Show the non-zero vectors $v_1,v_2,⋯,v_n$ are linearly independent if $(A−λI)^j+1v_j=0$, where $j=0,1,⋯,n$.A question about linear independenceUnique linear combination problemIs $span(v_1, . . . ,v_m)$ a linearly dependent or linearly independent set of vectors? Also, what will happen if we take span of span?Showing Linear dependencyHow to prove that $v_1,v_2$ are linearly independent if $Spv_1,v_2=Spw_1,w_2$ and $v_1,w_2$ are linearly independent?linear-Independency is retained over a linear MapShow that if Span$v_1,v_2,v_3$=$mathbbR^3$ then $v_1,v_2,v_3$ is linearly independent.Let $v_1,v_2,ldots,v_n$ a linear independent set of vectors, and let $w inlangle v_1,v_2,ldots,v_nrangle$, prove linear independenceThe length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.Proof of $j=1$ where $v_j in span(v_1, …,v_j-1)$

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Show the non-zero vectors $v_1,v_2,⋯,v_n$ are linearly independent if $(A−λI)^j+1v_j=0$, where $j=0,1,⋯,n$.


A question about linear independenceUnique linear combination problemIs $span(v_1, . . . ,v_m)$ a linearly dependent or linearly independent set of vectors? Also, what will happen if we take span of span?Showing Linear dependencyHow to prove that $v_1,v_2$ are linearly independent if $Spv_1,v_2=Spw_1,w_2$ and $v_1,w_2$ are linearly independent?linear-Independency is retained over a linear MapShow that if Span$v_1,v_2,v_3$=$mathbbR^3$ then $v_1,v_2,v_3$ is linearly independent.Let $v_1,v_2,ldots,v_n$ a linear independent set of vectors, and let $w inlangle v_1,v_2,ldots,v_nrangle$, prove linear independenceThe length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.Proof of $j=1$ where $v_j in span(v_1, …,v_j-1)$













1












$begingroup$


My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^j+1$ to both sides, we have
$$ 0 = (A - lambda I)^j+1(a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1) + (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$
And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?










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Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    $v_j=0$ is an obvious counterexample.
    $endgroup$
    – Kavi Rama Murthy
    yesterday






  • 1




    $begingroup$
    Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
    $endgroup$
    – Jens Schwaiger
    yesterday










  • $begingroup$
    I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
    $endgroup$
    – Yiorgos S. Smyrlis
    yesterday
















1












$begingroup$


My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^j+1$ to both sides, we have
$$ 0 = (A - lambda I)^j+1(a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1) + (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$
And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?










share|cite|improve this question









New contributor




Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    $v_j=0$ is an obvious counterexample.
    $endgroup$
    – Kavi Rama Murthy
    yesterday






  • 1




    $begingroup$
    Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
    $endgroup$
    – Jens Schwaiger
    yesterday










  • $begingroup$
    I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
    $endgroup$
    – Yiorgos S. Smyrlis
    yesterday














1












1








1





$begingroup$


My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^j+1$ to both sides, we have
$$ 0 = (A - lambda I)^j+1(a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1) + (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$
And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?










share|cite|improve this question









New contributor




Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is,
$$v_j = a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n.$$ Applying the operator $(A - lambda I)^j+1$ to both sides, we have
$$ 0 = (A - lambda I)^j+1(a_1v_1 + a_2v_2 + cdots + a_j-1v_j-1) + (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - lambda I)^j+1(a_j-1v_j-1 + a_j+1v_j+1 + cdots + a_nv_n).$$
And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?







linear-algebra vector-spaces vectors






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New contributor




Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday







Yejus













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asked yesterday









YejusYejus

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Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Yejus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    $v_j=0$ is an obvious counterexample.
    $endgroup$
    – Kavi Rama Murthy
    yesterday






  • 1




    $begingroup$
    Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
    $endgroup$
    – Jens Schwaiger
    yesterday










  • $begingroup$
    I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
    $endgroup$
    – Yiorgos S. Smyrlis
    yesterday

















  • $begingroup$
    $v_j=0$ is an obvious counterexample.
    $endgroup$
    – Kavi Rama Murthy
    yesterday






  • 1




    $begingroup$
    Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
    $endgroup$
    – Jens Schwaiger
    yesterday










  • $begingroup$
    I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
    $endgroup$
    – Yiorgos S. Smyrlis
    yesterday
















$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
yesterday




$begingroup$
$v_j=0$ is an obvious counterexample.
$endgroup$
– Kavi Rama Murthy
yesterday




1




1




$begingroup$
Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
yesterday




$begingroup$
Possibly the additional condition $(A−λI)^jv_jnot=0$ has to be imposed.
$endgroup$
– Jens Schwaiger
yesterday












$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
yesterday





$begingroup$
I imagine that what you mean is that: $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0.$$
$endgroup$
– Yiorgos S. Smyrlis
yesterday











1 Answer
1






active

oldest

votes


















0












$begingroup$

I use the assumption that
$$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
otherwise it is not true. For example
$$
lambda=0, quad
A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
$$

In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.



Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.



Assume that
$v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
$$
0ne (A-lambda I)^kv_k
=(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
$$






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    0












    $begingroup$

    I use the assumption that
    $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
    otherwise it is not true. For example
    $$
    lambda=0, quad
    A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
    $$

    In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.



    Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.



    Assume that
    $v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
    $$
    0ne (A-lambda I)^kv_k
    =(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
    $$






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      I use the assumption that
      $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
      otherwise it is not true. For example
      $$
      lambda=0, quad
      A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
      $$

      In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.



      Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.



      Assume that
      $v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
      $$
      0ne (A-lambda I)^kv_k
      =(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
      $$






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        I use the assumption that
        $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
        otherwise it is not true. For example
        $$
        lambda=0, quad
        A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
        $$

        In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.



        Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.



        Assume that
        $v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
        $$
        0ne (A-lambda I)^kv_k
        =(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
        $$






        share|cite|improve this answer









        $endgroup$



        I use the assumption that
        $$ (A-lambda I)^jv_jne 0quadtextandquad (A-lambda I)^j+1v_j=0,$$
        otherwise it is not true. For example
        $$
        lambda=0, quad
        A=left(beginarraycc 0 & 1\ 0 & 0endarrayright), quad v_j=binom01, quad jinmathbb N.
        $$

        In order to prove that $v_1,ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,ldots,v_k-1$.



        Clearly, for $k=1$, this holds, since $(A-lambda)v_1ne 0$, and hence $v_1ne 0$.



        Assume that
        $v_k=c_1v_1+cdots+c_k-1v_k-1$. Then
        $$
        0ne (A-lambda I)^kv_k
        =(A-lambda I)^kbig(c_1v_1+cdots+c_k-1v_k-1big)=0
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Yiorgos S. SmyrlisYiorgos S. Smyrlis

        63.4k1385164




        63.4k1385164




















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