Prove conditions for upper triangular matrix












-2












$begingroup$


In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:




$textbf{Conditions for upper-triangular matrix:}$



Suppose $T in mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:



(a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular



(b) $Tv_j in span(v_1,...,v_j)$ for each $j=1,...,n$



(c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$




In typical Sheldon Axler fashion, his proof begins as follows:



"The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"



While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$Rightarrow$(b) and (c)$Rightarrow$(b)?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$

















    -2












    $begingroup$


    In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:




    $textbf{Conditions for upper-triangular matrix:}$



    Suppose $T in mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:



    (a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular



    (b) $Tv_j in span(v_1,...,v_j)$ for each $j=1,...,n$



    (c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$




    In typical Sheldon Axler fashion, his proof begins as follows:



    "The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"



    While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$Rightarrow$(b) and (c)$Rightarrow$(b)?






    linear-algebra eigenvalues-eigenvectors







    share|cite|improve this question









    $endgroup$















      -2












      -2








      -2





      $begingroup$


      In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:




      $textbf{Conditions for upper-triangular matrix:}$



      Suppose $T in mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:



      (a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular



      (b) $Tv_j in span(v_1,...,v_j)$ for each $j=1,...,n$



      (c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$




      In typical Sheldon Axler fashion, his proof begins as follows:



      "The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"



      While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$Rightarrow$(b) and (c)$Rightarrow$(b)?






      linear-algebra eigenvalues-eigenvectors







      share|cite|improve this question









      $endgroup$




      In Sheldon Axler's Linear Algebra Done Right, I find the proof of definition 5.26 to be insufficiently rigorous. It states the following:




      $textbf{Conditions for upper-triangular matrix:}$



      Suppose $T in mathscr L(V)$ and $v_1,...,v_n$ is a basis of $V$. Then the following are equivalent:



      (a) the matrix of $T$ with respect to $v_1,...,v_n$ is upper-triangular



      (b) $Tv_j in span(v_1,...,v_j)$ for each $j=1,...,n$



      (c) $span(v_1,...,v_j)$ is invariant under $T$ for each $j=1,...,n$




      In typical Sheldon Axler fashion, his proof begins as follows:



      "The equivalence of (a) and (b) follows easily from the definitions and a moment's thought. Obviously (c) implies (b). Hence to complete the proof, we need only prove that (b) implies (c)......"



      While I do see why this is intuitively true, can anyone provide more truly rigorous (and less condescending) proofs of (a)$Rightarrow$(b) and (c)$Rightarrow$(b)?






      linear-algebra eigenvalues-eigenvectors




      linear-algebra eigenvalues-eigenvectors






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 16 '18 at 7:53









      jmarsjmars

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          $begingroup$

          $a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.



          $c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$






          share|cite|improve this answer









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            $begingroup$

            $a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.



            $c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$






            share|cite|improve this answer









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              0












              $begingroup$

              $a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.



              $c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.



                $c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$






                share|cite|improve this answer









                $endgroup$



                $a$ implies $b$ The $j$-column of the matrix of $T$ with respect to $v_1,...,v_n$ are the cordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$. Since the matrix of $T$ with respect to $(v_1,...,v_n)$ is upper-triangular, this implies that for $i>j$, the $i$-coordinates of $T(v_j)$ with respect to $(v_1,...,v_n)$ is $0$, this implies that $T(v_j)in span(v_1,...,v_j)$.



                $c$ implies $b$ if $Span(v_1,...,v_j)$ is invariant by $T$, it implies that $T(v_j)in span(v_1,...,v_j)$ since $v_jin (v_1,...,v_j)$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 16 '18 at 8:06









                Tsemo AristideTsemo Aristide

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