What is the projector decomposition of a symmetric matrix?












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Hi I am studying linear algebra and I bumped into projector decomposition of symmetric matrices and I just don’t know what it is can anyone of you help please?










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    Hi I am studying linear algebra and I bumped into projector decomposition of symmetric matrices and I just don’t know what it is can anyone of you help please?










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      $begingroup$


      Hi I am studying linear algebra and I bumped into projector decomposition of symmetric matrices and I just don’t know what it is can anyone of you help please?










      share|cite|improve this question









      $endgroup$




      Hi I am studying linear algebra and I bumped into projector decomposition of symmetric matrices and I just don’t know what it is can anyone of you help please?







      linear-algebra matrices






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      asked Jan 4 at 21:42









      Ali DahudAli Dahud

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          If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
          $$
          Ax = sum_{n=1}^{N}lambda_n P_nx
          $$

          where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
          $$
          P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
          $$



          These projections satisfy:
          $$
          P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
          P_nP_m = 0 mbox{ for } nne m \
          AP_n = lambda_n P_n \
          sum_{n=1}^{N} P_n = I,\
          A = sum_{n=1}^{N}lambda_n P_n.
          $$






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            $begingroup$

            If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
            $$
            Ax = sum_{n=1}^{N}lambda_n P_nx
            $$

            where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
            $$
            P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
            $$



            These projections satisfy:
            $$
            P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
            P_nP_m = 0 mbox{ for } nne m \
            AP_n = lambda_n P_n \
            sum_{n=1}^{N} P_n = I,\
            A = sum_{n=1}^{N}lambda_n P_n.
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
              $$
              Ax = sum_{n=1}^{N}lambda_n P_nx
              $$

              where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
              $$
              P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
              $$



              These projections satisfy:
              $$
              P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
              P_nP_m = 0 mbox{ for } nne m \
              AP_n = lambda_n P_n \
              sum_{n=1}^{N} P_n = I,\
              A = sum_{n=1}^{N}lambda_n P_n.
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
                $$
                Ax = sum_{n=1}^{N}lambda_n P_nx
                $$

                where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
                $$
                P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
                $$



                These projections satisfy:
                $$
                P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
                P_nP_m = 0 mbox{ for } nne m \
                AP_n = lambda_n P_n \
                sum_{n=1}^{N} P_n = I,\
                A = sum_{n=1}^{N}lambda_n P_n.
                $$






                share|cite|improve this answer









                $endgroup$



                If $A$ is a symmetric matrix over a real space or a Hermitian matrix over a complex space, then
                $$
                Ax = sum_{n=1}^{N}lambda_n P_nx
                $$

                where ${ lambda_1,cdots,lambda_N }$ are the distinct eigenvalues of $A$, and where $P_n$ is the orthogonal projection onto the eigenspace $mathcal{ker}(A-lambda_n I)$. For example, if ${ e_{n,1},e_{n,2},cdots,e_{n,k_n} }$ is an orthonormal basis of $mbox{ker}(A-lambda_n I)$, then the orthogona projection $P_n$ may be written as
                $$
                P_nx = sum_{j=1}^{k_n}langle x,e_{n,j}rangle e_{n,j}
                $$



                These projections satisfy:
                $$
                P_n^2 = P_n, \ P_n^*= P_n; (mbox{or } P_n^{T}=P_n mbox{ for real spaces}) \
                P_nP_m = 0 mbox{ for } nne m \
                AP_n = lambda_n P_n \
                sum_{n=1}^{N} P_n = I,\
                A = sum_{n=1}^{N}lambda_n P_n.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 22:15









                DisintegratingByPartsDisintegratingByParts

                59.2k42580




                59.2k42580






























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