How do I prove this equality holds? If A is similar to a diagonal matrix with diagonal $a_1,…,a_n$ then...












0












$begingroup$


I'm not quite sure how to prove whats written in the title. We have a field $F$ and a matrix $A$ that is similar to the diagonal matrix $D=operatorname{diag}(a_1,...,a_n)=begin{bmatrix} a_1& 0 &0&. . .& 0 \ 0& a_2 &0&...&0 \ . \. \.\0&0&0 &...&a_nend{bmatrix}$
I need to show that $(A-a_1I) cdot (A-a_2I) cdot ... cdot (A-a_nI)=0$.
I tried this for long time but I am not sure how to do this. I know there is a matrix $P$ such that $A=P^{-1}DP$, but I'm not sure how that helps. It was written that it is enough to prove this for n=2 (n is the number of rows and columns of A) and I was able to do this for n=2 but I don't know how that helps me to prove this generally. If anyone can provide any help (I'd love to get some hints first, and only then the full solution if I'm still stuck), It will help me a lot. Thanks!

**All I know about similar matrices is that there exists $P$ such that $A=P^{-1}BP$, so the solution should be quite basic, using that fact only.






linear-algebra matrices







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  • 1




    $begingroup$
    Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
    $endgroup$
    – dezdichado
    Dec 28 '18 at 21:01










  • $begingroup$
    @dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
    $endgroup$
    – Omer
    Dec 28 '18 at 21:07
















0












$begingroup$


I'm not quite sure how to prove whats written in the title. We have a field $F$ and a matrix $A$ that is similar to the diagonal matrix $D=operatorname{diag}(a_1,...,a_n)=begin{bmatrix} a_1& 0 &0&. . .& 0 \ 0& a_2 &0&...&0 \ . \. \.\0&0&0 &...&a_nend{bmatrix}$
I need to show that $(A-a_1I) cdot (A-a_2I) cdot ... cdot (A-a_nI)=0$.
I tried this for long time but I am not sure how to do this. I know there is a matrix $P$ such that $A=P^{-1}DP$, but I'm not sure how that helps. It was written that it is enough to prove this for n=2 (n is the number of rows and columns of A) and I was able to do this for n=2 but I don't know how that helps me to prove this generally. If anyone can provide any help (I'd love to get some hints first, and only then the full solution if I'm still stuck), It will help me a lot. Thanks!

**All I know about similar matrices is that there exists $P$ such that $A=P^{-1}BP$, so the solution should be quite basic, using that fact only.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
    $endgroup$
    – dezdichado
    Dec 28 '18 at 21:01










  • $begingroup$
    @dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
    $endgroup$
    – Omer
    Dec 28 '18 at 21:07














0












0








0





$begingroup$


I'm not quite sure how to prove whats written in the title. We have a field $F$ and a matrix $A$ that is similar to the diagonal matrix $D=operatorname{diag}(a_1,...,a_n)=begin{bmatrix} a_1& 0 &0&. . .& 0 \ 0& a_2 &0&...&0 \ . \. \.\0&0&0 &...&a_nend{bmatrix}$
I need to show that $(A-a_1I) cdot (A-a_2I) cdot ... cdot (A-a_nI)=0$.
I tried this for long time but I am not sure how to do this. I know there is a matrix $P$ such that $A=P^{-1}DP$, but I'm not sure how that helps. It was written that it is enough to prove this for n=2 (n is the number of rows and columns of A) and I was able to do this for n=2 but I don't know how that helps me to prove this generally. If anyone can provide any help (I'd love to get some hints first, and only then the full solution if I'm still stuck), It will help me a lot. Thanks!

**All I know about similar matrices is that there exists $P$ such that $A=P^{-1}BP$, so the solution should be quite basic, using that fact only.






linear-algebra matrices







share|cite|improve this question











$endgroup$




I'm not quite sure how to prove whats written in the title. We have a field $F$ and a matrix $A$ that is similar to the diagonal matrix $D=operatorname{diag}(a_1,...,a_n)=begin{bmatrix} a_1& 0 &0&. . .& 0 \ 0& a_2 &0&...&0 \ . \. \.\0&0&0 &...&a_nend{bmatrix}$
I need to show that $(A-a_1I) cdot (A-a_2I) cdot ... cdot (A-a_nI)=0$.
I tried this for long time but I am not sure how to do this. I know there is a matrix $P$ such that $A=P^{-1}DP$, but I'm not sure how that helps. It was written that it is enough to prove this for n=2 (n is the number of rows and columns of A) and I was able to do this for n=2 but I don't know how that helps me to prove this generally. If anyone can provide any help (I'd love to get some hints first, and only then the full solution if I'm still stuck), It will help me a lot. Thanks!

**All I know about similar matrices is that there exists $P$ such that $A=P^{-1}BP$, so the solution should be quite basic, using that fact only.






linear-algebra matrices




linear-algebra matrices






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edited Dec 28 '18 at 21:09







Omer

















asked Dec 28 '18 at 20:56









OmerOmer

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535110








  • 1




    $begingroup$
    Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
    $endgroup$
    – dezdichado
    Dec 28 '18 at 21:01










  • $begingroup$
    @dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
    $endgroup$
    – Omer
    Dec 28 '18 at 21:07














  • 1




    $begingroup$
    Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
    $endgroup$
    – dezdichado
    Dec 28 '18 at 21:01










  • $begingroup$
    @dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
    $endgroup$
    – Omer
    Dec 28 '18 at 21:07








1




1




$begingroup$
Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
$endgroup$
– dezdichado
Dec 28 '18 at 21:01




$begingroup$
Hint: Prove that similar matrices has the same characteristic polynomial and the fact a matrix satisfies its own characteristic polynomial.
$endgroup$
– dezdichado
Dec 28 '18 at 21:01












$begingroup$
@dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
$endgroup$
– Omer
Dec 28 '18 at 21:07




$begingroup$
@dezdichado Hi, I haven't learnt about characterisitc polynomial yet, all I know about similar matrices is that there exists P such that A=P^-1*B*P, and that's it. They said in the question that it is enough to show this for n=2 and than explain how that also implies it for all values of n. sorry for not mentioning that, I'll edit my post now.
$endgroup$
– Omer
Dec 28 '18 at 21:07










4 Answers
4






active

oldest

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0












$begingroup$

Hint:



First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.



Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
    $$
    p(B)=c_0I_n+c_1B+dots+c_kB^k=0
    $$
    (briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.



    Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Hint: PT
      $$
      A= P^{-1}begin{bmatrix}
      a_1 & & \
      &a_2 & \
      & & ddots \
      end{bmatrix}Pimplies
      A-a_{1}I= P^{-1}begin{bmatrix}
      0 & & \
      &a_{2}-a_{1} & \
      & & ddots
      end{bmatrix}P
      $$



      then multiply all the $A-a_{i}I$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.




        Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.




        [Make sure you see why Fact 1 is true.]



        On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.



        This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.






        share|cite|improve this answer











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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Hint:



          First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.



          Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$






          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            Hint:



            First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.



            Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$






            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              Hint:



              First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.



              Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$






              share|cite|improve this answer









              $endgroup$



              Hint:



              First check that $;A-alpha I=P^{-1}DP-alpha I=P^{-1}(D-alpha I)P$.



              Second, check that $;(D-alpha_1I)(D-alpha_2I)dots(D-alpha_nI)=0$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 28 '18 at 21:22









              BernardBernard

              124k741117




              124k741117























                  1












                  $begingroup$

                  Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
                  $$
                  p(B)=c_0I_n+c_1B+dots+c_kB^k=0
                  $$
                  (briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.



                  Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
                    $$
                    p(B)=c_0I_n+c_1B+dots+c_kB^k=0
                    $$
                    (briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.



                    Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
                      $$
                      p(B)=c_0I_n+c_1B+dots+c_kB^k=0
                      $$
                      (briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.



                      Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.






                      share|cite|improve this answer









                      $endgroup$



                      Prove that if $p(x)=c_0+c_1x+dots+c^kx^k$ is a polynomial and
                      $$
                      p(B)=c_0I_n+c_1B+dots+c_kB^k=0
                      $$
                      (briefly, $B$ satisfies $p$) then also $p(A)=0$, whenever $A$ is similar to $B$.



                      Next prove that $D=operatorname{diag}(a_1,dots,a_n)$ (the diagonal matrix with the given elements on the diagonal) satisfies $p(x)=(x-a_1)(x-a_2)dots(x-a_n)$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 28 '18 at 21:11









                      egregegreg

                      186k1486208




                      186k1486208























                          0












                          $begingroup$

                          Hint: PT
                          $$
                          A= P^{-1}begin{bmatrix}
                          a_1 & & \
                          &a_2 & \
                          & & ddots \
                          end{bmatrix}Pimplies
                          A-a_{1}I= P^{-1}begin{bmatrix}
                          0 & & \
                          &a_{2}-a_{1} & \
                          & & ddots
                          end{bmatrix}P
                          $$



                          then multiply all the $A-a_{i}I$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Hint: PT
                            $$
                            A= P^{-1}begin{bmatrix}
                            a_1 & & \
                            &a_2 & \
                            & & ddots \
                            end{bmatrix}Pimplies
                            A-a_{1}I= P^{-1}begin{bmatrix}
                            0 & & \
                            &a_{2}-a_{1} & \
                            & & ddots
                            end{bmatrix}P
                            $$



                            then multiply all the $A-a_{i}I$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Hint: PT
                              $$
                              A= P^{-1}begin{bmatrix}
                              a_1 & & \
                              &a_2 & \
                              & & ddots \
                              end{bmatrix}Pimplies
                              A-a_{1}I= P^{-1}begin{bmatrix}
                              0 & & \
                              &a_{2}-a_{1} & \
                              & & ddots
                              end{bmatrix}P
                              $$



                              then multiply all the $A-a_{i}I$.






                              share|cite|improve this answer









                              $endgroup$



                              Hint: PT
                              $$
                              A= P^{-1}begin{bmatrix}
                              a_1 & & \
                              &a_2 & \
                              & & ddots \
                              end{bmatrix}Pimplies
                              A-a_{1}I= P^{-1}begin{bmatrix}
                              0 & & \
                              &a_{2}-a_{1} & \
                              & & ddots
                              end{bmatrix}P
                              $$



                              then multiply all the $A-a_{i}I$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 28 '18 at 21:14









                              zimbra314zimbra314

                              640312




                              640312























                                  0












                                  $begingroup$

                                  Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.




                                  Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.




                                  [Make sure you see why Fact 1 is true.]



                                  On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.



                                  This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.




                                    Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.




                                    [Make sure you see why Fact 1 is true.]



                                    On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.



                                    This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.




                                      Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.




                                      [Make sure you see why Fact 1 is true.]



                                      On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.



                                      This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.






                                      share|cite|improve this answer











                                      $endgroup$



                                      Let $P$ be as in the OP and let $e_k$; $k=1,2,ldots, n$ denote the vector such that the $k$-th component is 1 and every other component is 0.




                                      Fact 1: Then on the one hand, the $P^{-1}e_k$s span $mathbb{F}^n$, so if there is a matrix $B$ such that $B(P^{-1}e_k)$ is 0 for all $k$, then $B$ must be the 0 matrix.




                                      [Make sure you see why Fact 1 is true.]



                                      On the other hand, $A(P^{-1} e_k) = P^{-1}DP (P^{-1}e_k) = P^{-1}D(PP^{-1})e_k$ $ = P^{-1}De_k = P^{-1} a_ke_k = a_kP^{-1}e_k$. Thus $(A-a_kI)(P^{-1}e_k) = 0$ for each such $k$.



                                      This implies $[prod_{l=1}^n (A-a_lI)](P^{-1}e_k) = 0$ for each such $k$. This and Fact 1 implies $[prod_l (A-a_lI)]$ is the 0 matrix.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 28 '18 at 21:21

























                                      answered Dec 28 '18 at 21:16









                                      MikeMike

                                      4,641512




                                      4,641512






























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                                          How to change which sound is reproduced for terminal bell?

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                                          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
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                                          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                                          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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                                          Can I use Tabulator js library in my java Spring + Thymeleaf project?

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                                          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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