True or False ( Invertibility of $A$)












0












$begingroup$


Let $A$ be $10 times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.



$A^2 + A = –I$



$A(–A–I) = I$
So, – $A$ – $I$ is inverse of A which is clearly $10 times 10$ matrix, since $A$ and $I$ are $10 times 10$ matrices.



Is this explanation enough to prove the statement true?






linear-algebra matrices







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$endgroup$








  • 1




    $begingroup$
    Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 17:27


















0












$begingroup$


Let $A$ be $10 times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.



$A^2 + A = –I$



$A(–A–I) = I$
So, – $A$ – $I$ is inverse of A which is clearly $10 times 10$ matrix, since $A$ and $I$ are $10 times 10$ matrices.



Is this explanation enough to prove the statement true?






linear-algebra matrices







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 17:27
















0












0








0





$begingroup$


Let $A$ be $10 times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.



$A^2 + A = –I$



$A(–A–I) = I$
So, – $A$ – $I$ is inverse of A which is clearly $10 times 10$ matrix, since $A$ and $I$ are $10 times 10$ matrices.



Is this explanation enough to prove the statement true?






linear-algebra matrices







share|cite|improve this question









$endgroup$




Let $A$ be $10 times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.



$A^2 + A = –I$



$A(–A–I) = I$
So, – $A$ – $I$ is inverse of A which is clearly $10 times 10$ matrix, since $A$ and $I$ are $10 times 10$ matrices.



Is this explanation enough to prove the statement true?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 16:55









MathsaddictMathsaddict

3459




3459








  • 1




    $begingroup$
    Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 17:27
















  • 1




    $begingroup$
    Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
    $endgroup$
    – Jack D'Aurizio
    Dec 9 '18 at 17:27










1




1




$begingroup$
Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 17:27






$begingroup$
Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $vneq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0notintext{Spec}A$).
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 17:27












2 Answers
2






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$begingroup$

I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Short Answer: Yes.



    Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






      active

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      3












      $begingroup$

      I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.






          share|cite|improve this answer









          $endgroup$



          I don't think you need to stress that the sizes of the matrices in sight are still $10times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 16:58









          GenericMathematicianGenericMathematician

          863




          863























              1












              $begingroup$

              Short Answer: Yes.



              Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Short Answer: Yes.



                Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Short Answer: Yes.



                  Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.






                  share|cite|improve this answer









                  $endgroup$



                  Short Answer: Yes.



                  Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y in R^{10times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 17:05









                  bruderjakob17bruderjakob17

                  1997




                  1997






























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