Show that [latex] R^{4}[/latex] and [latex]P_{3}[/latex] are isomorphic.

Consider the linear transformation [latex]T:R^{4}\to P_{3},[/latex] which is defined by [latex] T(a,b,c,d)=a x^{3}+b x^{2}+c x+d.[/latex] Obviously [latex]\dim\left(R^{4}\right)=\dim\left(P_{3}\right)=4.[/latex] Let, [latex]T(a,b,c,d)=\bar0[/latex] [latex]\begin{array}{c}{\Rightarrow a x^{3}+b x^{2}+c x+d=\bar 0}\\ {\Rightarrow a=0,\ b=0,\,c=0,\ d=0}\\ {\therefore\!\operatorname{ker}(T)=\{0\}.}\end{array}[/latex] [latex]\therefore T[/latex] is one-one and onto. [latex]\therefore R^{4}\cong P_{3}.[/latex] [latex]R^{4}[/latex] and [latex]P_{3}[/latex] are isomorphic.

Question 3.4: Show that R^4 and P3 are isomorphic....

Linear Transformation: Examples and Solutions [1470167]

Show that $R^{4}$ and $P_{3}$ are isomorphic.

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Consider the linear transformation $T:R^{4}\to P_{3},$ which is defined by

$T(a,b,c,d)=a x^{3}+b x^{2}+c x+d.$

Obviously $\dim\left(R^{4}\right)=\dim\left(P_{3}\right)=4.$

Let, $T(a,b,c,d)=\bar0$

$\begin{array}{c}{\Rightarrow a x^{3}+b x^{2}+c x+d=\bar 0}\\ {\Rightarrow a=0,\ b=0,\,c=0,\ d=0}\\ {\therefore\!\operatorname{ker}(T)=\{0\}.}\end{array}$

$\therefore T$ is one-one and onto.

$\therefore R^{4}\cong P_{3}.$

$R^{4}$ and $P_{3}$ are isomorphic.