14.2 find a basis for NS(A) and dim{NS(A)}

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In summary, the null space of the matrix A is one dimensional and a basis for it is given by the vector [1, -3, 1].
  • #1
karush
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$\tiny{370.14.2.}$
For the matrix
$A=\left[
\begin{array}{rrrr}
1&0&1\\0&1&3
\end{array}\right]$
find a basis for NS(A) and $\dim{NS(A)}$
-----------------------------------------------------------
altho it didn't say I assume the notation means Null Space of A
Reducing the augmented matrix for the system $$AX=0$$ to reduced row-echelon form.
$\left[
\begin{array}{ccc}
1 & 0 & 1 \\ 0 & 1 & 3
\end{array} \right]
\left[ \begin{array}{c}
x_{1} \\ x_{2} \\ x_{3}
\end{array} \right]
=\left[ \begin{array}{c}
0 \\ 0
\end{array}
\right]$
OK just seeing if I am going in the right direction
 
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  • #2
karush said:
$\tiny{370.14.2.}$
For the matrix
$A=\left[
\begin{array}{rrrr}
1&0&1\\0&1&3
\end{array}\right]$
find a basis for NS(A) and $\dim{NS(A)}$
-----------------------------------------------------------
altho it didn't say I assume the notation means Null Space of A
Reducing the augmented matrix for the system $$AX=0$$ to reduced row-echelon form.
$\left[
\begin{array}{ccc}
1 & 0 & 1 \\ 0 & 1 & 3
\end{array} \right]
\left[ \begin{array}{c}
x_{1} \\ x_{2} \\ x_{3}
\end{array} \right]
=\left[ \begin{array}{c}
0 \\ 0
\end{array}
\right]$
OK just seeing if I am going in the right direction
Yes, that is correct. Do you see that the matrix equation is equivalent to the two equations [tex]x_1+ x_3= 0[/tex] and [tex]x_2+ 3x_3= 0[/tex]? Do you see that [tex]x_1[/tex] and [tex]x_2[/tex] can be written in terms of [tex]x_3[/tex] so this null space is one dimensional?
 

FAQ: 14.2 find a basis for NS(A) and dim{NS(A)}

1. What is NS(A)?

NS(A) stands for the null space of matrix A. It is the set of all vectors that, when multiplied by matrix A, result in the zero vector.

2. How do you find a basis for NS(A)?

To find a basis for NS(A), you need to first reduce matrix A to its reduced row echelon form. Then, the columns corresponding to the leading variables will form the basis for NS(A).

3. Can there be more than one basis for NS(A)?

Yes, there can be multiple bases for NS(A). This is because any scalar multiple of a vector in the basis will still span the same space.

4. What is the significance of finding a basis for NS(A)?

Finding a basis for NS(A) allows us to understand the structure of the null space and determine the dimension of the null space. It also helps us to find solutions to linear systems and understand the linear independence of the columns of matrix A.

5. How is the dimension of NS(A) related to the rank of matrix A?

The dimension of NS(A) is equal to the number of free variables in the reduced row echelon form of matrix A. This is also equal to the difference between the number of columns and the rank of matrix A. In other words, dim{NS(A)} + rank(A) = number of columns in A.

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