ignore the augmented part).
The first two columns contain leading 1’s. Hence by the leading 1 method, we see that
\[\left\{ \,\begin{bmatrix}
1\\ 0\\ 1 \\0 \\0
\end{bmatrix}, \,
\begin{bmatrix}
-1\\ 1\\ -1 \\2\\0
\end{bmatrix} \, \right\}\] is a basis for the range $\calR(A)$.

(c) Find a basis of the row space of $A$.

Again by the reduced row echelon form matrix in $A$, we see that the set of the nonzero rows
\[\left\{ \,\begin{bmatrix}
1\\ 0\\ 1 \\1
\end{bmatrix}, \,
\begin{bmatrix}
0\\ 1\\ 1 \\1
\end{bmatrix} \, \right\}\] is a basis for the row space of $A$ by the row space method.

Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017

1. Vector Space of 2 by 2 Traceless Matrices
2. Find an Orthonormal Basis of the Given Two Dimensional Vector Space
3. Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?
4. Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix ←The current problem
5. Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$
6. Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors
7. Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less

The post Find Bases for the Null Space, Range, and the Row Space of a \times 4$ Matrix first appeared on Problems in Mathematics.

Let \[A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}.\]

(a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$.

(b) Find a basis for the null space of $A$.

(c) Find a basis for the range of $A$ that consists of columns of $A$. For each columns, $A_j$ of $A$ that does not appear in the basis, express $A_j$ as a linear combination of the basis vectors.

(d) Exhibit a basis for the row space of $A$.

Hint.

In part (c), you may use the following theorem.

In part (d), you may use the following theorem.

Solution.

(a) Find a matrix $B$ in reduced roe echelon form such that $B$ is row equivalent to the matrix $A$.

We apply the elementary row operations to the matrix $A$ and obtain
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}
\xrightarrow{\substack{R_2-2R_1\\ R_3-2R_1}}
\begin{bmatrix}
1 & 1 & 2 \\
0 &0 &0 \\
0 & 1 & 1
\end{bmatrix}
\xrightarrow{R_2 \leftrightarrow R_3}\\
\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form that is row equivalent to $A$.
Thus we set
\[B=\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1\\
0 &0 &0
\end{bmatrix}.\]

(b) Find a basis for the null space of $A$.

The null space $\calN(A)$ of the matrix is the set of solutions of the homogeneous system $A\mathbf{x}=\mathbf{0}$.
By part (a), the augmented matrix $[A\mid \mathbf{0}]$ is row equivalent to $[B \mid \mathbf{0}]$.
Thus, the solution $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ must satisfy
\[x_1=-x_3 \text{ and } x_2=-x_3,\] where $x_3$ is a free variable.
Thus the solutions are given by
\[\mathbf{x}=\begin{bmatrix}
-x_3 \\
-x_3 \\
x_3
\end{bmatrix}=x_3\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\] for any number $x_3$.
Hence we have
\begin{align*}
\calN(A)&=\{ \mathbf{x}\in \R^3 \mid \mathbf{x}=x_3\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix} \text{ for any } x_3\in \R \} \\
&=\Span\left\{ \quad\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\quad \right \}.
\end{align*}

From this, we deduce that the set
\[\left\{ \quad\begin{bmatrix}
-1 \\
-1 \\
1
\end{bmatrix}\quad \right \}\] is a basis for the null space of $A$.

(c) Find a basis for the range of $A$ that consists of columns of $A$. (Longer version)

Let us write
\[A=\begin{bmatrix}
1 & 1 & 2 \\
2 &2 &4 \\
2 & 3 & 5
\end{bmatrix}=[A_1, A_2, A_3],\] where $A_1, A_2, A_3$ are the column vectors of the matrix $A$.
The range $\calR(A)$ of $A$ is the same as the column space of $A$.
Thus, it suffices to find the maximal number of linearly independent vectors among the column vectors $A_1, A_2, A_3$.
Consider the linear combination
\[x_1A_1+x_2A_2+x_3A_3=\mathbf{0}. \tag{*} \] Then this is equivalent to the homogeneous system
\[A\mathbf{x}=\mathbf{0}\] and we already found the solutions in part (b):
\[x_1=-x_3 \text{ and } x_2=-x_3.\] This tells us the vectors $A_1, A_2, A_3$ are linearly dependent.
For example, $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of the system (*).
Thus, we have
\[\calR(A)=\Span(A_1, A_2, A_3)=\Span(A_1, A_2).\]

On the other hand, if we consider only $A_1$ and $A_2$, they are linearly independent.
(To see this, you just need to repeat the above argument without $A_3$. This amounts to just ignoring the third columns in the computations.)

Therefore, the set $\{A_1, A_2\}$ is a linearly independent spanning set for the range.
Hence a basis of the range consisting column vectors of $A$ is
\[\{A_1, A_2\}.\]

The only column vector which is not a basis vector is $A_3$.
We already found that $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of (*).
Thus we have
\[-A_1-A_2+A_3=\mathbf{0}.\] Solving this for $A_3$, we obtain the linear combination for $A_3$ of the basis vectors:
\[A_3=A_1+A_2.\]

(c) A shorter solution using the leading 1 method

Here is a shorter solution which uses the following theorem.

Theorem (leading-1 method): If $B$ is a matrix in reduced row echelon form that is row equivalent to $A$, then all the column vectors of $A$ whose corresponding columns in $B$ have leading 1’s form a basis of the range of $A$.

Looking at the matrix $B$, we see that the first and the second columns of $B$ have leading 1’s. Thus the first and the second column vectors of $A$ form a basis for the range of $A$.

(d) Exhibit a basis for the row space of $A$.

We use the following theorem.

Theorem (Row-space method): If $B$ is a matrix in (reduced) row echelon form that is row equivalent to $A$, then the nonzero rows of $B$ form a basis for the row space of $A$.

We have already found such $B$ in part (a), and the first and the second row vectors are nonzero. Thus they form a basis for the row space of $A$.
Hence a basis for the row space of $A$ is
\[\left\{\,\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix} \, \right \}.\]

Comment.

The longer solution of part (c) is essentially the proof of Theorem (leading 1 method).
It is good to know where the theorem came from, but when you solve a problem you may forget and just use the theorem like the shorter solution of (c).

The post Row Equivalent Matrix, Bases for the Null Space, Range, and Row Space of a Matrix first appeared on Problems in Mathematics.