Range and Null Space of a Matrix - Linear Algebra

Card 1 of 80

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

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

Tap to reveal answer

Answer

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 .

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

← Didn't Know|Knew It →

Question

Calculate the Null Space of the following Matrix.

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

Tap to reveal answer

Answer

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}$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 $\begin{bmatrix} 0\-1 \4 \end{bmatrix}$ ;

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 $\begin{bmatrix} 0\1 \end{bmatrix}$ ;

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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.

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 .

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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;

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 .

← Didn't Know|Knew It →

Question

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

What is the nullity of ?

Tap to reveal answer

Answer

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 .

← Didn't Know|Knew It →

Question

$C(-\infty , \infty)$ , 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 $C(-\infty , \infty)$ .

Tap to reveal answer

Answer

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 $C(-\infty, \infty)$ .

← Didn't Know|Knew It →

Question

$C(-\infty , \infty)$ , 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 $C(-\infty , \infty)$ .

Tap to reveal answer

Answer

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 $C (-\infty, \infty)$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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

.

← Didn't Know|Knew It →

Question

$C(-\infty , \infty)$ , 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 $C(-\infty , \infty)$ .

Tap to reveal answer

Answer

A subset of a vector space is a subspace of that vector space if and only if it meets two criteria. Both will be given and tested, letting

$S = \left \{ f(x)| f(x) = 2n x + n , n \in \mathbb{R}\right \}$ .

This can be rewritten as

$S = \left \{ f(x)| f(x) = n( 2 x +1) , n \in \mathbb{R}\right \}$

One criterion for to be a subspace is closure under addition; that is:

If $a, b \in S$ , then $a+b \in S$ .

Let $f, g \in S$ as defined. Then for some real :

It follows that $f+g \in S$ .

The second criterion for to be a subspace is closure under scalar multiplication; that is:

If $a \in S , c \in \mathbb{R}$ , then $ca \in S$

Let $h \in S$ as defined. Then for some real :

$= c \cdot p (2 x +1)$

It follows that $ch \in S$

, as defined, is a subspace of $C(-\infty , \infty)$ .

← Didn't Know|Knew It →

Question

$C(-\infty , \infty)$ , 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 $C(-\infty , \infty)$ .

Tap to reveal answer

Answer

Let $S = \left \{ f(x)| f(x) = |nx | , n \in \mathbb{R}\right \}$ .

We can show through counterexample that this is not a subspace of $C(-\infty , \infty)$ .

Let . This is an element of .

One condition for to be a subspace of a vector space is closure under scalar multiplication. Multiply by scalar . The product is the function

.

$g(x) \ne S$ .

This violates a criterion for a subspace, so is not a subspace of $C(-\infty , \infty)$ .

← Didn't Know|Knew It →

Question

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

Let be the set of all functions of the form

$f(x) = \left\{\begin{matrix} 0,& x< 0\ nx,& x \ge 0 \end{matrix}\right.$

True or false: is a subspace of $C(-\infty , \infty)$ .

Tap to reveal answer

Answer

A set is a subspace of a vector space if and only if two conditions hold, both of which are tested here.

The first condition is closure under addition - that is:

If $a , b \in S$ , then $a+ b \in S$

Let $f , g \in S$ as defined. Then for some $n, m \in \mathbb{R}$ ,

$f(x) = \left\{\begin{matrix} 0,& x< 0\ nx,& x \ge 0 \end{matrix}\right.$

and

$g(x) = \left\{\begin{matrix} 0,& x< 0\mx,& x \ge 0 \end{matrix}\right.$ .

$(f+ g )(x) = \left\{\begin{matrix} 0+ 0,& x< 0 x + mx,& x \ge 0 \end{matrix}\right.$

or

$(f+ g )(x) = \left\{\begin{matrix} 0 ,& x< 0\(n + m)x,& x \ge 0 \end{matrix}\right.$ ,

$f +g \in S$ . The first condition is met.

The second condition is closure under scalar multiplication - that is:

If $a \in S$ and is a scalar, then $ca \in S$

Let $h \in S$ as defined. Then for some $p \in \mathbb{R}$ ,

$h(x) = \left\{\begin{matrix} 0,& x< 0\ px,& x \ge 0 \end{matrix}\right.$

For any scalar ,

$c \cdot h(x) = \left\{\begin{matrix} c \cdot 0,& x< 0\ c \cdot px,& x \ge 0 \end{matrix}\right.$ ,

and

$(c h)(x) = \left\{\begin{matrix} 0,& x< 0\ (c p)x,& x \ge 0 \end{matrix}\right.$

$c h \in S$ . The second condition is met.

, as defined, is a subspace.

← Didn't Know|Knew It →

Question

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

Let be the set of all functions of the form

$f(x) = \left\{\begin{matrix} 2nx+2n , & x \le - 1 \ 0, & -1 < x < 1\ -nx+n & x \ge 1 \end{matrix}\right.$

for some real

True or false: is a subspace of $C(-\infty , \infty)$ .

Tap to reveal answer

Answer

A set is a subspace of a vector space if and only if two conditions hold, both of which are tested here.

The first condition is closure under addition - that is:

If $a , b \in S$ , then $a+ b \in S$

Let $f , g \in S$ as defined. Then for some $n, m \in \mathbb{R}$ ,

$f(x) = \left\{\begin{matrix} 2nx+2n , & x \le - 1 \ 0, & -1 < x < 1\ -nx+n & x \ge 1 \end{matrix}\right.$

and

$g(x) = \left\{\begin{matrix} 2mx+2m, & x \le - 1 \ 0, & -1 < x < 1\ -mx+m & x \ge 1 \end{matrix}\right.$

Then

$(f+g)(x) = \left\{\begin{matrix}( 2nx+2n)+ ( 2mx+2m) , & x \le - 1 \ 0+0 , & -1 < x < 1\ (-nx+n) + (-mx+m) & x \ge 1 \end{matrix}\right.$

or

$(f+g)(x) = \left\{\begin{matrix}( 2nx+2mx )+ ( 2n +2m) , & x \le - 1 \ 0+0 , & -1 < x < 1\ (-nx+(-mx)) + (n+m) & x \ge 1 \end{matrix}\right.$

or

$(f+g)(x) = \left\{\begin{matrix} 2(n + m)x+ 2 ( n +m) , & x \le - 1 \ 0 , & -1 < x < 1\ -(n +m)x + (n+m) & x \ge 1 \end{matrix}\right.$

$f +g \in S$ . The first condition is met.

The second condition is closure under scalar multiplication - that is:

If $a \in S$ and is a scalar, then $ca \in S$

Let $h \in S$ as defined. Then for some $p \in \mathbb{R}$ ,

$h(x) = \left\{\begin{matrix} 2px+2p , & x \le - 1 \ 0, & -1 < x < 1\ -px+p & x \ge 1 \end{matrix}\right.$

For any scalar ,

$c h(x) = \left\{\begin{matrix} c (2px+2p) , & x \le - 1 \ c \cdot 0, & -1 < x < 1\ c(-px+p) & x \ge 1 \end{matrix}\right.$

or

$(c h)(x) = \left\{\begin{matrix} 2(cp)x+2 (c p) , & x \le - 1 \ 0, & -1 < x < 1\ -(cp)x+cp & x \ge 1 \end{matrix}\right.$

$c h \in S$ . The second condition is met.

, as defined, is a subspace.

← Didn't Know|Knew It →

0

Knew It