Range and Null Space of a Matrix - Linear Algebra

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Find a basis for the range space of the transformation given by the matrix $A=\begin{bmatrix} 1& 2& 3 \ 0&0&2\ 0& 0& -1\end{bmatrix}$ .

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We can find a basis for 's range space first by finding a basis for the column space of its reduced row echelon form.

Using a calculator or row reduction, we obtain

$R=\begin{bmatrix} 1& 2& 0 \ 0& 0&1\ 0&0&0 \end{bmatrix}$

for the reduced row echelon form.

The second column in this matrix can be seen by inspection to be a linear combination of the first column, so it is not included in our basis for . Hence the first and the third columns form a basis for the column space of , and therefore the first and the third columns of form a basis for the range space of .

$\begin{Bmatrix} \begin{bmatrix} 1\ 0\ 0\ \end{bmatrix} _, \begin{bmatrix} 3\ 2\ -1\ \end{bmatrix}\end{Bmatrix}$

Calculate the Null Space of the following Matrix.

$A=\left[\begin{matrix} 1 & 1& 1& -1 \ 2 & 4&5 &6 \ 3& 9& 5& 4 \end{matrix}\right]$

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The first step is to create an augmented matrix having a column of zeros.

$\left[\begin{matrix} 1 & 1& 1& -1 \ 2 & 4&5 &6 \ 3& 9& 5& 4 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

The next step is to get this into RREF.

$2R_1-R_2\rightarrow R_2$

$\left[\begin{matrix} 1 & 1& 1& -1 \ 0 & -2&-3 &-8 \ 3& 9& 5& 4 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$3R_1-R_3\rightarrow R_3$

$\left[\begin{matrix} 1 & 1& 1& -1 \ 0 & -2&-3 &-8 \ 0& -6& -2& -7 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$-3R_2+R_3\rightarrow R_3$

$\left[\begin{matrix} 1 & 1& 1& -1 \ 0 & -2&-3 &-8 \ 0& 0& 7& 17 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$$\frac{1}{2}$R_2+R_1\rightarrow R_1$

$\left[\begin{matrix} 1 & 0& -$\frac{1}{2}$& -5 \ 0 & -2&-3 &-8 \ 0& 0& 7& 17 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

We can simplify to

$\left[\begin{matrix} 1 & 0& -$\frac{1}{2}$& -5 \ 0 & 1&$\frac{3}{2}$ &4 \ 0& 0& 1& $\frac{17}{7}$ \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$-$\frac{3}{2}$R_3+R_2\rightarrow R_2$

$\left[\begin{matrix} 1 & 0& -$\frac{1}{2}$& -5 \ 0 & 1&0 & \ $\frac{5}{14}$ \ 0& 0& 1& $\frac{17}{7}$ \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$$\frac{1}{2}$R_3+R_1\rightarrow R_1$

$\left[\begin{matrix} 1 & 0& 0& -$\frac{53}{14}$ \ 0 & 1&0 & \ $\frac{5}{14}$ \ 0& 0& 1& $\frac{17}{7}$ \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

This tells us the following.

$\x_1-$\frac{53}{14}$x_4=0 \ \ x_2+\frac{5}{14}$x_4=0 \ \ x_3+\frac{17}{7}$x_4=0$

$\x_1=\frac{53}{14}$x_4 \ \ x_2=-$\frac{5}{14}$x_4 \ \ x_3=-$\frac{17}{7}$x_4$

Now we need to write this as a linear combination.

$\begin{bmatrix} x_1\ x_2\ x_3\ x_4 \end{bmatrix} =x_4\begin{bmatrix} $\frac{53}{14}$\ \ -$\frac{5}{14}$\ \ -$\frac{17}{7}$ \ 1 \end{bmatrix}$

The null space is then

$N(A)=Span\begin{bmatrix} $\frac{53}{14}$\ \ -$\frac{5}{14}$\ \ -$\frac{17}{7}$ \ 1 \end{bmatrix}$

Find the null space of the matrix operator $T_M = \begin{bmatrix} 1 &1 & 0\ 0& -1& 1\ 9& 0& 1\end{bmatrix}$ .

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The null space of the operator is the set of solutions to the equation

$\begin{bmatrix} 1 &1 & 0\ 0& -1& 1\ 9& 0& 1\end{bmatrix} \begin{bmatrix} x_1\x_2\x_3 \end{bmatrix} = \begin{bmatrix} 0\0 \0 \end{bmatrix}$ .

We can solve the above system by row reducing our $3 \times 3$ matrix using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0 & 0\0 & 1& 0\ 0& 0&1 \end{bmatrix}\begin{bmatrix} x_1\x_2 \x_3 \end{bmatrix}= \begin{bmatrix} 0\0 \0 \end{bmatrix} \Longrightarrow \begin{align} x_1 &=0 \x_2 &=0 \ x_3&=0 \end{align}$

Hence the null space consists of only the zero vector. $\begin{Bmatrix} \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \end{Bmatrix}$

Find the null space of the matrix $A = \begin{bmatrix} 2 & 0& 0\1 &0 &0 \9 &8 &2 \end{bmatrix}$ .

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The null space of the matrix is the set of solutions to the equation

$\begin{bmatrix} 2 &0 & 0\ 1& 0& 0\ 9& 8& 2\end{bmatrix} \begin{bmatrix} x_1\x_2\x_3 \end{bmatrix} = \begin{bmatrix} 0\0 \0 \end{bmatrix}$ .

We can solve the above system by row reducing using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0 & 0\0 & 1& 1/4\ 0& 0&0 \end{bmatrix}\begin{bmatrix} x_1\x_2 \x_3 \end{bmatrix}= \begin{bmatrix} 0\0 \0 \end{bmatrix} \Longrightarrow \begin{align} x_1 &=0 \x_2 &=-(1/4) x_3 \ x_3&=x_3 \end{align} \Longrightarrow$

$X= \begin{bmatrix} 0\-1/4 \1 \end{bmatrix}t$

Multiplying this vector by gets rid of the fraction, and does not affect our answer, since there is an arbitrary constant behind it.

Hence the null space consists of all vectors spanned by $\begin{bmatrix} 0\-1 \4 \end{bmatrix}$ ;

$Span\begin{Bmatrix} \begin{bmatrix} 0\ -1\4 \end{bmatrix} \end{Bmatrix}$ .

Find the null space of the matrix $A = \begin{bmatrix} 2 & 0\2 &0 \end{bmatrix}$ .

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The null space of the matrix is the set of solutions to the equation

$\begin{bmatrix} 2 &0\ 2& 0\ \end{bmatrix} \begin{bmatrix} x_1\x_2 \end{bmatrix} = \begin{bmatrix} 0\0 \end{bmatrix}$ .

We can solve the above system by row reducing using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0 \0 &0\end{bmatrix}\begin{bmatrix} x_1\x_2 \end{bmatrix}= \begin{bmatrix} 0\0 \end{bmatrix} \Longrightarrow \begin{align} x_1 &=0 \x_2 &=x_2 \end{align} \Longrightarrow$

$X= \begin{bmatrix} 0\1 \end{bmatrix}t$

Hence the null space consists of all vectors spanned by $\begin{bmatrix} 0\1 \end{bmatrix}$ ;

$Span\begin{Bmatrix} \begin{bmatrix} 0\ 1\\end{bmatrix} \end{Bmatrix}$ .

Find a basis for the range space of the transformation given by the matrix $A=\begin{bmatrix} 1& 2& 3&4 \ -1&-3 &-2 &-5 \ 0& 1& 1& 3 \end{bmatrix}$ .

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We can find a basis for 's range space first by finding a basis for the column space of its reduced row echelon form.

Using a calculator or row reduction, we obtain

$\begin{bmatrix} 1& 0& 0&-3 \ 0& 1&0 & 2\ 0&0 &1 & 1 \end{bmatrix}$

for the reduced row echelon form.

The fourth column in this matrix can be seen by inspection to be a linear combination of the other three columns, so it is not included in our basis. Hence the first three columns form a basis for the column space of the reduced row echelon form of , and therefore the first three columns of form a basis for its range space.

$\begin{Bmatrix} \begin{bmatrix} 1\ -1\ 0\ \end{bmatrix} _, \begin{bmatrix} 2\ -3\ 1\ \end{bmatrix} _,\begin{bmatrix} 3\ -2\ 1\ \end{bmatrix} \end{Bmatrix}$ .

Find a basis for the range space of the transformation given by the matrix $A=\begin{bmatrix} 0& 1& 1&0\ 0&0&1&1\ 1&0&0&1\1&1&0&0\end{bmatrix}$ .

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We can find a basis for 's range space first by finding a basis for the column space of its reduced row echelon form.

Using a calculator or row reduction, we obtain

$R=\begin{bmatrix} 1& 0& 0&1\ 0& 1&0&-1\ 0&0&1&1\0&0&0&0 \end{bmatrix}$

for the reduced row echelon form.

The fourth column in this matrix can be seen by inspection to be a linear combination of the first three columns, so it is not included in our basis for . Hence the first three columns form a basis for the column space of , and therefore the first three columns of form a basis for the range space of .

$\begin{Bmatrix} \begin{bmatrix} 0\ 0\ 1\ 1\ \end{bmatrix}_,\begin{bmatrix} 1\ 0\ 0\ 1\ \end{bmatrix}_, \begin{bmatrix} 1\ 1\ 0\ 0\ \end{bmatrix} \end{Bmatrix}$

Find a basis for the null space of the matrix $A= \begin{bmatrix} 5& 2\ 1& 0 \end{bmatrix}$ .

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The null space of the matrix is the set of solutions to the equation

$\begin{bmatrix} 5 &2\ 1& 0 \end{bmatrix} \begin{bmatrix} x_1\x_2\end{bmatrix} = \begin{bmatrix} 0\0 \end{bmatrix}$ .

We can solve the above system by row reducing using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0\0 & 1 \end{bmatrix}\begin{bmatrix} x_1\x_2 \end{bmatrix}= \begin{bmatrix} 0\0 \end{bmatrix} \Longrightarrow \begin{align} x_1 &=0 \x_2 &=0 \end{align} \Longrightarrow$

$X= \begin{bmatrix} 0\0 \end{bmatrix}$

Hence a basis for the null space is just the zero vector;

$\begin{Bmatrix} \begin{bmatrix} 0\ 0\ \end{bmatrix} \end{Bmatrix}$ .

What is the largest possible rank of a $3 \times 5$ matrix?

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The rank is equal to the dimension of the row space and the column space (both spaces always have the same dimension). This matrix has three rows and five columns, which means the largest possible number of vectors in a basis for the row space of a $3 \times 5$ matrix is , so this is the largest possible rank.

What is the smallest possible nullity of a $3 \times 5$ matrix?

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According to the Rank + Nullity Theorem,

Since the matrix has columns, we can rearrange the equation to get

So to make the nullity as small as possible, we need to make the rank as large as possible.

The rank is equal to the dimension of the row space and the column space (both spaces always have the same dimension). This matrix has three rows and five columns, which means the largest possible number of vectors in a basis for the row space of a $3 \times 5$ matrix is , so this is the largest possible rank.

Hence the smallest possible nullity is .

What is the largest possible rank of a $7 \times 2$ matrix?

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The rank is equal to the dimension of the row space and the column space (both spaces always have the same dimension). This matrix has seven rows and two columns, which means the largest possible number of vectors in a basis for the column space of a $7 \times 2$ matrix is , so this is the largest possible rank.

What is the smallest possible nullity of a $7 \times 2$ matrix?

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According to the Rank + Nullity Theorem,

Since the matrix has columns, we can rearrange the equation to get

So to make the nullity as small as possible, we need to make the rank as large as possible.

The rank is equal to the dimension of the row space and the column space (both spaces always have the same dimension). This matrix has three rows and five columns, which means the largest possible number of vectors in a basis for the row space of a $7 \times 2$ matrix is , so this is the largest possible rank.

Hence the smallest possible nullity is .

A matrix with five rows and four columns has rank 3.

What is the nullity of ?

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The sum of the rank and the nullity of any matrix is always equal to to the number of columns in the matrix. Therefore, a matrix with four columns and rank 3, such as , must have as its nullity .

$C(-\infty , \infty)$ , the set of all continuous functions defined on $(-\infty, \infty)$ , is a vector space under the usual rules of addition and scalar multiplication.

True or false: The set of all functions of the form

,

where is a real number, is a subspace of $C(-\infty , \infty)$ .

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A subset of a vector space can be proved to not be a subspace of the space by showing that the zero of the space is not in .

Let be the subset in question, and let , the zero function, which is in $C(-\infty, \infty)$ . This cannot be expressed as for any . It if could then

, in which case ,

and

, in which case .

By contradiction, $0 \notin S$ . is not a subspace of $C(-\infty, \infty)$ .

$C(-\infty , \infty)$ , the set of all continuous functions defined on $(-\infty, \infty)$ , is a vector space under the usual rules of addition and scalar multiplication.

True or false: The set of all functions defined on $(-\infty, \infty)$ with inverses is a subspace of $C(-\infty , \infty)$ .

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Let be the set of all functions on $(-\infty, \infty)$ such that is defined.

A sufficient condition for to not be a subspace of $C (-\infty, \infty)$ is that the zero of the set - which here is the zero function - is not in . $f \notin S$ because the zero function - a constant function - does not have an inverse ( , violating a condition of an invertible function). It follows that is not a subspace of $C (-\infty, \infty)$ .

If is an $m \times n$ matrix, find

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Since a basis for the row space and the column space of a matrix have the same, number of vectors then their dimensions are the same, say .

By the rank-nullity theorem, we have , or same to say

.

.

Hence .

Finally, applying the rank-nullity theorem to the transpose of , we have

, or the same to say

.

(The row space dimension of is the same as its transpose.)

.

Adding all four of our findings together gives us

.

Find a basis for the range space of the transformation given by the matrix $A=\begin{bmatrix} 1& 2& 3 \ 0&0&2\ 0& 0& -1\end{bmatrix}$ .

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We can find a basis for 's range space first by finding a basis for the column space of its reduced row echelon form.

Using a calculator or row reduction, we obtain

$R=\begin{bmatrix} 1& 2& 0 \ 0& 0&1\ 0&0&0 \end{bmatrix}$

for the reduced row echelon form.

The second column in this matrix can be seen by inspection to be a linear combination of the first column, so it is not included in our basis for . Hence the first and the third columns form a basis for the column space of , and therefore the first and the third columns of form a basis for the range space of .

$\begin{Bmatrix} \begin{bmatrix} 1\ 0\ 0\ \end{bmatrix} _, \begin{bmatrix} 3\ 2\ -1\ \end{bmatrix}\end{Bmatrix}$

Calculate the Null Space of the following Matrix.

$A=\left[\begin{matrix} 1 & 1& 1& -1 \ 2 & 4&5 &6 \ 3& 9& 5& 4 \end{matrix}\right]$

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The first step is to create an augmented matrix having a column of zeros.

$\left[\begin{matrix} 1 & 1& 1& -1 \ 2 & 4&5 &6 \ 3& 9& 5& 4 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

The next step is to get this into RREF.

$2R_1-R_2\rightarrow R_2$

$\left[\begin{matrix} 1 & 1& 1& -1 \ 0 & -2&-3 &-8 \ 3& 9& 5& 4 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$3R_1-R_3\rightarrow R_3$

$\left[\begin{matrix} 1 & 1& 1& -1 \ 0 & -2&-3 &-8 \ 0& -6& -2& -7 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$-3R_2+R_3\rightarrow R_3$

$\left[\begin{matrix} 1 & 1& 1& -1 \ 0 & -2&-3 &-8 \ 0& 0& 7& 17 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$$\frac{1}{2}$R_2+R_1\rightarrow R_1$

$\left[\begin{matrix} 1 & 0& -$\frac{1}{2}$& -5 \ 0 & -2&-3 &-8 \ 0& 0& 7& 17 \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

We can simplify to

$\left[\begin{matrix} 1 & 0& -$\frac{1}{2}$& -5 \ 0 & 1&$\frac{3}{2}$ &4 \ 0& 0& 1& $\frac{17}{7}$ \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$-$\frac{3}{2}$R_3+R_2\rightarrow R_2$

$\left[\begin{matrix} 1 & 0& -$\frac{1}{2}$& -5 \ 0 & 1&0 & \ $\frac{5}{14}$ \ 0& 0& 1& $\frac{17}{7}$ \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

$$\frac{1}{2}$R_3+R_1\rightarrow R_1$

$\left[\begin{matrix} 1 & 0& 0& -$\frac{53}{14}$ \ 0 & 1&0 & \ $\frac{5}{14}$ \ 0& 0& 1& $\frac{17}{7}$ \end{matrix}\right|\left.\begin{matrix} 0\ 0\ 0 \end{matrix}\right]$

This tells us the following.

$\x_1-$\frac{53}{14}$x_4=0 \ \ x_2+\frac{5}{14}$x_4=0 \ \ x_3+\frac{17}{7}$x_4=0$

$\x_1=\frac{53}{14}$x_4 \ \ x_2=-$\frac{5}{14}$x_4 \ \ x_3=-$\frac{17}{7}$x_4$

Now we need to write this as a linear combination.

$\begin{bmatrix} x_1\ x_2\ x_3\ x_4 \end{bmatrix} =x_4\begin{bmatrix} $\frac{53}{14}$\ \ -$\frac{5}{14}$\ \ -$\frac{17}{7}$ \ 1 \end{bmatrix}$

The null space is then

$N(A)=Span\begin{bmatrix} $\frac{53}{14}$\ \ -$\frac{5}{14}$\ \ -$\frac{17}{7}$ \ 1 \end{bmatrix}$

Find the null space of the matrix operator $T_M = \begin{bmatrix} 1 &1 & 0\ 0& -1& 1\ 9& 0& 1\end{bmatrix}$ .

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The null space of the operator is the set of solutions to the equation

$\begin{bmatrix} 1 &1 & 0\ 0& -1& 1\ 9& 0& 1\end{bmatrix} \begin{bmatrix} x_1\x_2\x_3 \end{bmatrix} = \begin{bmatrix} 0\0 \0 \end{bmatrix}$ .

We can solve the above system by row reducing our $3 \times 3$ matrix using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0 & 0\0 & 1& 0\ 0& 0&1 \end{bmatrix}\begin{bmatrix} x_1\x_2 \x_3 \end{bmatrix}= \begin{bmatrix} 0\0 \0 \end{bmatrix} \Longrightarrow \begin{align} x_1 &=0 \x_2 &=0 \ x_3&=0 \end{align}$

Hence the null space consists of only the zero vector. $\begin{Bmatrix} \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \end{Bmatrix}$

Find the null space of the matrix $A = \begin{bmatrix} 2 & 0& 0\1 &0 &0 \9 &8 &2 \end{bmatrix}$ .

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The null space of the matrix is the set of solutions to the equation

$\begin{bmatrix} 2 &0 & 0\ 1& 0& 0\ 9& 8& 2\end{bmatrix} \begin{bmatrix} x_1\x_2\x_3 \end{bmatrix} = \begin{bmatrix} 0\0 \0 \end{bmatrix}$ .

We can solve the above system by row reducing using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0 & 0\0 & 1& 1/4\ 0& 0&0 \end{bmatrix}\begin{bmatrix} x_1\x_2 \x_3 \end{bmatrix}= \begin{bmatrix} 0\0 \0 \end{bmatrix} \Longrightarrow \begin{align} x_1 &=0 \x_2 &=-(1/4) x_3 \ x_3&=x_3 \end{align} \Longrightarrow$

$X= \begin{bmatrix} 0\-1/4 \1 \end{bmatrix}t$

Multiplying this vector by gets rid of the fraction, and does not affect our answer, since there is an arbitrary constant behind it.

Hence the null space consists of all vectors spanned by $\begin{bmatrix} 0\-1 \4 \end{bmatrix}$ ;

$Span\begin{Bmatrix} \begin{bmatrix} 0\ -1\4 \end{bmatrix} \end{Bmatrix}$ .