Prove that If A is a real skew-symmetric matrix, then I + A is invertible, where I is the identity matrix....












0












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This question already has an answer here:




  • Showing $A-I$ is invertible, when $A$ is a skew-symmetric matrix

    4 answers




In the Wikipedia article for skew symmetric matrixes the following propert is stated: If A is a real skew-symmetric matrix, then I + A is invertible, where I is the identity matrix.



However I haven’t been able to find a proof of this or why it does not apply to complex matrices.



https://en.m.wikipedia.org/wiki/Skew-symmetric_matrix



This is different from the one where I - A is invertsble because this is I + A






linear-algebra







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marked as duplicate by T. Bongers, darij grinberg, jgon, Trevor Gunn, Lord Shark the Unknown Nov 29 '18 at 6:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Can we use the fact that $iA$ is self-adjoint?
    $endgroup$
    – Oscar Lanzi
    Nov 29 '18 at 2:23










  • $begingroup$
    I don’t believe this is a duplicate because that covers I - A and this covers I + A
    $endgroup$
    – pdid
    Nov 29 '18 at 2:46








  • 1




    $begingroup$
    A is skew symmetric iff -A is
    $endgroup$
    – user25959
    Nov 29 '18 at 4:04










  • $begingroup$
    Not only that, even if you didn't know that, if you go through the argument in the linked question and answer, you'll see it goes through essentially unchanged for $I+A$.
    $endgroup$
    – jgon
    Nov 30 '18 at 5:51










  • $begingroup$
    Technically the linked q/a doesn't answer one of your questions, namely about why it's not true for complex matrices. The answer has two parts, one is, it is true if we generalize skew symmetry properly to the property of being skew-Hermitian. The other is the proof requires an inner product (which is why skew-Hermitian is the right notion), and a nondegenerate bilinear form is not enough. A cx for complex skew symmetric matrices is $A:=begin{pmatrix} 0 & i \ -i & 0end{pmatrix}$, since $A+1=begin{pmatrix}1 & i \ -i & 1 end{pmatrix}$, which has determinant $1-1=0$.
    $endgroup$
    – jgon
    Nov 30 '18 at 6:01
















0












$begingroup$



This question already has an answer here:




  • Showing $A-I$ is invertible, when $A$ is a skew-symmetric matrix

    4 answers




In the Wikipedia article for skew symmetric matrixes the following propert is stated: If A is a real skew-symmetric matrix, then I + A is invertible, where I is the identity matrix.



However I haven’t been able to find a proof of this or why it does not apply to complex matrices.



https://en.m.wikipedia.org/wiki/Skew-symmetric_matrix



This is different from the one where I - A is invertsble because this is I + A






linear-algebra







share|cite|improve this question











$endgroup$



marked as duplicate by T. Bongers, darij grinberg, jgon, Trevor Gunn, Lord Shark the Unknown Nov 29 '18 at 6:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Can we use the fact that $iA$ is self-adjoint?
    $endgroup$
    – Oscar Lanzi
    Nov 29 '18 at 2:23










  • $begingroup$
    I don’t believe this is a duplicate because that covers I - A and this covers I + A
    $endgroup$
    – pdid
    Nov 29 '18 at 2:46








  • 1




    $begingroup$
    A is skew symmetric iff -A is
    $endgroup$
    – user25959
    Nov 29 '18 at 4:04










  • $begingroup$
    Not only that, even if you didn't know that, if you go through the argument in the linked question and answer, you'll see it goes through essentially unchanged for $I+A$.
    $endgroup$
    – jgon
    Nov 30 '18 at 5:51










  • $begingroup$
    Technically the linked q/a doesn't answer one of your questions, namely about why it's not true for complex matrices. The answer has two parts, one is, it is true if we generalize skew symmetry properly to the property of being skew-Hermitian. The other is the proof requires an inner product (which is why skew-Hermitian is the right notion), and a nondegenerate bilinear form is not enough. A cx for complex skew symmetric matrices is $A:=begin{pmatrix} 0 & i \ -i & 0end{pmatrix}$, since $A+1=begin{pmatrix}1 & i \ -i & 1 end{pmatrix}$, which has determinant $1-1=0$.
    $endgroup$
    – jgon
    Nov 30 '18 at 6:01














0












0








0





$begingroup$



This question already has an answer here:




  • Showing $A-I$ is invertible, when $A$ is a skew-symmetric matrix

    4 answers




In the Wikipedia article for skew symmetric matrixes the following propert is stated: If A is a real skew-symmetric matrix, then I + A is invertible, where I is the identity matrix.



However I haven’t been able to find a proof of this or why it does not apply to complex matrices.



https://en.m.wikipedia.org/wiki/Skew-symmetric_matrix



This is different from the one where I - A is invertsble because this is I + A






linear-algebra







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Showing $A-I$ is invertible, when $A$ is a skew-symmetric matrix

    4 answers




In the Wikipedia article for skew symmetric matrixes the following propert is stated: If A is a real skew-symmetric matrix, then I + A is invertible, where I is the identity matrix.



However I haven’t been able to find a proof of this or why it does not apply to complex matrices.



https://en.m.wikipedia.org/wiki/Skew-symmetric_matrix



This is different from the one where I - A is invertsble because this is I + A





This question already has an answer here:




  • Showing $A-I$ is invertible, when $A$ is a skew-symmetric matrix

    4 answers







linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 4:06







pdid

















asked Nov 29 '18 at 2:17









pdidpdid

33




33




marked as duplicate by T. Bongers, darij grinberg, jgon, Trevor Gunn, Lord Shark the Unknown Nov 29 '18 at 6:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by T. Bongers, darij grinberg, jgon, Trevor Gunn, Lord Shark the Unknown Nov 29 '18 at 6:14


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Can we use the fact that $iA$ is self-adjoint?
    $endgroup$
    – Oscar Lanzi
    Nov 29 '18 at 2:23










  • $begingroup$
    I don’t believe this is a duplicate because that covers I - A and this covers I + A
    $endgroup$
    – pdid
    Nov 29 '18 at 2:46








  • 1




    $begingroup$
    A is skew symmetric iff -A is
    $endgroup$
    – user25959
    Nov 29 '18 at 4:04










  • $begingroup$
    Not only that, even if you didn't know that, if you go through the argument in the linked question and answer, you'll see it goes through essentially unchanged for $I+A$.
    $endgroup$
    – jgon
    Nov 30 '18 at 5:51










  • $begingroup$
    Technically the linked q/a doesn't answer one of your questions, namely about why it's not true for complex matrices. The answer has two parts, one is, it is true if we generalize skew symmetry properly to the property of being skew-Hermitian. The other is the proof requires an inner product (which is why skew-Hermitian is the right notion), and a nondegenerate bilinear form is not enough. A cx for complex skew symmetric matrices is $A:=begin{pmatrix} 0 & i \ -i & 0end{pmatrix}$, since $A+1=begin{pmatrix}1 & i \ -i & 1 end{pmatrix}$, which has determinant $1-1=0$.
    $endgroup$
    – jgon
    Nov 30 '18 at 6:01


















  • $begingroup$
    Can we use the fact that $iA$ is self-adjoint?
    $endgroup$
    – Oscar Lanzi
    Nov 29 '18 at 2:23










  • $begingroup$
    I don’t believe this is a duplicate because that covers I - A and this covers I + A
    $endgroup$
    – pdid
    Nov 29 '18 at 2:46








  • 1




    $begingroup$
    A is skew symmetric iff -A is
    $endgroup$
    – user25959
    Nov 29 '18 at 4:04










  • $begingroup$
    Not only that, even if you didn't know that, if you go through the argument in the linked question and answer, you'll see it goes through essentially unchanged for $I+A$.
    $endgroup$
    – jgon
    Nov 30 '18 at 5:51










  • $begingroup$
    Technically the linked q/a doesn't answer one of your questions, namely about why it's not true for complex matrices. The answer has two parts, one is, it is true if we generalize skew symmetry properly to the property of being skew-Hermitian. The other is the proof requires an inner product (which is why skew-Hermitian is the right notion), and a nondegenerate bilinear form is not enough. A cx for complex skew symmetric matrices is $A:=begin{pmatrix} 0 & i \ -i & 0end{pmatrix}$, since $A+1=begin{pmatrix}1 & i \ -i & 1 end{pmatrix}$, which has determinant $1-1=0$.
    $endgroup$
    – jgon
    Nov 30 '18 at 6:01
















$begingroup$
Can we use the fact that $iA$ is self-adjoint?
$endgroup$
– Oscar Lanzi
Nov 29 '18 at 2:23




$begingroup$
Can we use the fact that $iA$ is self-adjoint?
$endgroup$
– Oscar Lanzi
Nov 29 '18 at 2:23












$begingroup$
I don’t believe this is a duplicate because that covers I - A and this covers I + A
$endgroup$
– pdid
Nov 29 '18 at 2:46






$begingroup$
I don’t believe this is a duplicate because that covers I - A and this covers I + A
$endgroup$
– pdid
Nov 29 '18 at 2:46






1




1




$begingroup$
A is skew symmetric iff -A is
$endgroup$
– user25959
Nov 29 '18 at 4:04




$begingroup$
A is skew symmetric iff -A is
$endgroup$
– user25959
Nov 29 '18 at 4:04












$begingroup$
Not only that, even if you didn't know that, if you go through the argument in the linked question and answer, you'll see it goes through essentially unchanged for $I+A$.
$endgroup$
– jgon
Nov 30 '18 at 5:51




$begingroup$
Not only that, even if you didn't know that, if you go through the argument in the linked question and answer, you'll see it goes through essentially unchanged for $I+A$.
$endgroup$
– jgon
Nov 30 '18 at 5:51












$begingroup$
Technically the linked q/a doesn't answer one of your questions, namely about why it's not true for complex matrices. The answer has two parts, one is, it is true if we generalize skew symmetry properly to the property of being skew-Hermitian. The other is the proof requires an inner product (which is why skew-Hermitian is the right notion), and a nondegenerate bilinear form is not enough. A cx for complex skew symmetric matrices is $A:=begin{pmatrix} 0 & i \ -i & 0end{pmatrix}$, since $A+1=begin{pmatrix}1 & i \ -i & 1 end{pmatrix}$, which has determinant $1-1=0$.
$endgroup$
– jgon
Nov 30 '18 at 6:01




$begingroup$
Technically the linked q/a doesn't answer one of your questions, namely about why it's not true for complex matrices. The answer has two parts, one is, it is true if we generalize skew symmetry properly to the property of being skew-Hermitian. The other is the proof requires an inner product (which is why skew-Hermitian is the right notion), and a nondegenerate bilinear form is not enough. A cx for complex skew symmetric matrices is $A:=begin{pmatrix} 0 & i \ -i & 0end{pmatrix}$, since $A+1=begin{pmatrix}1 & i \ -i & 1 end{pmatrix}$, which has determinant $1-1=0$.
$endgroup$
– jgon
Nov 30 '18 at 6:01










1 Answer
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$begingroup$

Eigenvalues of real skew-symmetric are either $0$ or purely imaginary..



Suppose eigenvalues of $A$ are $lambda_1$, $lambda_2$, $cdots$, $lambda_n$.



So, eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$.



Since eigenvalues of $A$ are either $0$ or purely imaginary, all eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$ are not equal to zero.



Hence, $det(I+A)neq 0$. This implies, $I+A$ is invertible..



Note: If one of the eigenvalue of $A$ is $-1$, then one of the eigenvalue of $I+A$ is zero, and hence, $I+A$ is not invertible (as $det(I+A)=0$).



Take $$A=left(begin{array}{cc} 0 & i\
-i & 0
end{array}right).$$



Then eigenvalues of $A$ are $1$ and $-1$.



So, eigenvalues of $I+A$ are $2$ and $0$. Hence, $I+A$ is not invertible.






share|cite|improve this answer











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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Eigenvalues of real skew-symmetric are either $0$ or purely imaginary..



    Suppose eigenvalues of $A$ are $lambda_1$, $lambda_2$, $cdots$, $lambda_n$.



    So, eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$.



    Since eigenvalues of $A$ are either $0$ or purely imaginary, all eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$ are not equal to zero.



    Hence, $det(I+A)neq 0$. This implies, $I+A$ is invertible..



    Note: If one of the eigenvalue of $A$ is $-1$, then one of the eigenvalue of $I+A$ is zero, and hence, $I+A$ is not invertible (as $det(I+A)=0$).



    Take $$A=left(begin{array}{cc} 0 & i\
    -i & 0
    end{array}right).$$



    Then eigenvalues of $A$ are $1$ and $-1$.



    So, eigenvalues of $I+A$ are $2$ and $0$. Hence, $I+A$ is not invertible.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Eigenvalues of real skew-symmetric are either $0$ or purely imaginary..



      Suppose eigenvalues of $A$ are $lambda_1$, $lambda_2$, $cdots$, $lambda_n$.



      So, eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$.



      Since eigenvalues of $A$ are either $0$ or purely imaginary, all eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$ are not equal to zero.



      Hence, $det(I+A)neq 0$. This implies, $I+A$ is invertible..



      Note: If one of the eigenvalue of $A$ is $-1$, then one of the eigenvalue of $I+A$ is zero, and hence, $I+A$ is not invertible (as $det(I+A)=0$).



      Take $$A=left(begin{array}{cc} 0 & i\
      -i & 0
      end{array}right).$$



      Then eigenvalues of $A$ are $1$ and $-1$.



      So, eigenvalues of $I+A$ are $2$ and $0$. Hence, $I+A$ is not invertible.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Eigenvalues of real skew-symmetric are either $0$ or purely imaginary..



        Suppose eigenvalues of $A$ are $lambda_1$, $lambda_2$, $cdots$, $lambda_n$.



        So, eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$.



        Since eigenvalues of $A$ are either $0$ or purely imaginary, all eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$ are not equal to zero.



        Hence, $det(I+A)neq 0$. This implies, $I+A$ is invertible..



        Note: If one of the eigenvalue of $A$ is $-1$, then one of the eigenvalue of $I+A$ is zero, and hence, $I+A$ is not invertible (as $det(I+A)=0$).



        Take $$A=left(begin{array}{cc} 0 & i\
        -i & 0
        end{array}right).$$



        Then eigenvalues of $A$ are $1$ and $-1$.



        So, eigenvalues of $I+A$ are $2$ and $0$. Hence, $I+A$ is not invertible.






        share|cite|improve this answer











        $endgroup$



        Eigenvalues of real skew-symmetric are either $0$ or purely imaginary..



        Suppose eigenvalues of $A$ are $lambda_1$, $lambda_2$, $cdots$, $lambda_n$.



        So, eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$.



        Since eigenvalues of $A$ are either $0$ or purely imaginary, all eigenvalues of $I+A$ are $1+lambda_1$, $1+lambda_2$, $cdots$, $1+lambda_n$ are not equal to zero.



        Hence, $det(I+A)neq 0$. This implies, $I+A$ is invertible..



        Note: If one of the eigenvalue of $A$ is $-1$, then one of the eigenvalue of $I+A$ is zero, and hence, $I+A$ is not invertible (as $det(I+A)=0$).



        Take $$A=left(begin{array}{cc} 0 & i\
        -i & 0
        end{array}right).$$



        Then eigenvalues of $A$ are $1$ and $-1$.



        So, eigenvalues of $I+A$ are $2$ and $0$. Hence, $I+A$ is not invertible.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 5:47

























        answered Nov 29 '18 at 5:06









        GoldyGoldy

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