Question

2. Consider a linear transformation T: R³ → R³ defined by

  ([x         [ x + 4y +3z  

T  y     =      -5y - 4z 

    z])        5x + 10y + 7z ]

Note: T is a 3x1 matrix containing x, y, z respectively. T is equal to another matrix as shown above.

a) Find the matrix A for T

b) Find a basis for ker(T) and the dim(ker(T)). Then find dim(Im(T)), without finding a basis for Im(T). (Show all working)

c) Find a basis for Im(T)

Answer 1

$a:The\,\,matrix\,\,is\,\,of\,\,coefficients\,\,written\,\,in\,\,lines:\\A=\left[ \begin{matrix} 1& 4& 3\\ 0& -5& -4\\ 5& 10& 7\\\end{matrix} \right] \\b:\\A\left[ \begin{array}{c} x\\ y\\ z\\\end{array} \right] =0\Rightarrow \left\{ \begin{array}{c} x+4y+3z=0\\ -5y-4z=0\\ 5x+10y+7z=0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} x+4y-\frac{15}{4}y=0\\ 5x+10y-\frac{35}{4}y=0\\ z=-\frac{5}{4}y\\\end{array} \right. \Rightarrow \\\Rightarrow \left\{ \begin{array}{c} y=-4x\\ y=-4x\\ z=-\frac{5}{4}y\\\end{array} \right. \Rightarrow \left[ \begin{array}{c} x\\ y\\ z\\\end{array} \right] =t\left[ \begin{array}{c} 1\\ -4\\ 5\\\end{array} \right] \,\,-\,\,one\,\,vector\\dim\left( ker\left( T \right) \right) =1\\dim\left( \mathrm{Im}\left( T \right) \right) =3-dim\left( ker\left( T \right) \right) =3-1=2\\c:\\A\left[ \begin{array}{c} 1\\ 0\\ 0\\\end{array} \right] =\left[ \begin{array}{c} 1\\ 0\\ 5\\\end{array} \right] ,A\left[ \begin{array}{c} 0\\ 1\\ 0\\\end{array} \right] =\left[ \begin{array}{c} 4\\ -5\\ 10\\\end{array} \right] \\Two\,\,vectors\,\,\left[ \begin{array}{c} 1\\ 0\\ 5\\\end{array} \right] ,\left[ \begin{array}{c} 4\\ -5\\ 10\\\end{array} \right] \,\,belong\,\,to\,\,\mathrm{Im}\left( T \right) . \\These\,\,vectors\,\,are\,\,linearly\,\,independent\,\,\sin ce\,\,their\,\,coordinates\,\,are\,\,not\,\,proportional.\\Since\,\,dim\left( \mathrm{Im}\left( T \right) \right) =2, \left[ \begin{array}{c} 1\\ 0\\ 5\\\end{array} \right] ,\left[ \begin{array}{c} 4\\ -5\\ 10\\\end{array} \right] \,\,are\,\,the\,\,basis\,\,of\,\,\mathrm{Im}\left( T \right)$