# [SOLVED]14.3 Find a basis for NS(A) and dim{NS(A)}

#### karush

##### Well-known member
For the matrix
$A=\left[\begin{array}{rrrrr} 1&0&0&4&5\\ 0&1&0&3&2\\ 0&0&1&3&2\\ 0&0&0&0&0\end{array}\right]$
Find a basis for NS(A) and $\dim{NS(A)}$
$\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array}\right]= \left[\begin{array}{c} -4x_4-5x_5\\ -3x_4-2x_5\\ -3x_4-2x_5\\ x_4\\ x_5 \end{array}\right]$

ok I just did this but there is duplication in it

#### HallsofIvy

##### Well-known member
MHB Math Helper
"NS(A)" is the null space? If so then we are looking for $$\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}$$ such that $$\begin{bmatrix}1 & 0 & 0 & 4 & 5 \\ 0 & 1 & 0 & 3 & 2 \\ 0 & 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\x_4 \\ x_5\end{bmatrix}= \begin{bmatrix}x_1+ 4x_4+ 5x_5 \\ x_2+ 3x_4+ 2x_5 \\ x_3+ 3x_4+ 2x_5 \\ 0 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$.

Although I wouldn't have written the equations this way they do give, as you say, $$x_1= -4x_4- 5x_5$$, $$x_2= -3x_4- 2x_5$$, $$x_3= -3x_4- 2x_5$$, and 0= 0. If by "duplication" you mean $$x_2= -3x_4- 2x_5$$ and $$x_3= -3x_4- 2x_5$$, that just means that $$x_2= x_3$$ Since all of $$x_1$$, $$x_2$$, and $$x_3$$ depend upon $$x_4$$ and $$x_5$$ take them as parameters (and the null space is two dimensional).

In particular, taking $$x_4= 1$$ and $$x_5= 0$$, $$x_1= -4$$. $$x_2= -3$$, and $$x_3= -3$$. One vector in the null space is $$\begin{bmatrix}-4 \\ -3 \\ -3 \\ 1 \\ 0 \end{bmatrix}$$. Taking $$x_4= 0$$ and $$x_5= 1$$, $$x_1= -5$$, $$x_2= -2$$, and $$x_3= -2$$. Another vector in the null space is $$\begin{bmatrix}-5 \\ -2 \\ -2 \\ 0 \\ 1\end{bmatrix}$$. Since the null space is two dimensional and the those vectors are independent, they form a basis for the null space.

#### karush

##### Well-known member
$\left[ \begin{array}{c} - 5x_4 - 4x_5 \\ - 2x_4 - 3x_5\\ - 2x_4 - 3x_5 \\x_4 \\x_5 \end{array} \right] =\left[ \begin{array}{r} -4 \\-3 \\ -3 \\ 1 \\0 \end{array} \right]x_4 +\left[ \begin{array}{r} -5 \\ -2 \\ -2 \\ 0 \\ 1 \end{array} \right]x_5$
the basis for the null space is
$\left[ \begin{array}{r} -4 \\-3 \\ -3 \\ 1 \\0 \end{array} \right] ,\left[ \begin{array}{r} -5 \\ -2 \\ -2 \\ 0 \\ 1 \end{array} \right]$

kinda getin it

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