Question

Consider the following matrix: 1-5 -5 -1 3 A-1 5 5 2 -4 2-10-10 0 4 Give a set of vectors that spans each of im(A) and null(A) Number of Vectors: 1 Spanning set for im(A) Number of Vectors: 1 Spanning set for null(A)30

Answer 1

matrix is

1	-5	-5	-1	3
-1	5	5	2	-4
2	-10	-10	0	4

convert into Reduced Row Eschelon Form...

Add (1 * row1) to row2

1	-5	-5	-1	3
0	0	0	1	-1
2	-10	-10	0	4

Add (-2 * row1) to row3

1	-5	-5	-1	3
0	0	0	1	-1
0	0	0	2	-2

Add (-2 * row2) to row3

1	-5	-5	-1	3
0	0	0	1	-1
0	0	0	0	0

Add (1 * row2) to row1

1	-5	-5	0	2
0	0	0	1	-1
0	0	0	0	0

there are two pivot entry at first and fourth column

so image(A)= $\large {\color{Red} \begin{pmatrix}1\\ -1\\ 2\end{pmatrix},\begin{pmatrix}-1\\ 2\\ 0\end{pmatrix}}$

$\large \begin{pmatrix}1&-5&-5&0&2\\ 0&0&0&1&-1\\ 0&0&0&0&0\end{pmatrix}\begin{pmatrix}a\\ b\\ c\\ d\\ e\end{pmatrix}=0$

$\large a-5b-5c+2e=0....................a=5b+5c-2e$

$\large d-e=0....................................d=e$

$\large b=b$

$\large c=c$

$\large e=e$

$\large \begin{pmatrix}a\\ \:b\\ \:c\\ \:d\\ \:e\end{pmatrix}=b\begin{pmatrix}5\\ 1\\ 0\\ 0\\ 0\end{pmatrix}+c\begin{pmatrix}5\\ 0\\ 1\\ 0\\ 0\end{pmatrix}+e\begin{pmatrix}-2\\ 0\\ 0\\ 1\\ 1\end{pmatrix}$

null space(A)= $\large {\color{Red} \begin{pmatrix}5\\ \:1\\ \:0\\ \:0\\ \:0\end{pmatrix}\begin{pmatrix}5\\ \:0\\ \:1\\ \:0\\ \:0\end{pmatrix}\begin{pmatrix}-2\\ \:0\\ \:0\\ \:1\\ \:1\end{pmatrix}}$