Let $A$ be an $m \times n$ matrix.
Suppose that the nullspace of $A$ is a plane in $\R^3$ and the range is spanned by a nonzero vector $\mathbf{v}$ in $\R^5$. Determine $m$ and $n$. Also, find the rank and nullity of $A$.

Solution.

For an $m \times n$ matrix $A$, the nullspace consists of vectors $\mathbf{x}$ such that $A\mathbf{x}=\mathbf{0}$. Thus, such an $\mathbf{x}$ must be $n$-dimensional. Since the nullspace is a subspace in $\R^3$, we conclude that $n=3$.

The range of $A$ consists of vectors $\mathbf{y}$ such that $\mathbf{y}=A\mathbf{x}$ for some $\mathbf{x}\in \R^n$. Hence, $\mathbf{y}$ is $m$-dimensional. As the range is a subspace in $\R^5$, we conclude that $m=5$.

Since a plane is a $2$-dimensional subspace, the nullity of $A$ is $2$.

The range is spanned by a single vector $\mathbf{v}$. Thus, $\{\mathbf{v}\}$ is a basis for the range. Thus, the rank is $1$.

Here is another way to see this. By the rank-nullity theorem, we have
\[\text{rank of $A$ + nullity of $A = n$}.\] Since $n=3$ and the nullity is $2$, the rank is $1$.

The post In which $\R^k$, are the Nullspace and Range Subspaces? 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 $A$.
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 147

The post How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix 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.

Using the definition of the range of a matrix, describe the range of the matrix
\[A=\begin{bmatrix}
2 & 4 & 1 & -5 \\
1 &2 & 1 & -2 \\
1 & 2 & 0 & -3
\end{bmatrix}.\]

Solution.

By definition, the range $\calR(A)$ of the matrix $A$ is given by
\[\calR(A)=\left \{ \mathbf{b} \in \R^3 \quad \middle | \quad A\mathbf{x}=\mathbf{b} \text{ for some } \mathbf{x} \in \R^4 \right \}.\]

Thus, a vector $\mathbf{b}=\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}$ in $\R^3$ is in the range $\calR(A)$ if and only if the system $A\mathbf{x}=\mathbf{b}$ is consistent.
So, let us find the conditions on $\mathbf{b}$ so that the system is consistent.

To do this, we consider the augmented matrix of the system and reduce it as follows.
\begin{align*}
\left[\begin{array}{rrrr|r}
2 & 4 & 1 & -5 & b_1\\
1 &2 & 1 & -2 & b_2 \\
1 & 2 & 0 & -3 &b_3
\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
2 & 4 & 1 & -5 & b_1\\
1 & 2 & 0 & -3 &b_3
\end{array}\right] \xrightarrow[R_3-R_1]{R_2-2R_1}\\[6pt] \left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
0 & 0 & -1 & -1 & b_1-2b_2\\
0 & 0 & 0 & -1 & b_3-b_2
\end{array}\right] \xrightarrow{-R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
0 & 0 & 1 & 1 & -b_1+2b_2\\
0 & 0 & -1 & -1 & b_3-b_2
\end{array}\right]\\[6pt] \xrightarrow[R_3+R_2]{R_1-R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 0 & -3 & b_1-b_2 \\
0 & 0 & 1 & 1 & -b_1+2b_2\\
0 & 0 & 0 & 0 & -b_1 +b_2 +b_3
\end{array}\right].
\end{align*}

Note that if the $(3, 5)$-entry $-b_1+b_2+b_3$ is not zero, then the system $A\mathbf{x}=\mathbf{0}$ is inconsistent because this implies $0=1$.
On the other hand, if $-b_1+b_2+b_3=0$, then we see that the system is consistent.
Hence, the vector $\mathbf{b}$ is in the range $\calR(A)$ if and only if $-b_1+b_2+b_3=0$.

In summary, we have
\[\calR(A)=\left\{ \begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}\in \R^3 \quad \middle | \quad -b_1+b_2+b_3=0 \right \}.\]

Spanning set for the range

With a little bit additional computation, we can find the spanning set for the range as follows.

Thus, $\mathbf{b} \in \calR(A)$ if and only if
\begin{align*}
\mathbf{b}=\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}=\begin{bmatrix}
b_2+b_3 \\
b_2 \\
b_3
\end{bmatrix}=b_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+b_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}.
\end{align*}

In summary, we have
\begin{align*}
\calR(A)&=\left\{ \mathbf{b} \in \R^3 \quad \middle | \quad \mathbf{b}=b_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+b_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}\\[6pt] &=\Span\left\{ \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}.
\end{align*}
Hence, the spanning set is
\[ \left\{ \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}.\]

This spanning set is linearly independent, hence it’s a basis for the range.

Thus, the dimension of the range is $2$.

The post Describe the Range of the Matrix Using the Definition of the Range 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.

(a) Let $A=\begin{bmatrix}
1 & 3 & 0 & 0 \\
1 &3 & 1 & 2 \\
1 & 3 & 1 & 2
\end{bmatrix}$.
Find a basis for the range $\calR(A)$ of $A$ that consists of columns of $A$.

(b) Find the rank and nullity of the matrix $A$ in part (a).

Solution.

(a) Find a basis for the range $\calR(A)$

Note that the range $\calR(A)$ is the column space of $A$. Thus,
\[\calR(A)=\Span \{A_1, A_2, A_3, A_4\},\] where $A_i$ is the $i$-th column vector of $A$.

We use the “leading 1” method.
We reduce the matrix $A$ by elementary row operations as follows.
We have
\begin{align*}
A\xrightarrow{R_3-R_2} \begin{bmatrix}
1 & 3 & 0 & 0 \\
1 &3 & 1 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix} \xrightarrow{R_2-R_1}
\begin{bmatrix}
1 & 3 & 0 & 0 \\
0 &0 & 1 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}

The first and the third column contain the leading 1’s.
Thus, the set $\{A_1, A_3\}$ is a basis of the range $\calR(A)$, which consists of the first and the third column vectors of $A$.

(b) Find the rank and nullity of the matrix $A$

By part (a), we know that $\{A_1, A_3\}$ is a basis of the range of $A$.
Hence the dimension of the range is $2$.
Thus the rank of $A$, which is the dimension of the range $\calR(A)$, is $2$

Recall the rank-nullity theorem. Since $A$ is a $3\times 4$ matrix, we have
\[\text{rank of } A + \text{nullity of } A=4.\] Since we know that the rank of $A$ is $2$, it follows from the rank-nullity theorem that the nullity of $A$ is $2$.

Comment.

These are Quiz 7 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 7. Find a Basis of the Range, Rank, and Nullity of a 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.