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4. Find a basis for the kernel of each of the following linear transformations. 3 -6
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Solution:

small 4.left ( a ight )

small varphi _{1}:P_{3} ightarrow mathbb{R}

pi (p) = p (1

small mathrm{Ker}: varphi _{1}=left { pin P_{3}:pleft ( left ( 1 ight )=0 ight ) ight }

small dpi{100} small =left { a+bx+cx^{2}+dx^{3}:a+b+c+d=0 ight }

small =left { a+bx+cx^{2}+dx^{3}:a=-b-c -d ight }

small =left { -b-c-d+bx+cx^{2} +dx^{3} ight }

small =left { bx-b+cx^{2}-c+dx^{3}-d ight }=left { bleft ( x-1 ight )+cleft ( x^{2} -1 ight )+dleft ( x^{3} -1 ight ) ight }

span1( 1), (x2-1), (r -1))

Also, or (r _ 1) + β (12-1)+1(23-1):0

small Rightarrow alpha =eta=gamma =0

small herefore left { left ( x-1 ight ),: : : left ( x^{2} -1 ight ) ,: left ( x^{3} -1 ight ) ight } is linearly independent.

small herefore left { left ( x-1 ight ),: : : left ( x^{2} -1 ight ) ,: left ( x^{3} -1 ight ) ight } is a basis for the kernel.

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small left ( b ight )

small varphi _{2}:M_{2 imes 3} ightarrow M_{2 imes 3}

small varphi _{2}left ( B ight )=egin{pmatrix} 3 & -6 -1 &2 end{pmatrix}B

small mathrm{Ker}: varphi _{2}=left { Bin M_{2 imes 3} :varphi _{2}left ( B ight )=egin{pmatrix} 0 & 0 &0 0 & 0 &0 end{pmatrix} ight }

small =left { egin{pmatrix} a & c &e b &d & f end{pmatrix}in M_{2 imes 3} :egin{pmatrix} 3 &-6 -1 & 2 end{pmatrix}egin{pmatrix} a & c &e b& d & f end{pmatrix}=egin{pmatrix} 0 &0 & 0 0 & 0 & 0 end{pmatrix} ight }

small =left { egin{pmatrix} a & c &e b &d & f end{pmatrix}in M_{2 imes 3} :egin{pmatrix} 3a-6b & 3c-6d &3e-6f -a+2b& -c+2d & -e+2f end{pmatrix}=egin{pmatrix} 0 &0 & 0 0 & 0 & 0 end{pmatrix} ight }

small =left { egin{pmatrix} a & c &e b &d & f end{pmatrix}in M_{2 imes 3} :a=2b,: : c=2d,: : e=2f ight }

small =left { egin{pmatrix} 2b& 2d &2f b &d & f end{pmatrix} ight }

0 2 0 00 2 1 0 0

small =mathrm{span}left { egin{pmatrix} 2& 0 &0 1 &0 &0 end{pmatrix} ,egin{pmatrix} 0 & 2 &0 0 & 1 & 0 end{pmatrix},egin{pmatrix} 0 & 0 &2 0&0 & 1 end{pmatrix} ight }

Also, small left { egin{pmatrix} 2& 0 &0 1 &0 &0 end{pmatrix} ,egin{pmatrix} 0 & 2 &0 0 & 1 & 0 end{pmatrix},egin{pmatrix} 0 & 0 &2 0&0 & 1 end{pmatrix} ight }     is linearly independent.

small herefore left { egin{pmatrix} 2& 0 &0 1 &0 &0 end{pmatrix} ,egin{pmatrix} 0 & 2 &0 0 & 1 & 0 end{pmatrix},egin{pmatrix} 0 & 0 &2 0&0 & 1 end{pmatrix} ight } form a basis for the kernel.

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pi (p) = p (1

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span-1), (x2-1)

or (z _ 1) + β (2.2-1)=0

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0 2 0 00 2 1 0 0

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