Understanding the Vector Space of Polynomials of a Certain Type of Matrix [closed]












2














Let $A$ be a $4 times 4$ complex diagonal matrix with exactly three distint entries on its diagonal.



(1) What is the dimension of the vector space of polynomials of $A$?



(2) What is the dimension of the vector space of $4 times 4$ complex matrices that commute with $A$?



(3) If $B$ is a $4 times 4$ complex diagonal matrix with exactly three distinct entries on its diagonal, is it similar to a polynomial of $A$?



I am looking for explainations more so than the actual answers.










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closed as off-topic by Brahadeesh, KReiser, Alexander Gruber Nov 30 at 3:08


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, KReiser, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.













  • What are your thoughts on the problem? What have you tried?
    – Omnomnomnom
    Nov 29 at 20:38










  • Are you familiar with Jordan canonical form and with minimal polynomials?
    – Omnomnomnom
    Nov 29 at 20:38










  • @Omnomnomnom I have some familarity with Jordan canonical form.
    – LinearGuy
    Nov 29 at 20:40
















2














Let $A$ be a $4 times 4$ complex diagonal matrix with exactly three distint entries on its diagonal.



(1) What is the dimension of the vector space of polynomials of $A$?



(2) What is the dimension of the vector space of $4 times 4$ complex matrices that commute with $A$?



(3) If $B$ is a $4 times 4$ complex diagonal matrix with exactly three distinct entries on its diagonal, is it similar to a polynomial of $A$?



I am looking for explainations more so than the actual answers.










share|cite|improve this question















closed as off-topic by Brahadeesh, KReiser, Alexander Gruber Nov 30 at 3:08


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, KReiser, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.













  • What are your thoughts on the problem? What have you tried?
    – Omnomnomnom
    Nov 29 at 20:38










  • Are you familiar with Jordan canonical form and with minimal polynomials?
    – Omnomnomnom
    Nov 29 at 20:38










  • @Omnomnomnom I have some familarity with Jordan canonical form.
    – LinearGuy
    Nov 29 at 20:40














2












2








2


1





Let $A$ be a $4 times 4$ complex diagonal matrix with exactly three distint entries on its diagonal.



(1) What is the dimension of the vector space of polynomials of $A$?



(2) What is the dimension of the vector space of $4 times 4$ complex matrices that commute with $A$?



(3) If $B$ is a $4 times 4$ complex diagonal matrix with exactly three distinct entries on its diagonal, is it similar to a polynomial of $A$?



I am looking for explainations more so than the actual answers.










share|cite|improve this question















Let $A$ be a $4 times 4$ complex diagonal matrix with exactly three distint entries on its diagonal.



(1) What is the dimension of the vector space of polynomials of $A$?



(2) What is the dimension of the vector space of $4 times 4$ complex matrices that commute with $A$?



(3) If $B$ is a $4 times 4$ complex diagonal matrix with exactly three distinct entries on its diagonal, is it similar to a polynomial of $A$?



I am looking for explainations more so than the actual answers.







linear-algebra abstract-algebra linear-transformations






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edited Dec 4 at 20:22

























asked Nov 29 at 20:33









LinearGuy

1437




1437




closed as off-topic by Brahadeesh, KReiser, Alexander Gruber Nov 30 at 3:08


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, KReiser, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Brahadeesh, KReiser, Alexander Gruber Nov 30 at 3:08


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, KReiser, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.












  • What are your thoughts on the problem? What have you tried?
    – Omnomnomnom
    Nov 29 at 20:38










  • Are you familiar with Jordan canonical form and with minimal polynomials?
    – Omnomnomnom
    Nov 29 at 20:38










  • @Omnomnomnom I have some familarity with Jordan canonical form.
    – LinearGuy
    Nov 29 at 20:40


















  • What are your thoughts on the problem? What have you tried?
    – Omnomnomnom
    Nov 29 at 20:38










  • Are you familiar with Jordan canonical form and with minimal polynomials?
    – Omnomnomnom
    Nov 29 at 20:38










  • @Omnomnomnom I have some familarity with Jordan canonical form.
    – LinearGuy
    Nov 29 at 20:40
















What are your thoughts on the problem? What have you tried?
– Omnomnomnom
Nov 29 at 20:38




What are your thoughts on the problem? What have you tried?
– Omnomnomnom
Nov 29 at 20:38












Are you familiar with Jordan canonical form and with minimal polynomials?
– Omnomnomnom
Nov 29 at 20:38




Are you familiar with Jordan canonical form and with minimal polynomials?
– Omnomnomnom
Nov 29 at 20:38












@Omnomnomnom I have some familarity with Jordan canonical form.
– LinearGuy
Nov 29 at 20:40




@Omnomnomnom I have some familarity with Jordan canonical form.
– LinearGuy
Nov 29 at 20:40










1 Answer
1






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2














In order to make writing things easier, I will specifically consider the case where the repeated eigenvalue comes first. That is, we have
$$
A = pmatrix{lambda_1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3} = pmatrix{lambda_1 I_{2}\ & lambda_2 \ && lambda_3}
$$

where $I_2$ denotes a size $2$ identity matrix.



Hint for 1: Note that for any polynomial $p$,
$$
p(A) = pmatrix{p(lambda_1)I_{2} \ & p(lambda_2) \ && p(lambda_3)}
$$



Hint for 2: Verify that any block matrix of the form
$$
B = pmatrix{B_1\ & b_2 \ && b_3}
$$

will commute with $A$, where $B_1$ can be any $2 times 2$ matrix and the $b_i$ are scalars. Note too that these are the only matrices that commute with $A$.



Hint for 3: Every polynomial of $A$ is diagonal (and therefore diagonalizable). However, the matrix
$$
pmatrix{lambda_1&1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3}
$$

is neither diagonal nor diagonalizable.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    In order to make writing things easier, I will specifically consider the case where the repeated eigenvalue comes first. That is, we have
    $$
    A = pmatrix{lambda_1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3} = pmatrix{lambda_1 I_{2}\ & lambda_2 \ && lambda_3}
    $$

    where $I_2$ denotes a size $2$ identity matrix.



    Hint for 1: Note that for any polynomial $p$,
    $$
    p(A) = pmatrix{p(lambda_1)I_{2} \ & p(lambda_2) \ && p(lambda_3)}
    $$



    Hint for 2: Verify that any block matrix of the form
    $$
    B = pmatrix{B_1\ & b_2 \ && b_3}
    $$

    will commute with $A$, where $B_1$ can be any $2 times 2$ matrix and the $b_i$ are scalars. Note too that these are the only matrices that commute with $A$.



    Hint for 3: Every polynomial of $A$ is diagonal (and therefore diagonalizable). However, the matrix
    $$
    pmatrix{lambda_1&1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3}
    $$

    is neither diagonal nor diagonalizable.






    share|cite|improve this answer


























      2














      In order to make writing things easier, I will specifically consider the case where the repeated eigenvalue comes first. That is, we have
      $$
      A = pmatrix{lambda_1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3} = pmatrix{lambda_1 I_{2}\ & lambda_2 \ && lambda_3}
      $$

      where $I_2$ denotes a size $2$ identity matrix.



      Hint for 1: Note that for any polynomial $p$,
      $$
      p(A) = pmatrix{p(lambda_1)I_{2} \ & p(lambda_2) \ && p(lambda_3)}
      $$



      Hint for 2: Verify that any block matrix of the form
      $$
      B = pmatrix{B_1\ & b_2 \ && b_3}
      $$

      will commute with $A$, where $B_1$ can be any $2 times 2$ matrix and the $b_i$ are scalars. Note too that these are the only matrices that commute with $A$.



      Hint for 3: Every polynomial of $A$ is diagonal (and therefore diagonalizable). However, the matrix
      $$
      pmatrix{lambda_1&1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3}
      $$

      is neither diagonal nor diagonalizable.






      share|cite|improve this answer
























        2












        2








        2






        In order to make writing things easier, I will specifically consider the case where the repeated eigenvalue comes first. That is, we have
        $$
        A = pmatrix{lambda_1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3} = pmatrix{lambda_1 I_{2}\ & lambda_2 \ && lambda_3}
        $$

        where $I_2$ denotes a size $2$ identity matrix.



        Hint for 1: Note that for any polynomial $p$,
        $$
        p(A) = pmatrix{p(lambda_1)I_{2} \ & p(lambda_2) \ && p(lambda_3)}
        $$



        Hint for 2: Verify that any block matrix of the form
        $$
        B = pmatrix{B_1\ & b_2 \ && b_3}
        $$

        will commute with $A$, where $B_1$ can be any $2 times 2$ matrix and the $b_i$ are scalars. Note too that these are the only matrices that commute with $A$.



        Hint for 3: Every polynomial of $A$ is diagonal (and therefore diagonalizable). However, the matrix
        $$
        pmatrix{lambda_1&1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3}
        $$

        is neither diagonal nor diagonalizable.






        share|cite|improve this answer












        In order to make writing things easier, I will specifically consider the case where the repeated eigenvalue comes first. That is, we have
        $$
        A = pmatrix{lambda_1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3} = pmatrix{lambda_1 I_{2}\ & lambda_2 \ && lambda_3}
        $$

        where $I_2$ denotes a size $2$ identity matrix.



        Hint for 1: Note that for any polynomial $p$,
        $$
        p(A) = pmatrix{p(lambda_1)I_{2} \ & p(lambda_2) \ && p(lambda_3)}
        $$



        Hint for 2: Verify that any block matrix of the form
        $$
        B = pmatrix{B_1\ & b_2 \ && b_3}
        $$

        will commute with $A$, where $B_1$ can be any $2 times 2$ matrix and the $b_i$ are scalars. Note too that these are the only matrices that commute with $A$.



        Hint for 3: Every polynomial of $A$ is diagonal (and therefore diagonalizable). However, the matrix
        $$
        pmatrix{lambda_1&1 \ & lambda_1 \ && lambda_2 \ &&& lambda_3}
        $$

        is neither diagonal nor diagonalizable.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 at 20:46









        Omnomnomnom

        126k788176




        126k788176















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