is $det(A^2 + I)$ always non negative?












8














Obviously $det(A^2)$ is (casework), but is the above matrix non-negative? $det(A)det(A) geq 0$ as $det(A) > 0$ or $det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.






linear-algebra determinant







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    8














    Obviously $det(A^2)$ is (casework), but is the above matrix non-negative? $det(A)det(A) geq 0$ as $det(A) > 0$ or $det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.






    linear-algebra determinant







    share|cite|improve this question



























      8












      8








      8


      3





      Obviously $det(A^2)$ is (casework), but is the above matrix non-negative? $det(A)det(A) geq 0$ as $det(A) > 0$ or $det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.






      linear-algebra determinant







      share|cite|improve this question















      Obviously $det(A^2)$ is (casework), but is the above matrix non-negative? $det(A)det(A) geq 0$ as $det(A) > 0$ or $det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.






      linear-algebra determinant




      linear-algebra determinant






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 11:14









      Widawensen

      4,42121445




      4,42121445










      asked Dec 9 '18 at 22:25







      user624697





























          2 Answers
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          active

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          19














          I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so



          $overline{iA} = -iA; tag 1$



          then



          $det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
          $= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$






          share|cite|improve this answer

















          • 1




            Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
            – Widawensen
            Dec 11 '18 at 15:16












          • @Widawensen: thank you sir!
            – Robert Lewis
            Dec 11 '18 at 16:50



















          1














          It is possible also to use eigenvalues to prove the claim.
          Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.



          Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.



          As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.



          Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.



          $$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
          $$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$



          We have received once again a pair of conjugated complex numbers.



          The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.



          Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).






          share|cite|improve this answer























          • With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
            – Widawensen
            Dec 11 '18 at 11:50













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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          19














          I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so



          $overline{iA} = -iA; tag 1$



          then



          $det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
          $= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$






          share|cite|improve this answer

















          • 1




            Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
            – Widawensen
            Dec 11 '18 at 15:16












          • @Widawensen: thank you sir!
            – Robert Lewis
            Dec 11 '18 at 16:50
















          19














          I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so



          $overline{iA} = -iA; tag 1$



          then



          $det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
          $= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$






          share|cite|improve this answer

















          • 1




            Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
            – Widawensen
            Dec 11 '18 at 15:16












          • @Widawensen: thank you sir!
            – Robert Lewis
            Dec 11 '18 at 16:50














          19












          19








          19






          I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so



          $overline{iA} = -iA; tag 1$



          then



          $det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
          $= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$






          share|cite|improve this answer












          I assume $A$ is a real matrix; then $iA$ is purely imaginary, and so



          $overline{iA} = -iA; tag 1$



          then



          $det(I + A^2) = det ((I + iA)(I - iA)) = det(I + iA) det(I - iA)$
          $= det(I + iA)det(overline{I + iA)} = det(I + iA) overline{det(I + iA)} ge 0. tag 2$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 22:42









          Robert Lewis

          43.7k22963




          43.7k22963








          • 1




            Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
            – Widawensen
            Dec 11 '18 at 15:16












          • @Widawensen: thank you sir!
            – Robert Lewis
            Dec 11 '18 at 16:50














          • 1




            Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
            – Widawensen
            Dec 11 '18 at 15:16












          • @Widawensen: thank you sir!
            – Robert Lewis
            Dec 11 '18 at 16:50








          1




          1




          Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
          – Widawensen
          Dec 11 '18 at 15:16






          Excellent solution and very compact. Especially step $det(overline{M})=overline{det(M)}$ is very fast.
          – Widawensen
          Dec 11 '18 at 15:16














          @Widawensen: thank you sir!
          – Robert Lewis
          Dec 11 '18 at 16:50




          @Widawensen: thank you sir!
          – Robert Lewis
          Dec 11 '18 at 16:50











          1














          It is possible also to use eigenvalues to prove the claim.
          Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.



          Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.



          As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.



          Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.



          $$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
          $$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$



          We have received once again a pair of conjugated complex numbers.



          The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.



          Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).






          share|cite|improve this answer























          • With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
            – Widawensen
            Dec 11 '18 at 11:50


















          1














          It is possible also to use eigenvalues to prove the claim.
          Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.



          Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.



          As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.



          Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.



          $$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
          $$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$



          We have received once again a pair of conjugated complex numbers.



          The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.



          Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).






          share|cite|improve this answer























          • With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
            – Widawensen
            Dec 11 '18 at 11:50
















          1












          1








          1






          It is possible also to use eigenvalues to prove the claim.
          Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.



          Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.



          As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.



          Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.



          $$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
          $$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$



          We have received once again a pair of conjugated complex numbers.



          The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.



          Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).






          share|cite|improve this answer














          It is possible also to use eigenvalues to prove the claim.
          Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.



          Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.



          As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.



          Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $ge 1$) for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.



          $$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
          $$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$



          We have received once again a pair of conjugated complex numbers.



          The product of conjugated complex numbers is non-negative number ( $cc^*=vert cvert^2$) and the whole product is non-negative.



          Generally we can say even more: the determinant is mostly positive, except the case when matrix $A$ has $pm i $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 14:28

























          answered Dec 11 '18 at 11:10









          Widawensen

          4,42121445




          4,42121445












          • With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
            – Widawensen
            Dec 11 '18 at 11:50




















          • With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
            – Widawensen
            Dec 11 '18 at 11:50


















          With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
          – Widawensen
          Dec 11 '18 at 11:50






          With an argumentation similar to the presented one we could extend the claim even to the form that determinant of $A^{2m}+alpha I$, where $alphage 0$, is non-negative. It's worth to notice that even in the case of negative $alpha$ the sign of determinant depends solely on real eigenvalues of $A$ not the complex ones as they always come in conjugated pairs.
          – Widawensen
          Dec 11 '18 at 11:50




















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