Determinant of a large symmetric block matrix












1












$begingroup$


Consider a given matrix $Q in text{Mat}_N(mathbb{R})$, which is invertible, and $n geq 1$. I am looking for the determinant of the symmetric block matrix $I_n(Q)$ of total size $nN times nN$:
$$I_n(Q) = begin{pmatrix}Id_N & Q & cdots & Q \
Q^intercal & Id_N & cdots & Q \
vdots & vdots & ddots & vdots \
Q^intercal & Q^intercal & cdots & Id_N \
end{pmatrix}$$



If $Q$ is symmetric, I know that $det I_n(Q) = det((Id_N-Q)^{n-1}) det(Id_N+(n-1)Q)$, and I also have a formula for $I_n(Q)^{-1}$, but I couldn't derive a form for the non-symmetric case. Any help would be appreciated ! Thanks :)






linear-algebra matrices matrix-decomposition symmetric-matrices







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    1












    $begingroup$


    Consider a given matrix $Q in text{Mat}_N(mathbb{R})$, which is invertible, and $n geq 1$. I am looking for the determinant of the symmetric block matrix $I_n(Q)$ of total size $nN times nN$:
    $$I_n(Q) = begin{pmatrix}Id_N & Q & cdots & Q \
    Q^intercal & Id_N & cdots & Q \
    vdots & vdots & ddots & vdots \
    Q^intercal & Q^intercal & cdots & Id_N \
    end{pmatrix}$$



    If $Q$ is symmetric, I know that $det I_n(Q) = det((Id_N-Q)^{n-1}) det(Id_N+(n-1)Q)$, and I also have a formula for $I_n(Q)^{-1}$, but I couldn't derive a form for the non-symmetric case. Any help would be appreciated ! Thanks :)






    linear-algebra matrices matrix-decomposition symmetric-matrices







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Consider a given matrix $Q in text{Mat}_N(mathbb{R})$, which is invertible, and $n geq 1$. I am looking for the determinant of the symmetric block matrix $I_n(Q)$ of total size $nN times nN$:
      $$I_n(Q) = begin{pmatrix}Id_N & Q & cdots & Q \
      Q^intercal & Id_N & cdots & Q \
      vdots & vdots & ddots & vdots \
      Q^intercal & Q^intercal & cdots & Id_N \
      end{pmatrix}$$



      If $Q$ is symmetric, I know that $det I_n(Q) = det((Id_N-Q)^{n-1}) det(Id_N+(n-1)Q)$, and I also have a formula for $I_n(Q)^{-1}$, but I couldn't derive a form for the non-symmetric case. Any help would be appreciated ! Thanks :)






      linear-algebra matrices matrix-decomposition symmetric-matrices







      share|cite|improve this question









      $endgroup$




      Consider a given matrix $Q in text{Mat}_N(mathbb{R})$, which is invertible, and $n geq 1$. I am looking for the determinant of the symmetric block matrix $I_n(Q)$ of total size $nN times nN$:
      $$I_n(Q) = begin{pmatrix}Id_N & Q & cdots & Q \
      Q^intercal & Id_N & cdots & Q \
      vdots & vdots & ddots & vdots \
      Q^intercal & Q^intercal & cdots & Id_N \
      end{pmatrix}$$



      If $Q$ is symmetric, I know that $det I_n(Q) = det((Id_N-Q)^{n-1}) det(Id_N+(n-1)Q)$, and I also have a formula for $I_n(Q)^{-1}$, but I couldn't derive a form for the non-symmetric case. Any help would be appreciated ! Thanks :)






      linear-algebra matrices matrix-decomposition symmetric-matrices




      linear-algebra matrices matrix-decomposition symmetric-matrices






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      asked Jan 17 at 23:52









      seampseamp

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          $begingroup$

          Here is a partial answer. Let us consider the case where the number of block rows is $n=4$ first. The other cases are similar. We shall perform blockwise elementary row/column operations on the block matrix
          $$
          pmatrix{
          I&Q&Q&Q\
          Q^T&I&Q&Q\
          Q^T&Q^T&I&Q\
          Q^T&Q^T&Q^T&I}.
          $$
          First, for $i=n,n-1,ldots$ down to $2$, subtract the $i$-th block row by the one above it. We get
          $$
          pmatrix{
          I&Q&Q&Q\
          Q^T-I&I-Q&0&0\
          0&Q^T-I&I-Q&0\
          0&0&Q^T-I&I-Q}.
          $$
          Suppose $1$ is not an eigenvalue of $Q$. Let
          $$
          M=(I-Q)^{-1}(Q^T-I).
          $$
          Then for $j=n-1,n-2,ldots$ down to $1$, perform the block column operation $C_jleftarrow C_j-C_{j+1}M$ successively. When $n=4$, we will obtain, in three steps, the following matrices:
          begin{aligned}
          &pmatrix{
          I&Q&Q-QM&Q\
          Q^T-I&I-Q&0&0\
          0&Q^T-I&I-Q&0\
          0&0&0&I-Q},\
          &pmatrix{
          I&Q-QM+QM^2&Q-QM&Q\
          Q^T-I&I-Q&0&0\
          0&0&I-Q&0\
          0&0&0&I-Q},\
          &pmatrix{
          I-QM+QM^2-QM^3&Q-QM+QM^2&Q-QM&Q\
          0&I-Q&0&0\
          0&0&I-Q&0\
          0&0&0&I-Q}.
          end{aligned}
          Hence at the end we get a block upper triangular matrix. In general, if $1$ is not an eigenvalue of $Q$, the determinant of the original block matrix is given by
          $$
          detleft(I+Qleft[(-M)+(-M)^2+..+(-M)^{n-1}right]right)det(I-Q)^{n-1}.tag{1}
          $$
          Here we do not replace the square bracket term by $[(-M)-(-M)^n](I+M)^{-1}$ because $-1$ may be an eigenvalue of $M$ (in particular, this always occurs if the size $N$ of the matrix $Q$ is odd).



          Since the determinant of the original block matrix is a polynomial in the entries of $Q$, if a generic formula exists, we don't expect it to contain any matrix inverse term. Unfortunately I cannot manage to cancel out the inverse term $(I-Q)^{-1}$ inside the powers of $M$. Therefore formula $(1)$ cannot be further simplified at the moment and the requirement that $1$ is not an eigenvalue of $Q$ cannot be dropped yet.






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            $begingroup$

            Here is a partial answer. Let us consider the case where the number of block rows is $n=4$ first. The other cases are similar. We shall perform blockwise elementary row/column operations on the block matrix
            $$
            pmatrix{
            I&Q&Q&Q\
            Q^T&I&Q&Q\
            Q^T&Q^T&I&Q\
            Q^T&Q^T&Q^T&I}.
            $$
            First, for $i=n,n-1,ldots$ down to $2$, subtract the $i$-th block row by the one above it. We get
            $$
            pmatrix{
            I&Q&Q&Q\
            Q^T-I&I-Q&0&0\
            0&Q^T-I&I-Q&0\
            0&0&Q^T-I&I-Q}.
            $$
            Suppose $1$ is not an eigenvalue of $Q$. Let
            $$
            M=(I-Q)^{-1}(Q^T-I).
            $$
            Then for $j=n-1,n-2,ldots$ down to $1$, perform the block column operation $C_jleftarrow C_j-C_{j+1}M$ successively. When $n=4$, we will obtain, in three steps, the following matrices:
            begin{aligned}
            &pmatrix{
            I&Q&Q-QM&Q\
            Q^T-I&I-Q&0&0\
            0&Q^T-I&I-Q&0\
            0&0&0&I-Q},\
            &pmatrix{
            I&Q-QM+QM^2&Q-QM&Q\
            Q^T-I&I-Q&0&0\
            0&0&I-Q&0\
            0&0&0&I-Q},\
            &pmatrix{
            I-QM+QM^2-QM^3&Q-QM+QM^2&Q-QM&Q\
            0&I-Q&0&0\
            0&0&I-Q&0\
            0&0&0&I-Q}.
            end{aligned}
            Hence at the end we get a block upper triangular matrix. In general, if $1$ is not an eigenvalue of $Q$, the determinant of the original block matrix is given by
            $$
            detleft(I+Qleft[(-M)+(-M)^2+..+(-M)^{n-1}right]right)det(I-Q)^{n-1}.tag{1}
            $$
            Here we do not replace the square bracket term by $[(-M)-(-M)^n](I+M)^{-1}$ because $-1$ may be an eigenvalue of $M$ (in particular, this always occurs if the size $N$ of the matrix $Q$ is odd).



            Since the determinant of the original block matrix is a polynomial in the entries of $Q$, if a generic formula exists, we don't expect it to contain any matrix inverse term. Unfortunately I cannot manage to cancel out the inverse term $(I-Q)^{-1}$ inside the powers of $M$. Therefore formula $(1)$ cannot be further simplified at the moment and the requirement that $1$ is not an eigenvalue of $Q$ cannot be dropped yet.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Here is a partial answer. Let us consider the case where the number of block rows is $n=4$ first. The other cases are similar. We shall perform blockwise elementary row/column operations on the block matrix
              $$
              pmatrix{
              I&Q&Q&Q\
              Q^T&I&Q&Q\
              Q^T&Q^T&I&Q\
              Q^T&Q^T&Q^T&I}.
              $$
              First, for $i=n,n-1,ldots$ down to $2$, subtract the $i$-th block row by the one above it. We get
              $$
              pmatrix{
              I&Q&Q&Q\
              Q^T-I&I-Q&0&0\
              0&Q^T-I&I-Q&0\
              0&0&Q^T-I&I-Q}.
              $$
              Suppose $1$ is not an eigenvalue of $Q$. Let
              $$
              M=(I-Q)^{-1}(Q^T-I).
              $$
              Then for $j=n-1,n-2,ldots$ down to $1$, perform the block column operation $C_jleftarrow C_j-C_{j+1}M$ successively. When $n=4$, we will obtain, in three steps, the following matrices:
              begin{aligned}
              &pmatrix{
              I&Q&Q-QM&Q\
              Q^T-I&I-Q&0&0\
              0&Q^T-I&I-Q&0\
              0&0&0&I-Q},\
              &pmatrix{
              I&Q-QM+QM^2&Q-QM&Q\
              Q^T-I&I-Q&0&0\
              0&0&I-Q&0\
              0&0&0&I-Q},\
              &pmatrix{
              I-QM+QM^2-QM^3&Q-QM+QM^2&Q-QM&Q\
              0&I-Q&0&0\
              0&0&I-Q&0\
              0&0&0&I-Q}.
              end{aligned}
              Hence at the end we get a block upper triangular matrix. In general, if $1$ is not an eigenvalue of $Q$, the determinant of the original block matrix is given by
              $$
              detleft(I+Qleft[(-M)+(-M)^2+..+(-M)^{n-1}right]right)det(I-Q)^{n-1}.tag{1}
              $$
              Here we do not replace the square bracket term by $[(-M)-(-M)^n](I+M)^{-1}$ because $-1$ may be an eigenvalue of $M$ (in particular, this always occurs if the size $N$ of the matrix $Q$ is odd).



              Since the determinant of the original block matrix is a polynomial in the entries of $Q$, if a generic formula exists, we don't expect it to contain any matrix inverse term. Unfortunately I cannot manage to cancel out the inverse term $(I-Q)^{-1}$ inside the powers of $M$. Therefore formula $(1)$ cannot be further simplified at the moment and the requirement that $1$ is not an eigenvalue of $Q$ cannot be dropped yet.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Here is a partial answer. Let us consider the case where the number of block rows is $n=4$ first. The other cases are similar. We shall perform blockwise elementary row/column operations on the block matrix
                $$
                pmatrix{
                I&Q&Q&Q\
                Q^T&I&Q&Q\
                Q^T&Q^T&I&Q\
                Q^T&Q^T&Q^T&I}.
                $$
                First, for $i=n,n-1,ldots$ down to $2$, subtract the $i$-th block row by the one above it. We get
                $$
                pmatrix{
                I&Q&Q&Q\
                Q^T-I&I-Q&0&0\
                0&Q^T-I&I-Q&0\
                0&0&Q^T-I&I-Q}.
                $$
                Suppose $1$ is not an eigenvalue of $Q$. Let
                $$
                M=(I-Q)^{-1}(Q^T-I).
                $$
                Then for $j=n-1,n-2,ldots$ down to $1$, perform the block column operation $C_jleftarrow C_j-C_{j+1}M$ successively. When $n=4$, we will obtain, in three steps, the following matrices:
                begin{aligned}
                &pmatrix{
                I&Q&Q-QM&Q\
                Q^T-I&I-Q&0&0\
                0&Q^T-I&I-Q&0\
                0&0&0&I-Q},\
                &pmatrix{
                I&Q-QM+QM^2&Q-QM&Q\
                Q^T-I&I-Q&0&0\
                0&0&I-Q&0\
                0&0&0&I-Q},\
                &pmatrix{
                I-QM+QM^2-QM^3&Q-QM+QM^2&Q-QM&Q\
                0&I-Q&0&0\
                0&0&I-Q&0\
                0&0&0&I-Q}.
                end{aligned}
                Hence at the end we get a block upper triangular matrix. In general, if $1$ is not an eigenvalue of $Q$, the determinant of the original block matrix is given by
                $$
                detleft(I+Qleft[(-M)+(-M)^2+..+(-M)^{n-1}right]right)det(I-Q)^{n-1}.tag{1}
                $$
                Here we do not replace the square bracket term by $[(-M)-(-M)^n](I+M)^{-1}$ because $-1$ may be an eigenvalue of $M$ (in particular, this always occurs if the size $N$ of the matrix $Q$ is odd).



                Since the determinant of the original block matrix is a polynomial in the entries of $Q$, if a generic formula exists, we don't expect it to contain any matrix inverse term. Unfortunately I cannot manage to cancel out the inverse term $(I-Q)^{-1}$ inside the powers of $M$. Therefore formula $(1)$ cannot be further simplified at the moment and the requirement that $1$ is not an eigenvalue of $Q$ cannot be dropped yet.






                share|cite|improve this answer











                $endgroup$



                Here is a partial answer. Let us consider the case where the number of block rows is $n=4$ first. The other cases are similar. We shall perform blockwise elementary row/column operations on the block matrix
                $$
                pmatrix{
                I&Q&Q&Q\
                Q^T&I&Q&Q\
                Q^T&Q^T&I&Q\
                Q^T&Q^T&Q^T&I}.
                $$
                First, for $i=n,n-1,ldots$ down to $2$, subtract the $i$-th block row by the one above it. We get
                $$
                pmatrix{
                I&Q&Q&Q\
                Q^T-I&I-Q&0&0\
                0&Q^T-I&I-Q&0\
                0&0&Q^T-I&I-Q}.
                $$
                Suppose $1$ is not an eigenvalue of $Q$. Let
                $$
                M=(I-Q)^{-1}(Q^T-I).
                $$
                Then for $j=n-1,n-2,ldots$ down to $1$, perform the block column operation $C_jleftarrow C_j-C_{j+1}M$ successively. When $n=4$, we will obtain, in three steps, the following matrices:
                begin{aligned}
                &pmatrix{
                I&Q&Q-QM&Q\
                Q^T-I&I-Q&0&0\
                0&Q^T-I&I-Q&0\
                0&0&0&I-Q},\
                &pmatrix{
                I&Q-QM+QM^2&Q-QM&Q\
                Q^T-I&I-Q&0&0\
                0&0&I-Q&0\
                0&0&0&I-Q},\
                &pmatrix{
                I-QM+QM^2-QM^3&Q-QM+QM^2&Q-QM&Q\
                0&I-Q&0&0\
                0&0&I-Q&0\
                0&0&0&I-Q}.
                end{aligned}
                Hence at the end we get a block upper triangular matrix. In general, if $1$ is not an eigenvalue of $Q$, the determinant of the original block matrix is given by
                $$
                detleft(I+Qleft[(-M)+(-M)^2+..+(-M)^{n-1}right]right)det(I-Q)^{n-1}.tag{1}
                $$
                Here we do not replace the square bracket term by $[(-M)-(-M)^n](I+M)^{-1}$ because $-1$ may be an eigenvalue of $M$ (in particular, this always occurs if the size $N$ of the matrix $Q$ is odd).



                Since the determinant of the original block matrix is a polynomial in the entries of $Q$, if a generic formula exists, we don't expect it to contain any matrix inverse term. Unfortunately I cannot manage to cancel out the inverse term $(I-Q)^{-1}$ inside the powers of $M$. Therefore formula $(1)$ cannot be further simplified at the moment and the requirement that $1$ is not an eigenvalue of $Q$ cannot be dropped yet.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 18 at 23:12

























                answered Jan 18 at 1:31









                user1551user1551

                73k566128




                73k566128






























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