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

karush

Well-known member
Jan 31, 2012
2,648
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
Jan 29, 2012
1,151
"NS(A)" is the null space? If so then we are looking for [tex]\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}[/tex] such that [tex]\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}[/tex].

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

In particular, taking [tex]x_4= 1[/tex] and [tex]x_5= 0[/tex], [tex]x_1= -4[/tex]. [tex]x_2= -3[/tex], and [tex]x_3= -3[/tex]. One vector in the null space is [tex]\begin{bmatrix}-4 \\ -3 \\ -3 \\ 1 \\ 0 \end{bmatrix}[/tex]. Taking [tex]x_4= 0[/tex] and [tex]x_5= 1[/tex], [tex]x_1= -5[/tex], [tex]x_2= -2[/tex], and [tex]x_3= -2[/tex]. Another vector in the null space is [tex]\begin{bmatrix}-5 \\ -2 \\ -2 \\ 0 \\ 1\end{bmatrix}[/tex]. 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
Jan 31, 2012
2,648
$\left[ \begin{array}{c}
- 5x_4 - 4x_5 \\ - 2x_4 - 3x_5\\ - 2x_4 - 3x_5 \\x_4 \\x_5
\end{array} \right]
=\left[ \begin{array}{r} -4 \\-3 \\ -3 \\ 1 \\0
\end{array} \right]x_4
+\left[ \begin{array}{r} -5 \\ -2 \\ -2 \\ 0 \\ 1
\end{array} \right]x_5$
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
 
Last edited: