Decomposition of matrix into sum of product of vectors












3












$begingroup$


If I have an $n times n$ matrix $X$ that is symmetric and positive semi-definite, how do I prove that there exists vectors $v_1,ldots,v_n$ such that



$$X = sum_{i=1}^nv_iv_i^T$$



I know this is some form of factorization and know of how to find the eigenvectors of a matrix but am a little stuck on how to approach this. Any hints would be great!






linear-algebra eigenvalues-eigenvectors matrix-decomposition







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    If I have an $n times n$ matrix $X$ that is symmetric and positive semi-definite, how do I prove that there exists vectors $v_1,ldots,v_n$ such that



    $$X = sum_{i=1}^nv_iv_i^T$$



    I know this is some form of factorization and know of how to find the eigenvectors of a matrix but am a little stuck on how to approach this. Any hints would be great!






    linear-algebra eigenvalues-eigenvectors matrix-decomposition







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      If I have an $n times n$ matrix $X$ that is symmetric and positive semi-definite, how do I prove that there exists vectors $v_1,ldots,v_n$ such that



      $$X = sum_{i=1}^nv_iv_i^T$$



      I know this is some form of factorization and know of how to find the eigenvectors of a matrix but am a little stuck on how to approach this. Any hints would be great!






      linear-algebra eigenvalues-eigenvectors matrix-decomposition







      share|cite|improve this question











      $endgroup$




      If I have an $n times n$ matrix $X$ that is symmetric and positive semi-definite, how do I prove that there exists vectors $v_1,ldots,v_n$ such that



      $$X = sum_{i=1}^nv_iv_i^T$$



      I know this is some form of factorization and know of how to find the eigenvectors of a matrix but am a little stuck on how to approach this. Any hints would be great!






      linear-algebra eigenvalues-eigenvectors matrix-decomposition




      linear-algebra eigenvalues-eigenvectors matrix-decomposition






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 25 at 20:53









      mechanodroid

      28.6k62548




      28.6k62548










      asked Jan 25 at 20:14









      AnthonyAnthony

      35519




      35519






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.



          Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.



          Note that we have the equality
          $$V = sum_{i=1}^n lambda_iP_i$$



          Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
          $$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! This makes perfect sense
            $endgroup$
            – Anthony
            Jan 25 at 22:10










          • $begingroup$
            Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
            $endgroup$
            – Anthony
            Jan 26 at 20:59






          • 1




            $begingroup$
            @Anthony It was a typo, thanks for noticing.
            $endgroup$
            – mechanodroid
            Jan 26 at 21:45










          • $begingroup$
            sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
            $endgroup$
            – Anthony
            Jan 27 at 16:51






          • 1




            $begingroup$
            @Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
            $endgroup$
            – mechanodroid
            Jan 27 at 17:25





















          4












          $begingroup$

          Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.



            Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.



            Note that we have the equality
            $$V = sum_{i=1}^n lambda_iP_i$$



            Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
            $$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you! This makes perfect sense
              $endgroup$
              – Anthony
              Jan 25 at 22:10










            • $begingroup$
              Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
              $endgroup$
              – Anthony
              Jan 26 at 20:59






            • 1




              $begingroup$
              @Anthony It was a typo, thanks for noticing.
              $endgroup$
              – mechanodroid
              Jan 26 at 21:45










            • $begingroup$
              sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
              $endgroup$
              – Anthony
              Jan 27 at 16:51






            • 1




              $begingroup$
              @Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
              $endgroup$
              – mechanodroid
              Jan 27 at 17:25


















            1












            $begingroup$

            Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.



            Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.



            Note that we have the equality
            $$V = sum_{i=1}^n lambda_iP_i$$



            Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
            $$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you! This makes perfect sense
              $endgroup$
              – Anthony
              Jan 25 at 22:10










            • $begingroup$
              Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
              $endgroup$
              – Anthony
              Jan 26 at 20:59






            • 1




              $begingroup$
              @Anthony It was a typo, thanks for noticing.
              $endgroup$
              – mechanodroid
              Jan 26 at 21:45










            • $begingroup$
              sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
              $endgroup$
              – Anthony
              Jan 27 at 16:51






            • 1




              $begingroup$
              @Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
              $endgroup$
              – mechanodroid
              Jan 27 at 17:25
















            1












            1








            1





            $begingroup$

            Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.



            Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.



            Note that we have the equality
            $$V = sum_{i=1}^n lambda_iP_i$$



            Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
            $$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$






            share|cite|improve this answer











            $endgroup$



            Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.



            Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.



            Note that we have the equality
            $$V = sum_{i=1}^n lambda_iP_i$$



            Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
            $$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 26 at 21:45

























            answered Jan 25 at 20:50









            mechanodroidmechanodroid

            28.6k62548




            28.6k62548












            • $begingroup$
              Thank you! This makes perfect sense
              $endgroup$
              – Anthony
              Jan 25 at 22:10










            • $begingroup$
              Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
              $endgroup$
              – Anthony
              Jan 26 at 20:59






            • 1




              $begingroup$
              @Anthony It was a typo, thanks for noticing.
              $endgroup$
              – mechanodroid
              Jan 26 at 21:45










            • $begingroup$
              sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
              $endgroup$
              – Anthony
              Jan 27 at 16:51






            • 1




              $begingroup$
              @Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
              $endgroup$
              – mechanodroid
              Jan 27 at 17:25




















            • $begingroup$
              Thank you! This makes perfect sense
              $endgroup$
              – Anthony
              Jan 25 at 22:10










            • $begingroup$
              Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
              $endgroup$
              – Anthony
              Jan 26 at 20:59






            • 1




              $begingroup$
              @Anthony It was a typo, thanks for noticing.
              $endgroup$
              – mechanodroid
              Jan 26 at 21:45










            • $begingroup$
              sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
              $endgroup$
              – Anthony
              Jan 27 at 16:51






            • 1




              $begingroup$
              @Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
              $endgroup$
              – mechanodroid
              Jan 27 at 17:25


















            $begingroup$
            Thank you! This makes perfect sense
            $endgroup$
            – Anthony
            Jan 25 at 22:10




            $begingroup$
            Thank you! This makes perfect sense
            $endgroup$
            – Anthony
            Jan 25 at 22:10












            $begingroup$
            Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
            $endgroup$
            – Anthony
            Jan 26 at 20:59




            $begingroup$
            Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
            $endgroup$
            – Anthony
            Jan 26 at 20:59




            1




            1




            $begingroup$
            @Anthony It was a typo, thanks for noticing.
            $endgroup$
            – mechanodroid
            Jan 26 at 21:45




            $begingroup$
            @Anthony It was a typo, thanks for noticing.
            $endgroup$
            – mechanodroid
            Jan 26 at 21:45












            $begingroup$
            sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
            $endgroup$
            – Anthony
            Jan 27 at 16:51




            $begingroup$
            sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
            $endgroup$
            – Anthony
            Jan 27 at 16:51




            1




            1




            $begingroup$
            @Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
            $endgroup$
            – mechanodroid
            Jan 27 at 17:25






            $begingroup$
            @Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
            $endgroup$
            – mechanodroid
            Jan 27 at 17:25













            4












            $begingroup$

            Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).






                share|cite|improve this answer









                $endgroup$



                Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 25 at 20:43









                ScientificaScientifica

                6,82941335




                6,82941335






























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