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?
$begingroup$
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 ??
linear-algebra vector-spaces
edited yesterday
Maria Mazur
46.4k1160119
asked yesterday
jasminejasmine
1,872418
$endgroup$
add a comment |
$begingroup$
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 ??
linear-algebra vector-spaces
edited yesterday
Maria Mazur
46.4k1160119
asked yesterday
jasminejasmine
1,872418
$endgroup$
add a comment |
$begingroup$
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 ??
linear-algebra vector-spaces
edited yesterday
Maria Mazur
46.4k1160119
asked yesterday
jasminejasmine
1,872418
$endgroup$
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
linear-algebra vector-spaces
edited yesterday
Maria Mazur
46.4k1160119
asked yesterday
jasminejasmine
1,872418
edited yesterday
Maria Mazur
46.4k1160119
asked yesterday
jasminejasmine
1,872418
edited yesterday
Maria Mazur
46.4k1160119
edited yesterday
Maria Mazur
46.4k1160119
edited yesterday
Maria Mazur
46.4k1160119
46.4k1160119
asked yesterday
jasminejasmine
1,872418
asked yesterday
jasminejasmine
1,872418
asked yesterday
jasminejasmine
1,872418
1,872418
add a comment |
add a comment |
1 Answer
1
active
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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.
answered yesterday
Maria MazurMaria Mazur
46.4k1160119
$endgroup$
$begingroup$
oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
$endgroup$
– jasmine
yesterday
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
$begingroup$
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.
answered yesterday
Maria MazurMaria Mazur
46.4k1160119
$endgroup$
$begingroup$
oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
$endgroup$
– jasmine
yesterday
add a comment |
$begingroup$
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.
answered yesterday
Maria MazurMaria Mazur
46.4k1160119
$endgroup$
$begingroup$
oh i missed this counter example didn't came in my minds...and thanks u . i got its now @greedoid
$endgroup$
– jasmine
yesterday
add a comment |
$begingroup$
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.
answered yesterday
Maria MazurMaria Mazur
46.4k1160119
$endgroup$
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.
answered yesterday
Maria MazurMaria Mazur
46.4k1160119
answered yesterday
Maria MazurMaria Mazur
46.4k1160119
answered yesterday
Maria MazurMaria Mazur
46.4k1160119
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
add a comment |
$begingroup$
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
add a comment |
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