Linear transformations, prove that zero function belongs to vector space












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Let $ninmathbb{N}$ and $a_0,a_1,ldots,a_ninmathbb{R}$ such that $a_0< a_1dots < a_{n-1}< a_n$. Let $Esubset mathcal{F}([a_0,a_n],mathbb{R}) $ a vector subspace (with the usual operations of $mathcal{F}([a_0,a_n],mathbb{R})$), where begin{equation*} E: =left{ginmathcal{F}([a_0,a_n],mathbb{R});; text{such that} ;; forall iinleft{0,1,ldots,n-1right}; : ; grestriction{[a_i,a_{i+1}]}inmathcal{P}_1(mathbb{R}) right}end{equation*}
Prove that the zero function $thetainmathcal{F}([a_0,a_n],mathbb{R})$ belongs to $E$. Also prove that $dim(E) = n+1$.




First of all:begin{align*}mathcal{F}([a_0,a_n],mathbb{R})&;; &text{is the space of functions with domain in $[a_0,a_n]$ and codomain in}; mathbb{R} \ grestriction{[a_i,a_{i+1}]}&; ; &text{is the restriction of g to $[a_i,a_{i+1}]$ } end{align*}



For the first part, i've tried to prove that over any $[a_i,a_{i+1}]$ we can let $theta = 0 inmathcal{P}_1(mathbb{R})$ as the zero function, but i don't know how to proceed formally. To prove that $dim(E) = n + 1$ i think that maybe i can found a bijection on $E$, and then use the Rank-nullity theorem. How to proceed further?



Thanks in advance.






linear-algebra vector-spaces linear-transformations







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    $begingroup$



    Let $ninmathbb{N}$ and $a_0,a_1,ldots,a_ninmathbb{R}$ such that $a_0< a_1dots < a_{n-1}< a_n$. Let $Esubset mathcal{F}([a_0,a_n],mathbb{R}) $ a vector subspace (with the usual operations of $mathcal{F}([a_0,a_n],mathbb{R})$), where begin{equation*} E: =left{ginmathcal{F}([a_0,a_n],mathbb{R});; text{such that} ;; forall iinleft{0,1,ldots,n-1right}; : ; grestriction{[a_i,a_{i+1}]}inmathcal{P}_1(mathbb{R}) right}end{equation*}
    Prove that the zero function $thetainmathcal{F}([a_0,a_n],mathbb{R})$ belongs to $E$. Also prove that $dim(E) = n+1$.




    First of all:begin{align*}mathcal{F}([a_0,a_n],mathbb{R})&;; &text{is the space of functions with domain in $[a_0,a_n]$ and codomain in}; mathbb{R} \ grestriction{[a_i,a_{i+1}]}&; ; &text{is the restriction of g to $[a_i,a_{i+1}]$ } end{align*}



    For the first part, i've tried to prove that over any $[a_i,a_{i+1}]$ we can let $theta = 0 inmathcal{P}_1(mathbb{R})$ as the zero function, but i don't know how to proceed formally. To prove that $dim(E) = n + 1$ i think that maybe i can found a bijection on $E$, and then use the Rank-nullity theorem. How to proceed further?



    Thanks in advance.






    linear-algebra vector-spaces linear-transformations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      Let $ninmathbb{N}$ and $a_0,a_1,ldots,a_ninmathbb{R}$ such that $a_0< a_1dots < a_{n-1}< a_n$. Let $Esubset mathcal{F}([a_0,a_n],mathbb{R}) $ a vector subspace (with the usual operations of $mathcal{F}([a_0,a_n],mathbb{R})$), where begin{equation*} E: =left{ginmathcal{F}([a_0,a_n],mathbb{R});; text{such that} ;; forall iinleft{0,1,ldots,n-1right}; : ; grestriction{[a_i,a_{i+1}]}inmathcal{P}_1(mathbb{R}) right}end{equation*}
      Prove that the zero function $thetainmathcal{F}([a_0,a_n],mathbb{R})$ belongs to $E$. Also prove that $dim(E) = n+1$.




      First of all:begin{align*}mathcal{F}([a_0,a_n],mathbb{R})&;; &text{is the space of functions with domain in $[a_0,a_n]$ and codomain in}; mathbb{R} \ grestriction{[a_i,a_{i+1}]}&; ; &text{is the restriction of g to $[a_i,a_{i+1}]$ } end{align*}



      For the first part, i've tried to prove that over any $[a_i,a_{i+1}]$ we can let $theta = 0 inmathcal{P}_1(mathbb{R})$ as the zero function, but i don't know how to proceed formally. To prove that $dim(E) = n + 1$ i think that maybe i can found a bijection on $E$, and then use the Rank-nullity theorem. How to proceed further?



      Thanks in advance.






      linear-algebra vector-spaces linear-transformations







      share|cite|improve this question









      $endgroup$





      Let $ninmathbb{N}$ and $a_0,a_1,ldots,a_ninmathbb{R}$ such that $a_0< a_1dots < a_{n-1}< a_n$. Let $Esubset mathcal{F}([a_0,a_n],mathbb{R}) $ a vector subspace (with the usual operations of $mathcal{F}([a_0,a_n],mathbb{R})$), where begin{equation*} E: =left{ginmathcal{F}([a_0,a_n],mathbb{R});; text{such that} ;; forall iinleft{0,1,ldots,n-1right}; : ; grestriction{[a_i,a_{i+1}]}inmathcal{P}_1(mathbb{R}) right}end{equation*}
      Prove that the zero function $thetainmathcal{F}([a_0,a_n],mathbb{R})$ belongs to $E$. Also prove that $dim(E) = n+1$.




      First of all:begin{align*}mathcal{F}([a_0,a_n],mathbb{R})&;; &text{is the space of functions with domain in $[a_0,a_n]$ and codomain in}; mathbb{R} \ grestriction{[a_i,a_{i+1}]}&; ; &text{is the restriction of g to $[a_i,a_{i+1}]$ } end{align*}



      For the first part, i've tried to prove that over any $[a_i,a_{i+1}]$ we can let $theta = 0 inmathcal{P}_1(mathbb{R})$ as the zero function, but i don't know how to proceed formally. To prove that $dim(E) = n + 1$ i think that maybe i can found a bijection on $E$, and then use the Rank-nullity theorem. How to proceed further?



      Thanks in advance.






      linear-algebra vector-spaces linear-transformations




      linear-algebra vector-spaces linear-transformations






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      asked Jan 6 at 5:02









      Raúl AsteteRaúl Astete

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          Yes, for the first part the solution simply amounts to the observation that the zero function restricted to each interval is still a zero function (on that interval), which clearly belongs to $mathcal{P}_1(mathbb{R})$.



          For the second part, the key idea is that the values of such a function $gin Esubsetmathcal{F}([a_0,a_n],mathbb{R})$ at the points $a_0,a_1,ldots,a_n$ completely define this function, as we would simply connect each pair of consecutive points with the unique linear function (geometrically, with the unique straight line segment) passing thru them. To implement this idea, we can consider the linear transformation
          $$T:Etomathbb{R}^{n+1} quad text{via} quad T(g)=(g(a_0),g(a_1),ldots,g(a_n)).$$
          Show that $T$ is injective (its kernel is only the zero function) and that $T$ is surjective (because for any list of $n+1$ values there's a piecewise-linear function in $E$ that interpolates it), and you're done.






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            $begingroup$

            Yes, for the first part the solution simply amounts to the observation that the zero function restricted to each interval is still a zero function (on that interval), which clearly belongs to $mathcal{P}_1(mathbb{R})$.



            For the second part, the key idea is that the values of such a function $gin Esubsetmathcal{F}([a_0,a_n],mathbb{R})$ at the points $a_0,a_1,ldots,a_n$ completely define this function, as we would simply connect each pair of consecutive points with the unique linear function (geometrically, with the unique straight line segment) passing thru them. To implement this idea, we can consider the linear transformation
            $$T:Etomathbb{R}^{n+1} quad text{via} quad T(g)=(g(a_0),g(a_1),ldots,g(a_n)).$$
            Show that $T$ is injective (its kernel is only the zero function) and that $T$ is surjective (because for any list of $n+1$ values there's a piecewise-linear function in $E$ that interpolates it), and you're done.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Yes, for the first part the solution simply amounts to the observation that the zero function restricted to each interval is still a zero function (on that interval), which clearly belongs to $mathcal{P}_1(mathbb{R})$.



              For the second part, the key idea is that the values of such a function $gin Esubsetmathcal{F}([a_0,a_n],mathbb{R})$ at the points $a_0,a_1,ldots,a_n$ completely define this function, as we would simply connect each pair of consecutive points with the unique linear function (geometrically, with the unique straight line segment) passing thru them. To implement this idea, we can consider the linear transformation
              $$T:Etomathbb{R}^{n+1} quad text{via} quad T(g)=(g(a_0),g(a_1),ldots,g(a_n)).$$
              Show that $T$ is injective (its kernel is only the zero function) and that $T$ is surjective (because for any list of $n+1$ values there's a piecewise-linear function in $E$ that interpolates it), and you're done.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Yes, for the first part the solution simply amounts to the observation that the zero function restricted to each interval is still a zero function (on that interval), which clearly belongs to $mathcal{P}_1(mathbb{R})$.



                For the second part, the key idea is that the values of such a function $gin Esubsetmathcal{F}([a_0,a_n],mathbb{R})$ at the points $a_0,a_1,ldots,a_n$ completely define this function, as we would simply connect each pair of consecutive points with the unique linear function (geometrically, with the unique straight line segment) passing thru them. To implement this idea, we can consider the linear transformation
                $$T:Etomathbb{R}^{n+1} quad text{via} quad T(g)=(g(a_0),g(a_1),ldots,g(a_n)).$$
                Show that $T$ is injective (its kernel is only the zero function) and that $T$ is surjective (because for any list of $n+1$ values there's a piecewise-linear function in $E$ that interpolates it), and you're done.






                share|cite|improve this answer









                $endgroup$



                Yes, for the first part the solution simply amounts to the observation that the zero function restricted to each interval is still a zero function (on that interval), which clearly belongs to $mathcal{P}_1(mathbb{R})$.



                For the second part, the key idea is that the values of such a function $gin Esubsetmathcal{F}([a_0,a_n],mathbb{R})$ at the points $a_0,a_1,ldots,a_n$ completely define this function, as we would simply connect each pair of consecutive points with the unique linear function (geometrically, with the unique straight line segment) passing thru them. To implement this idea, we can consider the linear transformation
                $$T:Etomathbb{R}^{n+1} quad text{via} quad T(g)=(g(a_0),g(a_1),ldots,g(a_n)).$$
                Show that $T$ is injective (its kernel is only the zero function) and that $T$ is surjective (because for any list of $n+1$ values there's a piecewise-linear function in $E$ that interpolates it), and you're done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 5:51









                zipirovichzipirovich

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                11.3k11731






























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