Equivalency of simultaneously block diagonalizing two matrices and finding invariant subspaces












1












$begingroup$


@Victorliu specified in a comment for this question: "Block diagonalizing two matrices simultaneously is equivalent to finding invariant subspaces common to both matrices". There are two questions regarding this question:




  1. I am curious and want to know more in details about this equivalency. Is there any reference about this theorem/equivalency?

  2. There is a solution for this question by @JoonasIlmavirta, however, it is not that clear for finding such invariant subspaces. Would you please give more elaboration or give any reference about that.

  3. I am interested in reading papers (or any reference) about block diagonalization of a matrix in details. It would be great if you can suggest any reference.


Thanks.






linear-algebra vector-spaces linear-transformations numerical-linear-algebra







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    @Victorliu specified in a comment for this question: "Block diagonalizing two matrices simultaneously is equivalent to finding invariant subspaces common to both matrices". There are two questions regarding this question:




    1. I am curious and want to know more in details about this equivalency. Is there any reference about this theorem/equivalency?

    2. There is a solution for this question by @JoonasIlmavirta, however, it is not that clear for finding such invariant subspaces. Would you please give more elaboration or give any reference about that.

    3. I am interested in reading papers (or any reference) about block diagonalization of a matrix in details. It would be great if you can suggest any reference.


    Thanks.






    linear-algebra vector-spaces linear-transformations numerical-linear-algebra







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      @Victorliu specified in a comment for this question: "Block diagonalizing two matrices simultaneously is equivalent to finding invariant subspaces common to both matrices". There are two questions regarding this question:




      1. I am curious and want to know more in details about this equivalency. Is there any reference about this theorem/equivalency?

      2. There is a solution for this question by @JoonasIlmavirta, however, it is not that clear for finding such invariant subspaces. Would you please give more elaboration or give any reference about that.

      3. I am interested in reading papers (or any reference) about block diagonalization of a matrix in details. It would be great if you can suggest any reference.


      Thanks.






      linear-algebra vector-spaces linear-transformations numerical-linear-algebra







      share|cite|improve this question











      $endgroup$




      @Victorliu specified in a comment for this question: "Block diagonalizing two matrices simultaneously is equivalent to finding invariant subspaces common to both matrices". There are two questions regarding this question:




      1. I am curious and want to know more in details about this equivalency. Is there any reference about this theorem/equivalency?

      2. There is a solution for this question by @JoonasIlmavirta, however, it is not that clear for finding such invariant subspaces. Would you please give more elaboration or give any reference about that.

      3. I am interested in reading papers (or any reference) about block diagonalization of a matrix in details. It would be great if you can suggest any reference.


      Thanks.






      linear-algebra vector-spaces linear-transformations numerical-linear-algebra




      linear-algebra vector-spaces linear-transformations numerical-linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 7 at 2:47







      Amin

















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      AminAmin

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          $begingroup$

          I don't know of a reference, but here is a way to see the equivalence.
          When you write a matrix in block form, you decompose the underlying space(s) as a direct sum.
          You can also see writing a matrix in a given basis this way; the summands are one-dimensional, corresponding to the chosen basis vectors.



          If the domain is $mathbb R^n=D_1oplus D_2oplusdotsoplus D_k$ and the target is $mathbb R^l=T_1oplusdotsoplus T_m$ (as direct sums of linearly independent subspaces), then an $ltimes n$ matrix $A$ can be written as a $mtimes k$ block matrix.
          Since zero-dimensional spaces $D_i$ or $T_i$ should be excluded (they're just silly), we have $lgeq m$ and $ngeq k$ but there are no other constraints.



          It makes sense to say that the matrix is $A$ block-diagonal if $k=m$ and all off-diagonal blocks are zero.
          Observe that $k=m$ does not imply $n=l$.
          Considering $A$ as a mapping $mathbb R^ntomathbb R^l$, this is the same as requiring that $A(D_i)subset T_i$ for all $i$.
          If you had $A(D_i)notsubset T_i$, then $A(D_i)cap T_j$ would be non-zero for some $ineq j$, meaning that the block at $(j,i)$ is non-zero.



          If you have a square matrix, it is often most convenient to choose $D_i=T_i$ for every $i$, and this is what is typically meant by a block matrix.
          (Observe that if you change the basis of a matrix, you apply the same change on both sides of the matrix.)
          In this block structure block-diagonality means that $A(D_i)subset D_i$, which means that the space $D_i$ is an invariant subspace for $A$.
          That is, block-diagonialization amounts to finding subspaces $D_i$ so that the original space is the direct sum of them and $A(D_i)subset D_i$ for all $i$.



          If you have two matrices that are simultaneously block-diagonal, they both have to satisfy the block-diagonal assumption in the same basis.
          (Simultaneity means precisely that the same basis works for both.)
          That is, two $ntimes n$ matrices $A$ and $B$ are simultaneously block-diagonalized by the subspaces $D_1oplusdots D_k$ if and only if all the spaces are invariant for both:
          $A(D_i)subset D_i$ and $B(D_i)subset D_i$ for all $i$.






          share|cite|improve this answer









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            $begingroup$

            I don't know of a reference, but here is a way to see the equivalence.
            When you write a matrix in block form, you decompose the underlying space(s) as a direct sum.
            You can also see writing a matrix in a given basis this way; the summands are one-dimensional, corresponding to the chosen basis vectors.



            If the domain is $mathbb R^n=D_1oplus D_2oplusdotsoplus D_k$ and the target is $mathbb R^l=T_1oplusdotsoplus T_m$ (as direct sums of linearly independent subspaces), then an $ltimes n$ matrix $A$ can be written as a $mtimes k$ block matrix.
            Since zero-dimensional spaces $D_i$ or $T_i$ should be excluded (they're just silly), we have $lgeq m$ and $ngeq k$ but there are no other constraints.



            It makes sense to say that the matrix is $A$ block-diagonal if $k=m$ and all off-diagonal blocks are zero.
            Observe that $k=m$ does not imply $n=l$.
            Considering $A$ as a mapping $mathbb R^ntomathbb R^l$, this is the same as requiring that $A(D_i)subset T_i$ for all $i$.
            If you had $A(D_i)notsubset T_i$, then $A(D_i)cap T_j$ would be non-zero for some $ineq j$, meaning that the block at $(j,i)$ is non-zero.



            If you have a square matrix, it is often most convenient to choose $D_i=T_i$ for every $i$, and this is what is typically meant by a block matrix.
            (Observe that if you change the basis of a matrix, you apply the same change on both sides of the matrix.)
            In this block structure block-diagonality means that $A(D_i)subset D_i$, which means that the space $D_i$ is an invariant subspace for $A$.
            That is, block-diagonialization amounts to finding subspaces $D_i$ so that the original space is the direct sum of them and $A(D_i)subset D_i$ for all $i$.



            If you have two matrices that are simultaneously block-diagonal, they both have to satisfy the block-diagonal assumption in the same basis.
            (Simultaneity means precisely that the same basis works for both.)
            That is, two $ntimes n$ matrices $A$ and $B$ are simultaneously block-diagonalized by the subspaces $D_1oplusdots D_k$ if and only if all the spaces are invariant for both:
            $A(D_i)subset D_i$ and $B(D_i)subset D_i$ for all $i$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              I don't know of a reference, but here is a way to see the equivalence.
              When you write a matrix in block form, you decompose the underlying space(s) as a direct sum.
              You can also see writing a matrix in a given basis this way; the summands are one-dimensional, corresponding to the chosen basis vectors.



              If the domain is $mathbb R^n=D_1oplus D_2oplusdotsoplus D_k$ and the target is $mathbb R^l=T_1oplusdotsoplus T_m$ (as direct sums of linearly independent subspaces), then an $ltimes n$ matrix $A$ can be written as a $mtimes k$ block matrix.
              Since zero-dimensional spaces $D_i$ or $T_i$ should be excluded (they're just silly), we have $lgeq m$ and $ngeq k$ but there are no other constraints.



              It makes sense to say that the matrix is $A$ block-diagonal if $k=m$ and all off-diagonal blocks are zero.
              Observe that $k=m$ does not imply $n=l$.
              Considering $A$ as a mapping $mathbb R^ntomathbb R^l$, this is the same as requiring that $A(D_i)subset T_i$ for all $i$.
              If you had $A(D_i)notsubset T_i$, then $A(D_i)cap T_j$ would be non-zero for some $ineq j$, meaning that the block at $(j,i)$ is non-zero.



              If you have a square matrix, it is often most convenient to choose $D_i=T_i$ for every $i$, and this is what is typically meant by a block matrix.
              (Observe that if you change the basis of a matrix, you apply the same change on both sides of the matrix.)
              In this block structure block-diagonality means that $A(D_i)subset D_i$, which means that the space $D_i$ is an invariant subspace for $A$.
              That is, block-diagonialization amounts to finding subspaces $D_i$ so that the original space is the direct sum of them and $A(D_i)subset D_i$ for all $i$.



              If you have two matrices that are simultaneously block-diagonal, they both have to satisfy the block-diagonal assumption in the same basis.
              (Simultaneity means precisely that the same basis works for both.)
              That is, two $ntimes n$ matrices $A$ and $B$ are simultaneously block-diagonalized by the subspaces $D_1oplusdots D_k$ if and only if all the spaces are invariant for both:
              $A(D_i)subset D_i$ and $B(D_i)subset D_i$ for all $i$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                I don't know of a reference, but here is a way to see the equivalence.
                When you write a matrix in block form, you decompose the underlying space(s) as a direct sum.
                You can also see writing a matrix in a given basis this way; the summands are one-dimensional, corresponding to the chosen basis vectors.



                If the domain is $mathbb R^n=D_1oplus D_2oplusdotsoplus D_k$ and the target is $mathbb R^l=T_1oplusdotsoplus T_m$ (as direct sums of linearly independent subspaces), then an $ltimes n$ matrix $A$ can be written as a $mtimes k$ block matrix.
                Since zero-dimensional spaces $D_i$ or $T_i$ should be excluded (they're just silly), we have $lgeq m$ and $ngeq k$ but there are no other constraints.



                It makes sense to say that the matrix is $A$ block-diagonal if $k=m$ and all off-diagonal blocks are zero.
                Observe that $k=m$ does not imply $n=l$.
                Considering $A$ as a mapping $mathbb R^ntomathbb R^l$, this is the same as requiring that $A(D_i)subset T_i$ for all $i$.
                If you had $A(D_i)notsubset T_i$, then $A(D_i)cap T_j$ would be non-zero for some $ineq j$, meaning that the block at $(j,i)$ is non-zero.



                If you have a square matrix, it is often most convenient to choose $D_i=T_i$ for every $i$, and this is what is typically meant by a block matrix.
                (Observe that if you change the basis of a matrix, you apply the same change on both sides of the matrix.)
                In this block structure block-diagonality means that $A(D_i)subset D_i$, which means that the space $D_i$ is an invariant subspace for $A$.
                That is, block-diagonialization amounts to finding subspaces $D_i$ so that the original space is the direct sum of them and $A(D_i)subset D_i$ for all $i$.



                If you have two matrices that are simultaneously block-diagonal, they both have to satisfy the block-diagonal assumption in the same basis.
                (Simultaneity means precisely that the same basis works for both.)
                That is, two $ntimes n$ matrices $A$ and $B$ are simultaneously block-diagonalized by the subspaces $D_1oplusdots D_k$ if and only if all the spaces are invariant for both:
                $A(D_i)subset D_i$ and $B(D_i)subset D_i$ for all $i$.






                share|cite|improve this answer









                $endgroup$



                I don't know of a reference, but here is a way to see the equivalence.
                When you write a matrix in block form, you decompose the underlying space(s) as a direct sum.
                You can also see writing a matrix in a given basis this way; the summands are one-dimensional, corresponding to the chosen basis vectors.



                If the domain is $mathbb R^n=D_1oplus D_2oplusdotsoplus D_k$ and the target is $mathbb R^l=T_1oplusdotsoplus T_m$ (as direct sums of linearly independent subspaces), then an $ltimes n$ matrix $A$ can be written as a $mtimes k$ block matrix.
                Since zero-dimensional spaces $D_i$ or $T_i$ should be excluded (they're just silly), we have $lgeq m$ and $ngeq k$ but there are no other constraints.



                It makes sense to say that the matrix is $A$ block-diagonal if $k=m$ and all off-diagonal blocks are zero.
                Observe that $k=m$ does not imply $n=l$.
                Considering $A$ as a mapping $mathbb R^ntomathbb R^l$, this is the same as requiring that $A(D_i)subset T_i$ for all $i$.
                If you had $A(D_i)notsubset T_i$, then $A(D_i)cap T_j$ would be non-zero for some $ineq j$, meaning that the block at $(j,i)$ is non-zero.



                If you have a square matrix, it is often most convenient to choose $D_i=T_i$ for every $i$, and this is what is typically meant by a block matrix.
                (Observe that if you change the basis of a matrix, you apply the same change on both sides of the matrix.)
                In this block structure block-diagonality means that $A(D_i)subset D_i$, which means that the space $D_i$ is an invariant subspace for $A$.
                That is, block-diagonialization amounts to finding subspaces $D_i$ so that the original space is the direct sum of them and $A(D_i)subset D_i$ for all $i$.



                If you have two matrices that are simultaneously block-diagonal, they both have to satisfy the block-diagonal assumption in the same basis.
                (Simultaneity means precisely that the same basis works for both.)
                That is, two $ntimes n$ matrices $A$ and $B$ are simultaneously block-diagonalized by the subspaces $D_1oplusdots D_k$ if and only if all the spaces are invariant for both:
                $A(D_i)subset D_i$ and $B(D_i)subset D_i$ for all $i$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 10:20









                Joonas IlmavirtaJoonas Ilmavirta

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