True or False, diagonalization problem












1















Let $B_c= left{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)right}$, $T:mathbb{R}^4rightarrow mathbb{R}^4$ a linear operator such that begin{equation} detleft[T-Ilambdaright]_{B_{c}}=(2-lambda)^4qquad text{and}qquad T((0,0,0,1)) = (1,0,0,1)end{equation}
Is T diagonalizable?




I've tried to use the following reasoning: Since the algebraic multiplicity of the eigenvalue $2$ is $4$, we know that $T$ is diagonalizable if (and only if) the dimension of the eigenspace associated with the eigenvalue $2$ is $4$. But we also know that there's a vector in $mathbb{R}^4$ that don't belong to $S_{2}$ (because $(0,0,0,1)$ is not an eigenvector). Can I use this argument to show that $T$ is not diagonalizable? Thanks in advance






linear-algebra problem-solving diagonalization







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    Yes, that's fully correct.
    – egreg
    Dec 31 '18 at 23:44
















1















Let $B_c= left{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)right}$, $T:mathbb{R}^4rightarrow mathbb{R}^4$ a linear operator such that begin{equation} detleft[T-Ilambdaright]_{B_{c}}=(2-lambda)^4qquad text{and}qquad T((0,0,0,1)) = (1,0,0,1)end{equation}
Is T diagonalizable?




I've tried to use the following reasoning: Since the algebraic multiplicity of the eigenvalue $2$ is $4$, we know that $T$ is diagonalizable if (and only if) the dimension of the eigenspace associated with the eigenvalue $2$ is $4$. But we also know that there's a vector in $mathbb{R}^4$ that don't belong to $S_{2}$ (because $(0,0,0,1)$ is not an eigenvector). Can I use this argument to show that $T$ is not diagonalizable? Thanks in advance






linear-algebra problem-solving diagonalization







share|cite|improve this question


















  • 1




    Yes, that's fully correct.
    – egreg
    Dec 31 '18 at 23:44














1












1








1








Let $B_c= left{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)right}$, $T:mathbb{R}^4rightarrow mathbb{R}^4$ a linear operator such that begin{equation} detleft[T-Ilambdaright]_{B_{c}}=(2-lambda)^4qquad text{and}qquad T((0,0,0,1)) = (1,0,0,1)end{equation}
Is T diagonalizable?




I've tried to use the following reasoning: Since the algebraic multiplicity of the eigenvalue $2$ is $4$, we know that $T$ is diagonalizable if (and only if) the dimension of the eigenspace associated with the eigenvalue $2$ is $4$. But we also know that there's a vector in $mathbb{R}^4$ that don't belong to $S_{2}$ (because $(0,0,0,1)$ is not an eigenvector). Can I use this argument to show that $T$ is not diagonalizable? Thanks in advance






linear-algebra problem-solving diagonalization







share|cite|improve this question














Let $B_c= left{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)right}$, $T:mathbb{R}^4rightarrow mathbb{R}^4$ a linear operator such that begin{equation} detleft[T-Ilambdaright]_{B_{c}}=(2-lambda)^4qquad text{and}qquad T((0,0,0,1)) = (1,0,0,1)end{equation}
Is T diagonalizable?




I've tried to use the following reasoning: Since the algebraic multiplicity of the eigenvalue $2$ is $4$, we know that $T$ is diagonalizable if (and only if) the dimension of the eigenspace associated with the eigenvalue $2$ is $4$. But we also know that there's a vector in $mathbb{R}^4$ that don't belong to $S_{2}$ (because $(0,0,0,1)$ is not an eigenvector). Can I use this argument to show that $T$ is not diagonalizable? Thanks in advance






linear-algebra problem-solving diagonalization




linear-algebra problem-solving diagonalization






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asked Dec 31 '18 at 21:02









Raúl AsteteRaúl Astete

496




496








  • 1




    Yes, that's fully correct.
    – egreg
    Dec 31 '18 at 23:44














  • 1




    Yes, that's fully correct.
    – egreg
    Dec 31 '18 at 23:44








1




1




Yes, that's fully correct.
– egreg
Dec 31 '18 at 23:44




Yes, that's fully correct.
– egreg
Dec 31 '18 at 23:44










3 Answers
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1















An endomorphism φ of a finite dimensional vector space over a field F
is diagonalizable if and only if its minimal polynomial factors
completely over F into distinct linear factors. The fact that there is
only one factor X − λ for every eigenvalue




from https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)



I like to say that a matrix diagonalizes as some $P^{-1}AP = D$ if and only if the minimal polynomial is squarefree. They phrase that as "factors...into distinct linear factors"



If the minimal polynomial really were $lambda - 2,$ we would have $A-2I = 0,$ so that $A=2I,$ and everything would be an eigenvector. Since they give a vector that is not an eigenvector, that does it. The minimal polynomial is not linear, it is one of $(lambda-2)^2$ or $(lambda-2)^3$ or $(lambda-2)^4.$ Note that i am using the reverse order polynomial $det(lambda I - A).$ Oh, and $A$ is the square matrix defined by $T$






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    1














    Since $T((0,0,0,1)) neq 2 cdot (0,0,0,1)$ we know that $(0,0,0,1)$ isn't in the eigenspace $operatorname{Ker}(T-2I)$. The dimension of the eigenspace is therefore strictly less than the dimension of $mathbb{R}^4$.






    share|cite|improve this answer





























      1














      If diagonalized, the diagonal matrix would simply be $$D=2I_4$$
      As the result your matrix is $$B^{-1}DB=2I_4$$



      That contradicts $$A(1,0,0,0)^T=(1,0,0,1)^T$$






      share|cite|improve this answer





















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        3 Answers
        3






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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

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        1















        An endomorphism φ of a finite dimensional vector space over a field F
        is diagonalizable if and only if its minimal polynomial factors
        completely over F into distinct linear factors. The fact that there is
        only one factor X − λ for every eigenvalue




        from https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)



        I like to say that a matrix diagonalizes as some $P^{-1}AP = D$ if and only if the minimal polynomial is squarefree. They phrase that as "factors...into distinct linear factors"



        If the minimal polynomial really were $lambda - 2,$ we would have $A-2I = 0,$ so that $A=2I,$ and everything would be an eigenvector. Since they give a vector that is not an eigenvector, that does it. The minimal polynomial is not linear, it is one of $(lambda-2)^2$ or $(lambda-2)^3$ or $(lambda-2)^4.$ Note that i am using the reverse order polynomial $det(lambda I - A).$ Oh, and $A$ is the square matrix defined by $T$






        share|cite|improve this answer


























          1















          An endomorphism φ of a finite dimensional vector space over a field F
          is diagonalizable if and only if its minimal polynomial factors
          completely over F into distinct linear factors. The fact that there is
          only one factor X − λ for every eigenvalue




          from https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)



          I like to say that a matrix diagonalizes as some $P^{-1}AP = D$ if and only if the minimal polynomial is squarefree. They phrase that as "factors...into distinct linear factors"



          If the minimal polynomial really were $lambda - 2,$ we would have $A-2I = 0,$ so that $A=2I,$ and everything would be an eigenvector. Since they give a vector that is not an eigenvector, that does it. The minimal polynomial is not linear, it is one of $(lambda-2)^2$ or $(lambda-2)^3$ or $(lambda-2)^4.$ Note that i am using the reverse order polynomial $det(lambda I - A).$ Oh, and $A$ is the square matrix defined by $T$






          share|cite|improve this answer
























            1












            1








            1







            An endomorphism φ of a finite dimensional vector space over a field F
            is diagonalizable if and only if its minimal polynomial factors
            completely over F into distinct linear factors. The fact that there is
            only one factor X − λ for every eigenvalue




            from https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)



            I like to say that a matrix diagonalizes as some $P^{-1}AP = D$ if and only if the minimal polynomial is squarefree. They phrase that as "factors...into distinct linear factors"



            If the minimal polynomial really were $lambda - 2,$ we would have $A-2I = 0,$ so that $A=2I,$ and everything would be an eigenvector. Since they give a vector that is not an eigenvector, that does it. The minimal polynomial is not linear, it is one of $(lambda-2)^2$ or $(lambda-2)^3$ or $(lambda-2)^4.$ Note that i am using the reverse order polynomial $det(lambda I - A).$ Oh, and $A$ is the square matrix defined by $T$






            share|cite|improve this answer













            An endomorphism φ of a finite dimensional vector space over a field F
            is diagonalizable if and only if its minimal polynomial factors
            completely over F into distinct linear factors. The fact that there is
            only one factor X − λ for every eigenvalue




            from https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)



            I like to say that a matrix diagonalizes as some $P^{-1}AP = D$ if and only if the minimal polynomial is squarefree. They phrase that as "factors...into distinct linear factors"



            If the minimal polynomial really were $lambda - 2,$ we would have $A-2I = 0,$ so that $A=2I,$ and everything would be an eigenvector. Since they give a vector that is not an eigenvector, that does it. The minimal polynomial is not linear, it is one of $(lambda-2)^2$ or $(lambda-2)^3$ or $(lambda-2)^4.$ Note that i am using the reverse order polynomial $det(lambda I - A).$ Oh, and $A$ is the square matrix defined by $T$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 31 '18 at 21:22









            Will JagyWill Jagy

            102k5100199




            102k5100199























                1














                Since $T((0,0,0,1)) neq 2 cdot (0,0,0,1)$ we know that $(0,0,0,1)$ isn't in the eigenspace $operatorname{Ker}(T-2I)$. The dimension of the eigenspace is therefore strictly less than the dimension of $mathbb{R}^4$.






                share|cite|improve this answer


























                  1














                  Since $T((0,0,0,1)) neq 2 cdot (0,0,0,1)$ we know that $(0,0,0,1)$ isn't in the eigenspace $operatorname{Ker}(T-2I)$. The dimension of the eigenspace is therefore strictly less than the dimension of $mathbb{R}^4$.






                  share|cite|improve this answer
























                    1












                    1








                    1






                    Since $T((0,0,0,1)) neq 2 cdot (0,0,0,1)$ we know that $(0,0,0,1)$ isn't in the eigenspace $operatorname{Ker}(T-2I)$. The dimension of the eigenspace is therefore strictly less than the dimension of $mathbb{R}^4$.






                    share|cite|improve this answer












                    Since $T((0,0,0,1)) neq 2 cdot (0,0,0,1)$ we know that $(0,0,0,1)$ isn't in the eigenspace $operatorname{Ker}(T-2I)$. The dimension of the eigenspace is therefore strictly less than the dimension of $mathbb{R}^4$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 31 '18 at 21:31









                    mephistolotlmephistolotl

                    4701413




                    4701413























                        1














                        If diagonalized, the diagonal matrix would simply be $$D=2I_4$$
                        As the result your matrix is $$B^{-1}DB=2I_4$$



                        That contradicts $$A(1,0,0,0)^T=(1,0,0,1)^T$$






                        share|cite|improve this answer


























                          1














                          If diagonalized, the diagonal matrix would simply be $$D=2I_4$$
                          As the result your matrix is $$B^{-1}DB=2I_4$$



                          That contradicts $$A(1,0,0,0)^T=(1,0,0,1)^T$$






                          share|cite|improve this answer
























                            1












                            1








                            1






                            If diagonalized, the diagonal matrix would simply be $$D=2I_4$$
                            As the result your matrix is $$B^{-1}DB=2I_4$$



                            That contradicts $$A(1,0,0,0)^T=(1,0,0,1)^T$$






                            share|cite|improve this answer












                            If diagonalized, the diagonal matrix would simply be $$D=2I_4$$
                            As the result your matrix is $$B^{-1}DB=2I_4$$



                            That contradicts $$A(1,0,0,0)^T=(1,0,0,1)^T$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 31 '18 at 21:31









                            Mohammad Riazi-KermaniMohammad Riazi-Kermani

                            41.4k42061




                            41.4k42061






























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