Null Space of A: Find Rank & Dim.

In summary, the basis for the null space of A is given by the three linearly independent vectors: $\mathbf v_1=\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}, \mathbf v_2=\begin{pmatrix} 0 \\ -1 \\ 1 \\ -4 \\ 1 \\ 0 \\ 0\end{pmatrix}, \mathbf v_3=\begin{pmatrix} 0 \\ -1 \\ 3 \\ -2 \\ 0 \\ -3 \\ 1\end{pmatrix}$. Therefore, the dimension of the null space of A is 3
  • #1
karush
Gold Member
MHB
3,269
5
Let
$$\left[\begin{array}{rrrrrrr}
1 & 0 & -1 & 0 & 1 & 0 & 3\\
0 & 1 & 0 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 4 & 0 & 2\\
0 & 0 & 0 & 0 & 0 & 1 & 3
\end{array}\right]$$
Find a basis for the null space of A, the dimension of the null space of A, and the rank of A.ok following an book example I did this $Ax=b$
$$\left[ \begin{array}{ccccccc}
1 & 0 & -1 & 0 & 1 & 0 & 3 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 4 & 0 & 2 \\
0 & 0 & 0 & 0 & 0 & 1 & 3
\end{array} \right]
\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \\ x_{7}
\end{array} \right]
=\left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0
\end{array} \right]$$
which would result in
$$\begin{array}{rrrrrrr}
x_1 & &-x_3 & &x_5 & &3x_7=0 \\
&x_2 & & &x_5 & &x_7 =0\\
& & & x_4 & 4x_5 & &2x_7=0 \\
& & & & & x_6 &3x_7=0
\end{array}$$ hopefully so far !
 
Physics news on Phys.org
  • #2
It is clear (by inspection, working from the last equation upwards) that you can let $x_7,x_5,x_1$ be arbitrary, then $x_2,x_3,x_4,x_6$ can be expressed in terms of them as:
$$x_6\ =\ -3x_7 \\ x_4\ =\ -4x_5-2x_7 \\ x_2\ =\ -x_5-x_7 \\ x_3\ =\ x_1+x_5+3x_7.$$
Hence:
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \\ x_7\end{pmatrix}\ =\ x_1\underbrace{\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}}_{\mathbf v_1} + x_5\underbrace{\begin{pmatrix} 0 \\ -1 \\ 1 \\ -4 \\ 1 \\ 0 \\ 0\end{pmatrix}}_{\mathbf v_2} + x_7\underbrace{\begin{pmatrix} 0 \\ -1 \\ 3 \\ -2 \\ 0 \\ -3 \\ 1\end{pmatrix}}_{\mathbf v_3}.$$
Verify that $\mathbf v_1,\mathbf v_2,\mathbf v_3$ are linearly independent and thus form a basis for the nullspace of $A$; hence $A$ has nullity $3$. The rank of $A$ can then be found from the formula $r(A) = \dim(A)-n(A)$.
 
  • #3
You have 4 equations in 7 unknowns so the null space is 7- 4= 3 dimensional. You will need 3 basis vectors.
 
  • #4
Ok that was very helpful

Not sure if I would know the linear independence of these
 
  • #5
You prove those vectors are independent using the definitions of "independent" or "dependent".
Suppose the three vectors given by Olinguito were NOT independent. Then there would exist numbers, a, b, and c, not all zero, such that
[tex]a\begin{pmatrix}1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}+[/tex][tex] b\begin{pmatrix}0 \\ -1 \\ 1 \\ -4 \\ 1 \\ 0 \\ 0 \end{pmatrix}+[/tex][tex] c\begin{pmatrix} 0 \\ -1 \\ 3 \\ -2 \\ 0 \\ -3 \\ 1 \end{pmatrix}[/tex][tex]=[/tex][tex] \begin{pmatrix}a \\ -b- c \\ a+ b+ 3c \\ -4b- 2c \\ b \\ -3c \\ c\end{pmatrix}[/tex][tex]= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}[/tex].

That gives 7 equations: a= 0, -b- c= 0, a+ b+ 3c= 0, -4b- 2c= 0, b= 0, -3c= 0, c= 0. Clearly the only a, b, and c that satisfy all 7 equations are a= b= c= 0. Therefore the three vectors are independent.
 
Last edited:

Related to Null Space of A: Find Rank & Dim.

1. What is the null space of a matrix?

The null space of a matrix A is the set of all vectors x that, when multiplied by A, equal the zero vector. In other words, it is the set of all solutions to the equation Ax = 0.

2. How do you find the rank of a matrix?

The rank of a matrix is equal to the number of linearly independent columns or rows in the matrix. To find the rank, you can perform row operations on the matrix to reduce it to row-echelon form, and then count the number of non-zero rows.

3. How do you find the dimension of the null space of a matrix?

The dimension of the null space of a matrix is equal to the number of free variables in the reduced row-echelon form of the matrix. This can be found by counting the number of columns without a leading 1 in the reduced row-echelon form.

4. Can the rank of a matrix be greater than its dimension?

No, the rank of a matrix cannot be greater than its dimension. The rank is always less than or equal to the number of rows or columns in the matrix, while the dimension of the null space is always less than or equal to the number of columns.

5. How does the null space of a matrix relate to its solutions?

The null space of a matrix contains all possible solutions to the equation Ax = 0. This means that any vector in the null space can be multiplied by the matrix A to equal the zero vector. Additionally, the null space can also provide insight into the number of solutions for a system of equations represented by the matrix A.

Similar threads

  • Linear and Abstract Algebra
Replies
3
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
7
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
930
  • Linear and Abstract Algebra
Replies
4
Views
2K
  • Linear and Abstract Algebra
Replies
19
Views
2K
Back
Top