Determinant of $(I+uv^*)$












2














Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?






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    2














    Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?






    linear-algebra determinant numerical-linear-algebra







    share|cite|improve this question

























      2












      2








      2


      1





      Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?






      linear-algebra determinant numerical-linear-algebra







      share|cite|improve this question













      Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?






      linear-algebra determinant numerical-linear-algebra




      linear-algebra determinant numerical-linear-algebra






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      asked Nov 21 at 21:33









      Studying Optimization

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          Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
          $$
          {displaystyle det left({A} +{uv} ^{textsf {T}}right)
          =left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
          $$



          In your case, let $A=I$.



          The proof is essentially based on the following observation:
          $$
          {displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
          $$




          [Added later:]
          Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
          $$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
          $$
          (v^Tu,0,cdots,0).
          $$
          It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
          $$
          (1+v^Tu,1,cdots,1).
          $$
          But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.



          1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$






          share|cite|improve this answer































            3














            Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
            $$
            det(uv^*+I)=v^*u+1
            $$






            share|cite|improve this answer





















            • Very clever short proof.
              – Jean Marie
              Dec 9 at 23:27






            • 1




              @JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
              – egreg
              Dec 9 at 23:47





















            2














            You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$






            share|cite|improve this answer





















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              3 Answers
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              3














              Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
              $$
              {displaystyle det left({A} +{uv} ^{textsf {T}}right)
              =left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
              $$



              In your case, let $A=I$.



              The proof is essentially based on the following observation:
              $$
              {displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
              $$




              [Added later:]
              Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
              $$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
              $$
              (v^Tu,0,cdots,0).
              $$
              It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
              $$
              (1+v^Tu,1,cdots,1).
              $$
              But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.



              1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$






              share|cite|improve this answer




























                3














                Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
                $$
                {displaystyle det left({A} +{uv} ^{textsf {T}}right)
                =left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
                $$



                In your case, let $A=I$.



                The proof is essentially based on the following observation:
                $$
                {displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
                $$




                [Added later:]
                Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
                $$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
                $$
                (v^Tu,0,cdots,0).
                $$
                It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
                $$
                (1+v^Tu,1,cdots,1).
                $$
                But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.



                1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$






                share|cite|improve this answer


























                  3












                  3








                  3






                  Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
                  $$
                  {displaystyle det left({A} +{uv} ^{textsf {T}}right)
                  =left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
                  $$



                  In your case, let $A=I$.



                  The proof is essentially based on the following observation:
                  $$
                  {displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
                  $$




                  [Added later:]
                  Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
                  $$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
                  $$
                  (v^Tu,0,cdots,0).
                  $$
                  It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
                  $$
                  (1+v^Tu,1,cdots,1).
                  $$
                  But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.



                  1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$






                  share|cite|improve this answer














                  Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
                  $$
                  {displaystyle det left({A} +{uv} ^{textsf {T}}right)
                  =left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
                  $$



                  In your case, let $A=I$.



                  The proof is essentially based on the following observation:
                  $$
                  {displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
                  $$




                  [Added later:]
                  Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
                  $$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
                  $$
                  (v^Tu,0,cdots,0).
                  $$
                  It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
                  $$
                  (1+v^Tu,1,cdots,1).
                  $$
                  But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.



                  1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 21 at 22:06

























                  answered Nov 21 at 21:41









                  user587192

                  1,486112




                  1,486112























                      3














                      Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
                      $$
                      det(uv^*+I)=v^*u+1
                      $$






                      share|cite|improve this answer





















                      • Very clever short proof.
                        – Jean Marie
                        Dec 9 at 23:27






                      • 1




                        @JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
                        – egreg
                        Dec 9 at 23:47


















                      3














                      Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
                      $$
                      det(uv^*+I)=v^*u+1
                      $$






                      share|cite|improve this answer





















                      • Very clever short proof.
                        – Jean Marie
                        Dec 9 at 23:27






                      • 1




                        @JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
                        – egreg
                        Dec 9 at 23:47
















                      3












                      3








                      3






                      Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
                      $$
                      det(uv^*+I)=v^*u+1
                      $$






                      share|cite|improve this answer












                      Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
                      $$
                      det(uv^*+I)=v^*u+1
                      $$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 21 at 21:52









                      egreg

                      177k1484198




                      177k1484198












                      • Very clever short proof.
                        – Jean Marie
                        Dec 9 at 23:27






                      • 1




                        @JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
                        – egreg
                        Dec 9 at 23:47




















                      • Very clever short proof.
                        – Jean Marie
                        Dec 9 at 23:27






                      • 1




                        @JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
                        – egreg
                        Dec 9 at 23:47


















                      Very clever short proof.
                      – Jean Marie
                      Dec 9 at 23:27




                      Very clever short proof.
                      – Jean Marie
                      Dec 9 at 23:27




                      1




                      1




                      @JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
                      – egreg
                      Dec 9 at 23:47






                      @JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
                      – egreg
                      Dec 9 at 23:47













                      2














                      You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$






                      share|cite|improve this answer


























                        2














                        You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$






                        share|cite|improve this answer
























                          2












                          2








                          2






                          You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$






                          share|cite|improve this answer












                          You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 21 at 21:49









                          Fabian

                          19.3k3674




                          19.3k3674






























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