Linear Algebra - Range and Null Space of a Matrix

Calculate the Null Space of the following Matrix.

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

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

There is no Null Space

$N(A)=Span\begin{bmatrix} \frac{53}{14}\ \ 0\ \ -\frac{17}{7} \ 1 \end{bmatrix}$

$N(A)=Span\begin{bmatrix} 0\ \ -\frac{5}{14}\ \ -\frac{17}{7} \ 0 \end{bmatrix}$

$N(A)=Span\begin{bmatrix} 0\ \ 0\ \ 0 \ 1 \end{bmatrix}$

Explanation

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 $T_M = \begin{bmatrix} 1 &1 & 0\ 0& -1& 1\ 9& 0& 1\end{bmatrix}$ .

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

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

$Span\begin{Bmatrix} \begin{bmatrix} 1\0 \2 \end{bmatrix} \end{Bmatrix}$

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

None of the other answers

Explanation

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

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

True

False

Explanation

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

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

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

$(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.

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

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

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

$Span\begin{Bmatrix} \begin{bmatrix} 1\ -2\0 \ \end{bmatrix} \end{Bmatrix}$

None of the other answers

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

Explanation

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

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

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

True

False

Explanation

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

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

What is the nullity of ?

Explanation

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 defined on $(-\infty, \infty)$ with inverses is a subspace of $C(-\infty , \infty)$ .

False

True

Explanation

Let be the set of all functions on $(-\infty, \infty)$ such that $f^{-1}$ 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 . 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 $C (-\infty, \infty)$ .

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

True

False

Explanation

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

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

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

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

$f +g \in S$ . 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 ,

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

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

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

, as defined, is a subspace.

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

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

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

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

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

$\mathbb{R}^2$

Explanation

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

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