Question

If A=[(1, 2,1)

(2, 0, 0)

(0, 5, 0)]

A: R3->R3

1) Find the row reduced echelon form of A

2) Find the image of A

3) Find a nonzero vector in ker(A)

Answer 1

Solution:

$\small 1)$

A R R 3

1 2 1 R2-2RI 0 1 1 1 0 (-) ax(-) R2x A 2 1/2 -5/2 0 -4 -2 0 1 R1-2R2, R3-5R2 0 R2-R3 0 5 0 0 0

1 0 0 0 1 0 0 1

2)

$\small \mathrm{Im}A=\left \{ Ax:x\in \mathbb{R}^{3} \right \}$

  

10 0 yeR3 0 1 0 y 0 0 1

  

3 y =

  

0 0 y1 0,y, R 1

  

0 - Span 1

$\small 3)$

Since $\small \left \{ \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix},\begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix},\begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix} \right \}$ is  linearly independent ,  $\small \left \{ \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix},\begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix},\begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix} \right \}$ form a basis for image of  $\small A$ .

$\small \therefore \mathrm{dim\: Im}A=3$

So, by rank-nullity theorem

$\small n= \mathrm{dim\: Im}A+\mathrm{dim\: ker}A$

33dim kerA

$\small \therefore \mathrm{dim\: ker}A=0$

$\small \therefore$ there is no nonzero vector in  $\small ker\left ( A \right )$ .