Is S is a subspace of$ mathbbP[x]$?Diagonalisation and characteristic polynomialProve that the linear space of polynomials with root $alpha in mathbbR$ is a subspace of $mathbbR[x]_n$Conclusions of The Fundamental Theorem of Algebra over $BbbC$.Prove that every subspace of $V$ invariant under $T$ has even dimension.Show that $T$ has an invariant subspace of dimension $j$ for each $j=1,2,ldots dim V$.Dimension of invariant subspace proof checkEvery linear operator on $mathbbR^5$ has an invariant 3-dimensional subspaceNotation in this proof: If $X$ a finite dimensional NVS over $mathbb C$ then $T$ has a nontrivial inveriant subspace.Prove that $A in mathbbC^m times m$ has $m$ eigenvalues when counted with algebraic multiplicity.consider $mathbbP_n[x]$ and $mathbbP[x]$ over $mathbbR $ Is the following statement is True/ false?

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Is S is a subspace of$ mathbbP[x]$?


Diagonalisation and characteristic polynomialProve that the linear space of polynomials with root $alpha in mathbbR$ is a subspace of $mathbbR[x]_n$Conclusions of The Fundamental Theorem of Algebra over $BbbC$.Prove that every subspace of $V$ invariant under $T$ has even dimension.Show that $T$ has an invariant subspace of dimension $j$ for each $j=1,2,ldots dim V$.Dimension of invariant subspace proof checkEvery linear operator on $mathbbR^5$ has an invariant 3-dimensional subspaceNotation in this proof: If $X$ a finite dimensional NVS over $mathbb C$ then $T$ has a nontrivial inveriant subspace.Prove that $A in mathbbC^m times m$ has $m$ eigenvalues when counted with algebraic multiplicity.consider $mathbbP_n[x]$ and $mathbbP[x]$ over $mathbbR $ Is the following statement is True/ false?













0












$begingroup$



Consider $mathbbP_n[x]$ and $mathbbP[x] $ over $mathbbR.$



Is $ S= p(x) mid p(x) in mathbbP[x] text has degree 3$ is a subspace of $mathbbP[x]$?




My attempt : i thinks yes because by fundamental theorem of algebra every odd degree polynomial has atleast one root that mean $p(x) = x^3 +ax +b =0$ which implies $p(x)=0$



Is its true ??










share|cite|improve this question











$endgroup$
















    0












    $begingroup$



    Consider $mathbbP_n[x]$ and $mathbbP[x] $ over $mathbbR.$



    Is $ S= p(x) mid p(x) in mathbbP[x] text has degree 3$ is a subspace of $mathbbP[x]$?




    My attempt : i thinks yes because by fundamental theorem of algebra every odd degree polynomial has atleast one root that mean $p(x) = x^3 +ax +b =0$ which implies $p(x)=0$



    Is its true ??










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$



      Consider $mathbbP_n[x]$ and $mathbbP[x] $ over $mathbbR.$



      Is $ S= p(x) mid p(x) in mathbbP[x] text has degree 3$ is a subspace of $mathbbP[x]$?




      My attempt : i thinks yes because by fundamental theorem of algebra every odd degree polynomial has atleast one root that mean $p(x) = x^3 +ax +b =0$ which implies $p(x)=0$



      Is its true ??










      share|cite|improve this question











      $endgroup$





      Consider $mathbbP_n[x]$ and $mathbbP[x] $ over $mathbbR.$



      Is $ S= p(x) mid p(x) in mathbbP[x] text has degree 3$ is a subspace of $mathbbP[x]$?




      My attempt : i thinks yes because by fundamental theorem of algebra every odd degree polynomial has atleast one root that mean $p(x) = x^3 +ax +b =0$ which implies $p(x)=0$



      Is its true ??







      linear-algebra vector-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      Maria Mazur

      46.4k1160119




      46.4k1160119










      asked yesterday









      jasminejasmine

      1,872418




      1,872418




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Unfortunately it is not true. One of the requarements is that $0$ is in subspace and that is not true in your case. Polynomial $0$ is not $3$.rd degree polynomial.



          Or other reason, it is not closed for the operation $+$. Say $p(x)=-x^3+2$ and $q(x)=x^3$ are in $S$ but their sum $p(x)+q(x)=2$ is not.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
            $endgroup$
            – jasmine
            yesterday










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          $begingroup$

          Unfortunately it is not true. One of the requarements is that $0$ is in subspace and that is not true in your case. Polynomial $0$ is not $3$.rd degree polynomial.



          Or other reason, it is not closed for the operation $+$. Say $p(x)=-x^3+2$ and $q(x)=x^3$ are in $S$ but their sum $p(x)+q(x)=2$ is not.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
            $endgroup$
            – jasmine
            yesterday















          1












          $begingroup$

          Unfortunately it is not true. One of the requarements is that $0$ is in subspace and that is not true in your case. Polynomial $0$ is not $3$.rd degree polynomial.



          Or other reason, it is not closed for the operation $+$. Say $p(x)=-x^3+2$ and $q(x)=x^3$ are in $S$ but their sum $p(x)+q(x)=2$ is not.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
            $endgroup$
            – jasmine
            yesterday













          1












          1








          1





          $begingroup$

          Unfortunately it is not true. One of the requarements is that $0$ is in subspace and that is not true in your case. Polynomial $0$ is not $3$.rd degree polynomial.



          Or other reason, it is not closed for the operation $+$. Say $p(x)=-x^3+2$ and $q(x)=x^3$ are in $S$ but their sum $p(x)+q(x)=2$ is not.






          share|cite|improve this answer









          $endgroup$



          Unfortunately it is not true. One of the requarements is that $0$ is in subspace and that is not true in your case. Polynomial $0$ is not $3$.rd degree polynomial.



          Or other reason, it is not closed for the operation $+$. Say $p(x)=-x^3+2$ and $q(x)=x^3$ are in $S$ but their sum $p(x)+q(x)=2$ is not.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Maria MazurMaria Mazur

          46.4k1160119




          46.4k1160119











          • $begingroup$
            oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
            $endgroup$
            – jasmine
            yesterday
















          • $begingroup$
            oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
            $endgroup$
            – jasmine
            yesterday















          $begingroup$
          oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
          $endgroup$
          – jasmine
          yesterday




          $begingroup$
          oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
          $endgroup$
          – jasmine
          yesterday

















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