A question of non-singularity

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Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?



(A) $A + 2I$



(B) $B$



(C) $B + 2I$



(D) $A$



I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.










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  • I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
    – Robert Lewis
    3 hours ago















up vote
2
down vote

favorite
1












Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?



(A) $A + 2I$



(B) $B$



(C) $B + 2I$



(D) $A$



I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.










share|cite|improve this question























  • I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
    – Robert Lewis
    3 hours ago













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?



(A) $A + 2I$



(B) $B$



(C) $B + 2I$



(D) $A$



I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.










share|cite|improve this question















Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I
denotes the identity matrix. Which of the following matrices must be
nonsingular?



(A) $A + 2I$



(B) $B$



(C) $B + 2I$



(D) $A$



I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.







linear-algebra matrices determinant






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share|cite|improve this question













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edited 3 hours ago









greedoid

30.6k94184




30.6k94184










asked 3 hours ago









Kc17

245




245











  • I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
    – Robert Lewis
    3 hours ago

















  • I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
    – Robert Lewis
    3 hours ago
















I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
– Robert Lewis
3 hours ago





I edited your post a little to make the $LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers!
– Robert Lewis
3 hours ago











3 Answers
3






active

oldest

votes

















up vote
4
down vote













Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.






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    up vote
    2
    down vote













    We have



    $B^2 + AB + 2I = 0, tag 1$



    so



    $(B + A)B = B^2 + AB = -2I, tag 2$



    or



    $left ( -dfrac12(B + A) right )B = I, tag 3$



    that is,



    $B^-1 = -dfrac12(B + A), tag 4$



    and the correct answer is (B).






    share|cite|improve this answer






















    • @StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
      – Robert Lewis
      2 hours ago






    • 1




      @StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
      – Robert Lewis
      2 hours ago










    • How do you exclude the other cases?
      – egreg
      50 mins ago

















    up vote
    0
    down vote













    I'm assuming you're over the real numbers.



    Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
    $$
    0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
    $$

    a contradiction. Therefore $B$ is nonsingular.



    In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.



    The matrix $A$ can be the zero matrix, because
    $$
    beginbmatrix
    0 & 2 \
    -1 & 0
    endbmatrix^2=-2I
    $$



    Also $A+2I$ can be the zero matrix: the identity to satisfy would be
    $$
    B^2-2B+2I=0
    $$

    and the matrix
    beginbmatrix
    2 & -1 \
    2 & 0
    endbmatrix

    satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.



    Can $B+2I=0$?






    share|cite|improve this answer




















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      3 Answers
      3






      active

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      3 Answers
      3






      active

      oldest

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      active

      oldest

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      up vote
      4
      down vote













      Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.






      share|cite|improve this answer
























        up vote
        4
        down vote













        Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.






        share|cite|improve this answer






















          up vote
          4
          down vote










          up vote
          4
          down vote









          Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.






          share|cite|improve this answer












          Since $$det (B+A)cdot det (B) = det (B^2+AB) = det (-2I) =-2$$ we see that $det (B)ne 0$ so $B$ is invertibile.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          greedoid

          30.6k94184




          30.6k94184




















              up vote
              2
              down vote













              We have



              $B^2 + AB + 2I = 0, tag 1$



              so



              $(B + A)B = B^2 + AB = -2I, tag 2$



              or



              $left ( -dfrac12(B + A) right )B = I, tag 3$



              that is,



              $B^-1 = -dfrac12(B + A), tag 4$



              and the correct answer is (B).






              share|cite|improve this answer






















              • @StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
                – Robert Lewis
                2 hours ago






              • 1




                @StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
                – Robert Lewis
                2 hours ago










              • How do you exclude the other cases?
                – egreg
                50 mins ago














              up vote
              2
              down vote













              We have



              $B^2 + AB + 2I = 0, tag 1$



              so



              $(B + A)B = B^2 + AB = -2I, tag 2$



              or



              $left ( -dfrac12(B + A) right )B = I, tag 3$



              that is,



              $B^-1 = -dfrac12(B + A), tag 4$



              and the correct answer is (B).






              share|cite|improve this answer






















              • @StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
                – Robert Lewis
                2 hours ago






              • 1




                @StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
                – Robert Lewis
                2 hours ago










              • How do you exclude the other cases?
                – egreg
                50 mins ago












              up vote
              2
              down vote










              up vote
              2
              down vote









              We have



              $B^2 + AB + 2I = 0, tag 1$



              so



              $(B + A)B = B^2 + AB = -2I, tag 2$



              or



              $left ( -dfrac12(B + A) right )B = I, tag 3$



              that is,



              $B^-1 = -dfrac12(B + A), tag 4$



              and the correct answer is (B).






              share|cite|improve this answer














              We have



              $B^2 + AB + 2I = 0, tag 1$



              so



              $(B + A)B = B^2 + AB = -2I, tag 2$



              or



              $left ( -dfrac12(B + A) right )B = I, tag 3$



              that is,



              $B^-1 = -dfrac12(B + A), tag 4$



              and the correct answer is (B).







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 2 hours ago

























              answered 3 hours ago









              Robert Lewis

              39.7k22459




              39.7k22459











              • @StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
                – Robert Lewis
                2 hours ago






              • 1




                @StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
                – Robert Lewis
                2 hours ago










              • How do you exclude the other cases?
                – egreg
                50 mins ago
















              • @StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
                – Robert Lewis
                2 hours ago






              • 1




                @StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
                – Robert Lewis
                2 hours ago










              • How do you exclude the other cases?
                – egreg
                50 mins ago















              @StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
              – Robert Lewis
              2 hours ago




              @StubbornAtom: Oops! Typo; corrected. Thanks. Cheers!
              – Robert Lewis
              2 hours ago




              1




              1




              @StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
              – Robert Lewis
              2 hours ago




              @StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers!
              – Robert Lewis
              2 hours ago












              How do you exclude the other cases?
              – egreg
              50 mins ago




              How do you exclude the other cases?
              – egreg
              50 mins ago










              up vote
              0
              down vote













              I'm assuming you're over the real numbers.



              Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
              $$
              0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
              $$

              a contradiction. Therefore $B$ is nonsingular.



              In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.



              The matrix $A$ can be the zero matrix, because
              $$
              beginbmatrix
              0 & 2 \
              -1 & 0
              endbmatrix^2=-2I
              $$



              Also $A+2I$ can be the zero matrix: the identity to satisfy would be
              $$
              B^2-2B+2I=0
              $$

              and the matrix
              beginbmatrix
              2 & -1 \
              2 & 0
              endbmatrix

              satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.



              Can $B+2I=0$?






              share|cite|improve this answer
























                up vote
                0
                down vote













                I'm assuming you're over the real numbers.



                Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
                $$
                0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
                $$

                a contradiction. Therefore $B$ is nonsingular.



                In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.



                The matrix $A$ can be the zero matrix, because
                $$
                beginbmatrix
                0 & 2 \
                -1 & 0
                endbmatrix^2=-2I
                $$



                Also $A+2I$ can be the zero matrix: the identity to satisfy would be
                $$
                B^2-2B+2I=0
                $$

                and the matrix
                beginbmatrix
                2 & -1 \
                2 & 0
                endbmatrix

                satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.



                Can $B+2I=0$?






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  I'm assuming you're over the real numbers.



                  Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
                  $$
                  0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
                  $$

                  a contradiction. Therefore $B$ is nonsingular.



                  In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.



                  The matrix $A$ can be the zero matrix, because
                  $$
                  beginbmatrix
                  0 & 2 \
                  -1 & 0
                  endbmatrix^2=-2I
                  $$



                  Also $A+2I$ can be the zero matrix: the identity to satisfy would be
                  $$
                  B^2-2B+2I=0
                  $$

                  and the matrix
                  beginbmatrix
                  2 & -1 \
                  2 & 0
                  endbmatrix

                  satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.



                  Can $B+2I=0$?






                  share|cite|improve this answer












                  I'm assuming you're over the real numbers.



                  Suppose $B$ is singular and take $vne0$ such that $Bv=0$; then
                  $$
                  0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v,
                  $$

                  a contradiction. Therefore $B$ is nonsingular.



                  In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.



                  The matrix $A$ can be the zero matrix, because
                  $$
                  beginbmatrix
                  0 & 2 \
                  -1 & 0
                  endbmatrix^2=-2I
                  $$



                  Also $A+2I$ can be the zero matrix: the identity to satisfy would be
                  $$
                  B^2-2B+2I=0
                  $$

                  and the matrix
                  beginbmatrix
                  2 & -1 \
                  2 & 0
                  endbmatrix

                  satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.



                  Can $B+2I=0$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 51 mins ago









                  egreg

                  169k1283191




                  169k1283191



























                       

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