Determinant of a matrix of the form $M = I + xx^t$












5












$begingroup$


$$M=begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \ x_2x_1 & 1+x_2^2 &...&x_2x_n \...&...& &...& \x_nx_1 & x_nx_2& ...&1+x_n^2&end{pmatrix}.$$



So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.






linear-algebra determinant







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$endgroup$








  • 1




    $begingroup$
    Can you determine the characteristic polynomial of $M-I$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 20:24










  • $begingroup$
    Could you give me some more clue about the polynomial?
    $endgroup$
    – Yibei He
    Jan 22 at 20:30
















5












$begingroup$


$$M=begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \ x_2x_1 & 1+x_2^2 &...&x_2x_n \...&...& &...& \x_nx_1 & x_nx_2& ...&1+x_n^2&end{pmatrix}.$$



So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.






linear-algebra determinant







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Can you determine the characteristic polynomial of $M-I$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 20:24










  • $begingroup$
    Could you give me some more clue about the polynomial?
    $endgroup$
    – Yibei He
    Jan 22 at 20:30














5












5








5


1



$begingroup$


$$M=begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \ x_2x_1 & 1+x_2^2 &...&x_2x_n \...&...& &...& \x_nx_1 & x_nx_2& ...&1+x_n^2&end{pmatrix}.$$



So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.






linear-algebra determinant







share|cite|improve this question











$endgroup$




$$M=begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \ x_2x_1 & 1+x_2^2 &...&x_2x_n \...&...& &...& \x_nx_1 & x_nx_2& ...&1+x_n^2&end{pmatrix}.$$



So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.






linear-algebra determinant




linear-algebra determinant






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share|cite|improve this question













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edited Jan 22 at 23:02









Silverfish

1,0271223




1,0271223










asked Jan 22 at 20:21









Yibei HeYibei He

3139




3139








  • 1




    $begingroup$
    Can you determine the characteristic polynomial of $M-I$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 20:24










  • $begingroup$
    Could you give me some more clue about the polynomial?
    $endgroup$
    – Yibei He
    Jan 22 at 20:30














  • 1




    $begingroup$
    Can you determine the characteristic polynomial of $M-I$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 20:24










  • $begingroup$
    Could you give me some more clue about the polynomial?
    $endgroup$
    – Yibei He
    Jan 22 at 20:30








1




1




$begingroup$
Can you determine the characteristic polynomial of $M-I$?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:24




$begingroup$
Can you determine the characteristic polynomial of $M-I$?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 20:24












$begingroup$
Could you give me some more clue about the polynomial?
$endgroup$
– Yibei He
Jan 22 at 20:30




$begingroup$
Could you give me some more clue about the polynomial?
$endgroup$
– Yibei He
Jan 22 at 20:30










3 Answers
3






active

oldest

votes


















8












$begingroup$

You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).



For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have



$$
M = I + x x^T
$$



hence, in view of the aforementioned lemma,



$$
det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
$$






share|cite|improve this answer











$endgroup$





















    7












    $begingroup$

    Consider the matrix
    $$N=M-I.$$
    It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
    has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
    $$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
    Now
    $$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
      $$
      M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
      x_nmathbf{x}&1+x_n^2}.
      $$
      Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
      $$
      detleft(M_nright)=detpmatrix{M_{n-1}& 0\
      x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
      x_nmathbf{x}&x_n^2}.
      $$
      The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
      $$
      detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
      mathbf{x}&1}.
      $$
      The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        8












        $begingroup$

        You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).



        For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have



        $$
        M = I + x x^T
        $$



        hence, in view of the aforementioned lemma,



        $$
        det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
        $$






        share|cite|improve this answer











        $endgroup$


















          8












          $begingroup$

          You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).



          For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have



          $$
          M = I + x x^T
          $$



          hence, in view of the aforementioned lemma,



          $$
          det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
          $$






          share|cite|improve this answer











          $endgroup$
















            8












            8








            8





            $begingroup$

            You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).



            For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have



            $$
            M = I + x x^T
            $$



            hence, in view of the aforementioned lemma,



            $$
            det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
            $$






            share|cite|improve this answer











            $endgroup$



            You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).



            For a column vector $x = (x_1,...,x_n)$, and an $ntimes n $ identity matrix $I$, we have



            $$
            M = I + x x^T
            $$



            hence, in view of the aforementioned lemma,



            $$
            det M = (1 + x^T I x)det I = 1 + |x|^2 = 1 + x_1^2 + cdots +x_n^2.
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 23 at 11:37









            Rodrigo de Azevedo

            13k41960




            13k41960










            answered Jan 22 at 20:47









            HaykHayk

            2,6271214




            2,6271214























                7












                $begingroup$

                Consider the matrix
                $$N=M-I.$$
                It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
                has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
                $$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
                Now
                $$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$






                share|cite|improve this answer









                $endgroup$


















                  7












                  $begingroup$

                  Consider the matrix
                  $$N=M-I.$$
                  It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
                  has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
                  $$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
                  Now
                  $$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$






                  share|cite|improve this answer









                  $endgroup$
















                    7












                    7








                    7





                    $begingroup$

                    Consider the matrix
                    $$N=M-I.$$
                    It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
                    has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
                    $$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
                    Now
                    $$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$






                    share|cite|improve this answer









                    $endgroup$



                    Consider the matrix
                    $$N=M-I.$$
                    It has rank $le1$, so its characteristic polynomial $P(lambda)=det(lambda I-N)$
                    has the form $lambda^n-alambda^{n-1}$. But $a$ is the trace of $N$, so
                    $$P(lambda)=lambda^n-(x_1^2+cdots+x_n^2)lambda^{n-1}.$$
                    Now
                    $$det M=det(I+N)=(-1)^nP(-1)=1+x_1^2+cdots+x_n^2.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 22 at 20:35









                    Lord Shark the UnknownLord Shark the Unknown

                    106k1161133




                    106k1161133























                        2












                        $begingroup$

                        Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
                        $$
                        M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
                        x_nmathbf{x}&1+x_n^2}.
                        $$
                        Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
                        $$
                        detleft(M_nright)=detpmatrix{M_{n-1}& 0\
                        x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
                        x_nmathbf{x}&x_n^2}.
                        $$
                        The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
                        $$
                        detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
                        mathbf{x}&1}.
                        $$
                        The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
                          $$
                          M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
                          x_nmathbf{x}&1+x_n^2}.
                          $$
                          Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
                          $$
                          detleft(M_nright)=detpmatrix{M_{n-1}& 0\
                          x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
                          x_nmathbf{x}&x_n^2}.
                          $$
                          The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
                          $$
                          detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
                          mathbf{x}&1}.
                          $$
                          The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
                            $$
                            M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
                            x_nmathbf{x}&1+x_n^2}.
                            $$
                            Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
                            $$
                            detleft(M_nright)=detpmatrix{M_{n-1}& 0\
                            x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
                            x_nmathbf{x}&x_n^2}.
                            $$
                            The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
                            $$
                            detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
                            mathbf{x}&1}.
                            $$
                            The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.






                            share|cite|improve this answer









                            $endgroup$



                            Denote by $M_n$ the matrix we want to compute the determinant. Letting $mathbf{x}=(x_1,dots,x_{n-1})$, we express $M_n$ as a block matrix:
                            $$
                            M_n=pmatrix{M_{n-1}& x_nmathbf x^t\
                            x_nmathbf{x}&1+x_n^2}.
                            $$
                            Since $pmatrix{ x_nmathbf x^t\1+x_n^2}=pmatrix{ 0\1}+pmatrix{ mathbf x_nx^t\x_n^2}$, it follows by multi-linearity of the determinant that
                            $$
                            detleft(M_nright)=detpmatrix{M_{n-1}& 0\
                            x_nmathbf{x}&1}+detpmatrix{M_{n-1}& x_nmathbf x^t\
                            x_nmathbf{x}&x_n^2}.
                            $$
                            The first term of the right hand side is $detleft(M_{n-1}right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that
                            $$
                            detleft(M_{n}right)=detleft(M_{n-1}right)+x_n^2detpmatrix{M_{n-1}& mathbf x^t\
                            mathbf{x}&1}.
                            $$
                            The last determinant is one: one can make the linear combinations $C_ileftarrow x_iC_n$ to see this.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 22 at 21:46









                            Davide GiraudoDavide Giraudo

                            127k17154268




                            127k17154268






























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