Let V= R^3
W={(x1, x2, x3)| x1-x2 =x3}. Show that W is a
subspace of V. Further,find a basis for W and hence,find the dimension of W.
Let V= R^3
"W={(x_1, x_2, x_3)| x_1-x_2 =x_3}" . Show that W is a
subspace of V.
Further,find a basis for W and hence,find the dimension of W.
To show that W is a subspace of V we need to show the following:
let "r= (r_1, r_2, r_3)" , "q=(q_1, q_2,q_3) \\isin W" and "\\alpha\\isin" field
"\\implies q_1-q_2=q_3" and "r_1- r_2= r_3"
if "(0, 0,0)\\isin" W then "0-0=0"
therefore, W is not empty
"r+q=(r_1, r_2, r_3) +(q_1, q_2,q_3)={ (r_1+q_1), (r_2+q_2),(r_3+q_3)}"
since "(r_1+q_1)-(r_2+q_2)=(r_1-r_2) +(q_1-q_2)= r_3+ q_3"
"\\implies r+q\\isin W"
"\\alpha r=\\alpha (r_1, r_2, r_3)=r= (\\alpha r_1, \\alpha r_2,\\alpha r_3)"
then "\\alpha r_1-\\alpha r_2=\\alpha(r_1-r_2)=\\alpha r_3"
"\\implies \\alpha r\\isin W"
therefore W is a subspace of V
"r=\\begin{pmatrix}\n r_1 \\\\\n r_2 \\\\r_3 \n\\end{pmatrix} =\\begin{pmatrix}\n r_1 \\\\\n r_2 \\\\r_1-r_2 \n\\end{pmatrix} = r_1\\begin{pmatrix}\n 1 \\\\\n 0 \\\\1\n\\end{pmatrix}+r_2\\begin{pmatrix}\n 0 \\\\\n 0 \\\\-1\n\\end{pmatrix}"
"\\therefore ,\n\n\\begin{bmatrix}\n \\begin{pmatrix}\n 1 \\\\\n 0 \\\\1\n\\end{pmatrix} & \\begin{pmatrix}\n 0 \\\\\n 0 \\\\-1 \\end{pmatrix}\n \n\\end{bmatrix}" is a Basis of W
The Dimension of W is 2
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