Mixing
Determinant of matrix of submatrices
$begingroup$
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
edited Jan 5 at 21:11
Davide Giraudo
126k16150261
asked Jan 5 at 19:37
avan1235avan1235
3156
$endgroup$
add a comment |
$begingroup$
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
edited Jan 5 at 21:11
Davide Giraudo
126k16150261
asked Jan 5 at 19:37
avan1235avan1235
3156
$endgroup$
add a comment |
$begingroup$
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
edited Jan 5 at 21:11
Davide Giraudo
126k16150261
asked Jan 5 at 19:37
avan1235avan1235
3156
$endgroup$
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
edited Jan 5 at 21:11
Davide Giraudo
126k16150261
asked Jan 5 at 19:37
avan1235avan1235
3156
edited Jan 5 at 21:11
Davide Giraudo
126k16150261
asked Jan 5 at 19:37
avan1235avan1235
3156
edited Jan 5 at 21:11
Davide Giraudo
126k16150261
edited Jan 5 at 21:11
Davide Giraudo
126k16150261
edited Jan 5 at 21:11
Davide Giraudo
126k16150261
126k16150261
asked Jan 5 at 19:37
avan1235avan1235
3156
asked Jan 5 at 19:37
avan1235avan1235
3156
asked Jan 5 at 19:37
avan1235avan1235
3156
3156
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 5 at 19:47
user1551user1551
72.4k566127
$endgroup$
add a comment |
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$begingroup$
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.
answered Jan 5 at 19:47
user1551user1551
72.4k566127
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 19:47
user1551user1551
72.4k566127
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 5 at 19:47
user1551user1551
72.4k566127
$endgroup$
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.
answered Jan 5 at 19:47
user1551user1551
72.4k566127
answered Jan 5 at 19:47
user1551user1551
72.4k566127
answered Jan 5 at 19:47
user1551user1551
72.4k566127
answered Jan 5 at 19:47
user1551user1551
72.4k566127
72.4k566127
add a comment |
add a comment |
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