Determinant of matrix of submatrices












1














Can you check my solution to:



Task



Having two matrices $X,Yinmathbb{R}^{n,n}$ where $x,yinmathbb{R}$ and matrices are defined as



$
X=begin{bmatrix}
x & 0 &0 & dots & 0 \
x & x & 0& dots & 0\
x & x & x& dots & 0\
vdots&vdots&vdots&vdots&vdots \
x & x & x & dots & x \
end{bmatrix}
$ and $
Y=begin{bmatrix}
0 & dots & 0 &0 & y \
0 & dots & 0 & y & 0\
0 & dots & y & 0& 0\
vdots&vdots&vdots&vdots&vdots \
y & dots & 0 & 0 & 0 \
end{bmatrix}
$



find the $det_{2n}begin{bmatrix}
xI_n & Y \
-Y & X \
end{bmatrix}
$.



Solution:



We can observe that for matrices $A,B,C,Dinmathbb{R}^{n,n}$ when matrix $A$ is invertible then



$begin{bmatrix}
A & B \
C & D \
end{bmatrix}=begin{bmatrix}
A & 0 \
C & I_{n} \
end{bmatrix}cdotbegin{bmatrix}
I_n & A^{-1}B \
0 & D-CA^{-1}B \
end{bmatrix}
$



so



$det_{2n}begin{bmatrix}
A & B \
C & D \
end{bmatrix}=det_{n}(A) cdotdet_{n}(D-CA^{-1}B)$



We see that $xI_n$ is invertible matrix when $xneq 0$ and then we state that



$det_{2n}begin{bmatrix}
xI_n & Y \
-Y & X \
end{bmatrix}=det_{n}(xI_n) cdotdet_{n}(X+Y(xI_n)^{-1}Y)
$



and now let's observe that $det_{n}(xI_n)=x^n$ and



$
X+Y(xI_n)^{-1}Y=X+frac{1}{x}YI_n^{-1}Y=X+frac{1}{x}YI_{n}Y
=X+frac{1}{x}Y^2=X+frac{y^2}{x}I_n
$



so $det_n (X+Y(xI_n)^{-1}Y)=det_{n}left(X+frac{y^2}{x}I_nright) = (x+frac{y^2}{x})^n$



and in the end we can write that $det_{2n}begin{bmatrix}
xI_n & Y \
-Y & X \
end{bmatrix}=x^n (x+frac{y^2}{x})^n = (x^2+y^2)^n
$ when $xneq 0$.



In the case $x=0$ we see that
$det_{2n}begin{bmatrix}
xI_n & Y \
-Y & X \
end{bmatrix}=det_{2n}begin{bmatrix}
0 & Y \
-Y & 0 \
end{bmatrix} =(-1)^n det_{2n}begin{bmatrix}
Y & 0 \
0 & -Y \
end{bmatrix} =(-1)^n cdot y^n cdot (-y)^n = (-1)^ncdot (-1)^n cdot y^{2n} = y^{2n}$



Can you maybe find any simpler solution to this task?






matrices determinant block-matrices







share|cite|improve this question





























    1














    Can you check my solution to:



    Task



    Having two matrices $X,Yinmathbb{R}^{n,n}$ where $x,yinmathbb{R}$ and matrices are defined as



    $
    X=begin{bmatrix}
    x & 0 &0 & dots & 0 \
    x & x & 0& dots & 0\
    x & x & x& dots & 0\
    vdots&vdots&vdots&vdots&vdots \
    x & x & x & dots & x \
    end{bmatrix}
    $     and $
    Y=begin{bmatrix}
    0 & dots & 0 &0 & y \
    0 & dots & 0 & y & 0\
    0 & dots & y & 0& 0\
    vdots&vdots&vdots&vdots&vdots \
    y & dots & 0 & 0 & 0 \
    end{bmatrix}
    $



    find the $det_{2n}begin{bmatrix}
    xI_n & Y \
    -Y & X \
    end{bmatrix}
    $    .



    Solution:



    We can observe that for matrices $A,B,C,Dinmathbb{R}^{n,n}$ when matrix $A$ is invertible then



    $begin{bmatrix}
    A & B \
    C & D \
    end{bmatrix}=begin{bmatrix}
    A & 0 \
    C & I_{n} \
    end{bmatrix}cdotbegin{bmatrix}
    I_n & A^{-1}B \
    0 & D-CA^{-1}B \
    end{bmatrix}
    $



    so



    $det_{2n}begin{bmatrix}
    A & B \
    C & D \
    end{bmatrix}=det_{n}(A) cdotdet_{n}(D-CA^{-1}B)$



    We see that $xI_n$ is invertible matrix when $xneq 0$ and then we state that



    $det_{2n}begin{bmatrix}
    xI_n & Y \
    -Y & X \
    end{bmatrix}=det_{n}(xI_n) cdotdet_{n}(X+Y(xI_n)^{-1}Y)
    $



    and now let's observe that $det_{n}(xI_n)=x^n$ and



    $
    X+Y(xI_n)^{-1}Y=X+frac{1}{x}YI_n^{-1}Y=X+frac{1}{x}YI_{n}Y
    =X+frac{1}{x}Y^2=X+frac{y^2}{x}I_n
    $



    so $det_n (X+Y(xI_n)^{-1}Y)=det_{n}left(X+frac{y^2}{x}I_nright) = (x+frac{y^2}{x})^n$



    and in the end we can write that $det_{2n}begin{bmatrix}
    xI_n & Y \
    -Y & X \
    end{bmatrix}=x^n (x+frac{y^2}{x})^n = (x^2+y^2)^n
    $     when $xneq 0$.



    In the case $x=0$ we see that
    $det_{2n}begin{bmatrix}
    xI_n & Y \
    -Y & X \
    end{bmatrix}=det_{2n}begin{bmatrix}
    0 & Y \
    -Y & 0 \
    end{bmatrix} =(-1)^n det_{2n}begin{bmatrix}
    Y & 0 \
    0 & -Y \
    end{bmatrix} =(-1)^n cdot y^n cdot (-y)^n = (-1)^ncdot (-1)^n cdot y^{2n} = y^{2n}$



    Can you maybe find any simpler solution to this task?






    matrices determinant block-matrices







    share|cite|improve this question



























      1












      1








      1


      0





      Can you check my solution to:



      Task



      Having two matrices $X,Yinmathbb{R}^{n,n}$ where $x,yinmathbb{R}$ and matrices are defined as



      $
      X=begin{bmatrix}
      x & 0 &0 & dots & 0 \
      x & x & 0& dots & 0\
      x & x & x& dots & 0\
      vdots&vdots&vdots&vdots&vdots \
      x & x & x & dots & x \
      end{bmatrix}
      $       and $
      Y=begin{bmatrix}
      0 & dots & 0 &0 & y \
      0 & dots & 0 & y & 0\
      0 & dots & y & 0& 0\
      vdots&vdots&vdots&vdots&vdots \
      y & dots & 0 & 0 & 0 \
      end{bmatrix}
      $



      find the $det_{2n}begin{bmatrix}
      xI_n & Y \
      -Y & X \
      end{bmatrix}
      $      .



      Solution:



      We can observe that for matrices $A,B,C,Dinmathbb{R}^{n,n}$ when matrix $A$ is invertible then



      $begin{bmatrix}
      A & B \
      C & D \
      end{bmatrix}=begin{bmatrix}
      A & 0 \
      C & I_{n} \
      end{bmatrix}cdotbegin{bmatrix}
      I_n & A^{-1}B \
      0 & D-CA^{-1}B \
      end{bmatrix}
      $



      so



      $det_{2n}begin{bmatrix}
      A & B \
      C & D \
      end{bmatrix}=det_{n}(A) cdotdet_{n}(D-CA^{-1}B)$



      We see that $xI_n$ is invertible matrix when $xneq 0$ and then we state that



      $det_{2n}begin{bmatrix}
      xI_n & Y \
      -Y & X \
      end{bmatrix}=det_{n}(xI_n) cdotdet_{n}(X+Y(xI_n)^{-1}Y)
      $



      and now let's observe that $det_{n}(xI_n)=x^n$ and



      $
      X+Y(xI_n)^{-1}Y=X+frac{1}{x}YI_n^{-1}Y=X+frac{1}{x}YI_{n}Y
      =X+frac{1}{x}Y^2=X+frac{y^2}{x}I_n
      $



      so $det_n (X+Y(xI_n)^{-1}Y)=det_{n}left(X+frac{y^2}{x}I_nright) = (x+frac{y^2}{x})^n$



      and in the end we can write that $det_{2n}begin{bmatrix}
      xI_n & Y \
      -Y & X \
      end{bmatrix}=x^n (x+frac{y^2}{x})^n = (x^2+y^2)^n
      $       when $xneq 0$.



      In the case $x=0$ we see that
      $det_{2n}begin{bmatrix}
      xI_n & Y \
      -Y & X \
      end{bmatrix}=det_{2n}begin{bmatrix}
      0 & Y \
      -Y & 0 \
      end{bmatrix} =(-1)^n det_{2n}begin{bmatrix}
      Y & 0 \
      0 & -Y \
      end{bmatrix} =(-1)^n cdot y^n cdot (-y)^n = (-1)^ncdot (-1)^n cdot y^{2n} = y^{2n}$



      Can you maybe find any simpler solution to this task?






      matrices determinant block-matrices







      share|cite|improve this question















      Can you check my solution to:



      Task



      Having two matrices $X,Yinmathbb{R}^{n,n}$ where $x,yinmathbb{R}$ and matrices are defined as



      $
      X=begin{bmatrix}
      x & 0 &0 & dots & 0 \
      x & x & 0& dots & 0\
      x & x & x& dots & 0\
      vdots&vdots&vdots&vdots&vdots \
      x & x & x & dots & x \
      end{bmatrix}
      $       and $
      Y=begin{bmatrix}
      0 & dots & 0 &0 & y \
      0 & dots & 0 & y & 0\
      0 & dots & y & 0& 0\
      vdots&vdots&vdots&vdots&vdots \
      y & dots & 0 & 0 & 0 \
      end{bmatrix}
      $



      find the $det_{2n}begin{bmatrix}
      xI_n & Y \
      -Y & X \
      end{bmatrix}
      $      .



      Solution:



      We can observe that for matrices $A,B,C,Dinmathbb{R}^{n,n}$ when matrix $A$ is invertible then



      $begin{bmatrix}
      A & B \
      C & D \
      end{bmatrix}=begin{bmatrix}
      A & 0 \
      C & I_{n} \
      end{bmatrix}cdotbegin{bmatrix}
      I_n & A^{-1}B \
      0 & D-CA^{-1}B \
      end{bmatrix}
      $



      so



      $det_{2n}begin{bmatrix}
      A & B \
      C & D \
      end{bmatrix}=det_{n}(A) cdotdet_{n}(D-CA^{-1}B)$



      We see that $xI_n$ is invertible matrix when $xneq 0$ and then we state that



      $det_{2n}begin{bmatrix}
      xI_n & Y \
      -Y & X \
      end{bmatrix}=det_{n}(xI_n) cdotdet_{n}(X+Y(xI_n)^{-1}Y)
      $



      and now let's observe that $det_{n}(xI_n)=x^n$ and



      $
      X+Y(xI_n)^{-1}Y=X+frac{1}{x}YI_n^{-1}Y=X+frac{1}{x}YI_{n}Y
      =X+frac{1}{x}Y^2=X+frac{y^2}{x}I_n
      $



      so $det_n (X+Y(xI_n)^{-1}Y)=det_{n}left(X+frac{y^2}{x}I_nright) = (x+frac{y^2}{x})^n$



      and in the end we can write that $det_{2n}begin{bmatrix}
      xI_n & Y \
      -Y & X \
      end{bmatrix}=x^n (x+frac{y^2}{x})^n = (x^2+y^2)^n
      $       when $xneq 0$.



      In the case $x=0$ we see that
      $det_{2n}begin{bmatrix}
      xI_n & Y \
      -Y & X \
      end{bmatrix}=det_{2n}begin{bmatrix}
      0 & Y \
      -Y & 0 \
      end{bmatrix} =(-1)^n det_{2n}begin{bmatrix}
      Y & 0 \
      0 & -Y \
      end{bmatrix} =(-1)^n cdot y^n cdot (-y)^n = (-1)^ncdot (-1)^n cdot y^{2n} = y^{2n}$



      Can you maybe find any simpler solution to this task?






      matrices determinant block-matrices




      matrices determinant block-matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Davide Giraudo

      125k16150261




      125k16150261










      asked 2 days ago









      avan1235avan1235

      1926




      1926






















          1 Answer
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          Your derivation of the main result (that the determinant is $(x^2+y^2)^n$) is correct, but your verification is not. You should have $detpmatrix{0&Y\ -Y&0}=det(Y)^2=y^{2n}$. E.g. if you interchange the first and the last column of $pmatrix{0&Y\ -Y&0}$, you get a factor of $-1$. However, the entry $-y$ also contains a minus sign. So, the two minus signs cancel out each other when you compute the determinant. The same holds if you interchange the $j$-th and the $(2n-j)$-th columns of the matrix. Therefore the final determinant does not carry any minus sign.






          share|cite|improve this answer





















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            active

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            3














            Your derivation of the main result (that the determinant is $(x^2+y^2)^n$) is correct, but your verification is not. You should have $detpmatrix{0&Y\ -Y&0}=det(Y)^2=y^{2n}$. E.g. if you interchange the first and the last column of $pmatrix{0&Y\ -Y&0}$, you get a factor of $-1$. However, the entry $-y$ also contains a minus sign. So, the two minus signs cancel out each other when you compute the determinant. The same holds if you interchange the $j$-th and the $(2n-j)$-th columns of the matrix. Therefore the final determinant does not carry any minus sign.






            share|cite|improve this answer


























              3














              Your derivation of the main result (that the determinant is $(x^2+y^2)^n$) is correct, but your verification is not. You should have $detpmatrix{0&Y\ -Y&0}=det(Y)^2=y^{2n}$. E.g. if you interchange the first and the last column of $pmatrix{0&Y\ -Y&0}$, you get a factor of $-1$. However, the entry $-y$ also contains a minus sign. So, the two minus signs cancel out each other when you compute the determinant. The same holds if you interchange the $j$-th and the $(2n-j)$-th columns of the matrix. Therefore the final determinant does not carry any minus sign.






              share|cite|improve this answer
























                3












                3








                3






                Your derivation of the main result (that the determinant is $(x^2+y^2)^n$) is correct, but your verification is not. You should have $detpmatrix{0&Y\ -Y&0}=det(Y)^2=y^{2n}$. E.g. if you interchange the first and the last column of $pmatrix{0&Y\ -Y&0}$, you get a factor of $-1$. However, the entry $-y$ also contains a minus sign. So, the two minus signs cancel out each other when you compute the determinant. The same holds if you interchange the $j$-th and the $(2n-j)$-th columns of the matrix. Therefore the final determinant does not carry any minus sign.






                share|cite|improve this answer












                Your derivation of the main result (that the determinant is $(x^2+y^2)^n$) is correct, but your verification is not. You should have $detpmatrix{0&Y\ -Y&0}=det(Y)^2=y^{2n}$. E.g. if you interchange the first and the last column of $pmatrix{0&Y\ -Y&0}$, you get a factor of $-1$. However, the entry $-y$ also contains a minus sign. So, the two minus signs cancel out each other when you compute the determinant. The same holds if you interchange the $j$-th and the $(2n-j)$-th columns of the matrix. Therefore the final determinant does not carry any minus sign.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                user1551user1551

                71.8k566125




                71.8k566125






























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