Matrix proof : Show that if I − ED is nonsingular, then I − DE is nonsingular











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I is the identity matrix.



Let D ∈ $R^{ntimes m}$ and E ∈ $R^{mtimes n}$



Show that if I − ED is nonsingular, then I − DE is nonsingular
and



$(I − DE)^{−1} = I + D(I − ED)^{−1}E$



i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.



Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.






linear-algebra matrices







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  • This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
    – Jean Marie
    Nov 23 at 15:14

















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I is the identity matrix.



Let D ∈ $R^{ntimes m}$ and E ∈ $R^{mtimes n}$



Show that if I − ED is nonsingular, then I − DE is nonsingular
and



$(I − DE)^{−1} = I + D(I − ED)^{−1}E$



i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.



Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.






linear-algebra matrices







share|cite|improve this question






















  • This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
    – Jean Marie
    Nov 23 at 15:14















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I is the identity matrix.



Let D ∈ $R^{ntimes m}$ and E ∈ $R^{mtimes n}$



Show that if I − ED is nonsingular, then I − DE is nonsingular
and



$(I − DE)^{−1} = I + D(I − ED)^{−1}E$



i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.



Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.






linear-algebra matrices







share|cite|improve this question













I is the identity matrix.



Let D ∈ $R^{ntimes m}$ and E ∈ $R^{mtimes n}$



Show that if I − ED is nonsingular, then I − DE is nonsingular
and



$(I − DE)^{−1} = I + D(I − ED)^{−1}E$



i don't get how to this first of all i don't get how i could show the first proof. Because DE and ED are different dimensions i don't get how to you could ever compare them.



Second i dont get how to show that the equation is true because the I-ED is inbetween the E and D and with matrices you can't just multiple in any order i thought so i don't get how to get rid of it.






linear-algebra matrices




linear-algebra matrices






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asked Nov 22 at 15:06









Danielvanheuven

457




457












  • This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
    – Jean Marie
    Nov 23 at 15:14




















  • This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
    – Jean Marie
    Nov 23 at 15:14


















This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
– Jean Marie
Nov 23 at 15:14






This is a consequence of $det(I-DE)=det(I-ED)$ which is known as Sylvester's identity, math.stackexchange.com/q/17831
– Jean Marie
Nov 23 at 15:14












2 Answers
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Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
$$
(I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
$$
The product is the identity, which proves the claim.






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    We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
    $$ (I-DE) v = 0 .$$
    Consider $w = E vinmathbb{R}^m$.
    $$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
    so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.



    The form of the inverse is verified with simple matrix algebra, see daw's answer.






    share|cite|improve this answer





















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      2 Answers
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      Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
      $$
      (I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
      $$
      The product is the identity, which proves the claim.






      share|cite|improve this answer

























        up vote
        2
        down vote













        Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
        $$
        (I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
        $$
        The product is the identity, which proves the claim.






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
          $$
          (I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
          $$
          The product is the identity, which proves the claim.






          share|cite|improve this answer












          Compute the product of the matrix $I-DE$ and its wannabe inverse $I+D(I-ED)^{-1}E$:
          $$
          (I+D(I-ED)^{-1}E)(I-DE)=I-DE + D(I-ED)^{-1}E(I-DE)\=I-DE + D(I-ED)^{-1}(I-ED)E=I-DE+DE=I.
          $$
          The product is the identity, which proves the claim.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 15:12









          daw

          24k1544




          24k1544






















              up vote
              1
              down vote













              We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
              $$ (I-DE) v = 0 .$$
              Consider $w = E vinmathbb{R}^m$.
              $$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
              so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.



              The form of the inverse is verified with simple matrix algebra, see daw's answer.






              share|cite|improve this answer

























                up vote
                1
                down vote













                We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
                $$ (I-DE) v = 0 .$$
                Consider $w = E vinmathbb{R}^m$.
                $$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
                so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.



                The form of the inverse is verified with simple matrix algebra, see daw's answer.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
                  $$ (I-DE) v = 0 .$$
                  Consider $w = E vinmathbb{R}^m$.
                  $$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
                  so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.



                  The form of the inverse is verified with simple matrix algebra, see daw's answer.






                  share|cite|improve this answer












                  We can show that $I-DE$ is nonsingular by contradiction. If it is singular, there is a vector $vinmathbb{R}^n$ such that
                  $$ (I-DE) v = 0 .$$
                  Consider $w = E vinmathbb{R}^m$.
                  $$(I-ED)w = E v - EDE v = E (I-DE) v = 0$$
                  so $I-ED$ is also singular (contradiction). Therefore, $I-DE$ is not singular.



                  The form of the inverse is verified with simple matrix algebra, see daw's answer.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 15:24









                  fqq

                  29717




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