determinant of a block-diagonal matrix
$begingroup$
Question:
I can do the first part, where $A$ is a $1times1$ matrix.
$$Det(C)=Sigma_i (-1)^{i+j} C_{ij}Det(F_{ij})$$
$F_{ij}$ is the co-factor matrix (a matrix that includes all the elements of $C$ except for the ith row and jth column)
If we set $j=1$ then the only value for $i$ for which $C_{ij}neq0$ is 1. we also know that $C_{11}=Det(A)$ and therfoer the co-factor matrix is $F_{11}B$.
Our expression above becomes $Det(C)=Det(A)det(B)$ as required.
The second bit, however, I couldn't do via induction. I did find one method however:
$$
C=begin{bmatrix}
A & O \
O & I_m
end{bmatrix}
begin{bmatrix}
I_n & O \
O & B
end{bmatrix}
$$
where $I_n$ is the $ntimes n$ identity matrix.
$$
Det(C)=Det
(begin{bmatrix}
A & O \
O & I_m
end{bmatrix})
Det(
begin{bmatrix}
I_n & O \
O & B
end{bmatrix})
$$
Which is clearly (using similar logic to above) $Det(C)=Det(A)det(B)$
Can anyone say if 1. my proof is valid and 2. show me how to solve this via induction.
Thank you
linear-algebra matrices determinant
edited May 24 '17 at 15:57
StubbornAtom
6,02311239
asked May 24 '17 at 15:53
Toby PeterkenToby Peterken
1496
$endgroup$
add a comment |
$begingroup$
Question:
I can do the first part, where $A$ is a $1times1$ matrix.
$$Det(C)=Sigma_i (-1)^{i+j} C_{ij}Det(F_{ij})$$
$F_{ij}$ is the co-factor matrix (a matrix that includes all the elements of $C$ except for the ith row and jth column)
If we set $j=1$ then the only value for $i$ for which $C_{ij}neq0$ is 1. we also know that $C_{11}=Det(A)$ and therfoer the co-factor matrix is $F_{11}B$.
Our expression above becomes $Det(C)=Det(A)det(B)$ as required.
The second bit, however, I couldn't do via induction. I did find one method however:
$$
C=begin{bmatrix}
A & O \
O & I_m
end{bmatrix}
begin{bmatrix}
I_n & O \
O & B
end{bmatrix}
$$
where $I_n$ is the $ntimes n$ identity matrix.
$$
Det(C)=Det
(begin{bmatrix}
A & O \
O & I_m
end{bmatrix})
Det(
begin{bmatrix}
I_n & O \
O & B
end{bmatrix})
$$
Which is clearly (using similar logic to above) $Det(C)=Det(A)det(B)$
Can anyone say if 1. my proof is valid and 2. show me how to solve this via induction.
Thank you
linear-algebra matrices determinant
edited May 24 '17 at 15:57
StubbornAtom
6,02311239
asked May 24 '17 at 15:53
Toby PeterkenToby Peterken
1496
$endgroup$
$begingroup$
I am too lazy to write an answer, but if you apply Laplace formula to the first row of a matrix with $A$ of dimension $n+1$, you get the sum of $n+1$ element containing the determinants of block matrices with block of dimensions $n$ and $m$, where by hypotesis the product rule applies.
$endgroup$
– enzotib
May 24 '17 at 16:16
$begingroup$
We haven't covered the Laplace formula and so, although I can look it up, I don't think that its what the question wants Edit: Laplace is just the the co-factor expansion I did, i didn't know what it was called. Although I thought of that and can't see how.
$endgroup$
– Toby Peterken
May 24 '17 at 16:18
add a comment |
$begingroup$
Question:
I can do the first part, where $A$ is a $1times1$ matrix.
$$Det(C)=Sigma_i (-1)^{i+j} C_{ij}Det(F_{ij})$$
$F_{ij}$ is the co-factor matrix (a matrix that includes all the elements of $C$ except for the ith row and jth column)
If we set $j=1$ then the only value for $i$ for which $C_{ij}neq0$ is 1. we also know that $C_{11}=Det(A)$ and therfoer the co-factor matrix is $F_{11}B$.
Our expression above becomes $Det(C)=Det(A)det(B)$ as required.
The second bit, however, I couldn't do via induction. I did find one method however:
$$
C=begin{bmatrix}
A & O \
O & I_m
end{bmatrix}
begin{bmatrix}
I_n & O \
O & B
end{bmatrix}
$$
where $I_n$ is the $ntimes n$ identity matrix.
$$
Det(C)=Det
(begin{bmatrix}
A & O \
O & I_m
end{bmatrix})
Det(
begin{bmatrix}
I_n & O \
O & B
end{bmatrix})
$$
Which is clearly (using similar logic to above) $Det(C)=Det(A)det(B)$
Can anyone say if 1. my proof is valid and 2. show me how to solve this via induction.
Thank you
linear-algebra matrices determinant
edited May 24 '17 at 15:57
StubbornAtom
6,02311239
asked May 24 '17 at 15:53
Toby PeterkenToby Peterken
1496
$endgroup$
Question:
I can do the first part, where $A$ is a $1times1$ matrix.
$$Det(C)=Sigma_i (-1)^{i+j} C_{ij}Det(F_{ij})$$
$F_{ij}$ is the co-factor matrix (a matrix that includes all the elements of $C$ except for the ith row and jth column)
If we set $j=1$ then the only value for $i$ for which $C_{ij}neq0$ is 1. we also know that $C_{11}=Det(A)$ and therfoer the co-factor matrix is $F_{11}B$.
Our expression above becomes $Det(C)=Det(A)det(B)$ as required.
The second bit, however, I couldn't do via induction. I did find one method however:
$$
C=begin{bmatrix}
A & O \
O & I_m
end{bmatrix}
begin{bmatrix}
I_n & O \
O & B
end{bmatrix}
$$
where $I_n$ is the $ntimes n$ identity matrix.
$$
Det(C)=Det
(begin{bmatrix}
A & O \
O & I_m
end{bmatrix})
Det(
begin{bmatrix}
I_n & O \
O & B
end{bmatrix})
$$
Which is clearly (using similar logic to above) $Det(C)=Det(A)det(B)$
Can anyone say if 1. my proof is valid and 2. show me how to solve this via induction.
Thank you
linear-algebra matrices determinant
linear-algebra matrices determinant
edited May 24 '17 at 15:57
StubbornAtom
6,02311239
asked May 24 '17 at 15:53
Toby PeterkenToby Peterken
1496
edited May 24 '17 at 15:57
StubbornAtom
6,02311239
asked May 24 '17 at 15:53
Toby PeterkenToby Peterken
1496
edited May 24 '17 at 15:57
StubbornAtom
6,02311239
edited May 24 '17 at 15:57
StubbornAtom
6,02311239
edited May 24 '17 at 15:57
StubbornAtom
6,02311239
6,02311239
asked May 24 '17 at 15:53
Toby PeterkenToby Peterken
1496
asked May 24 '17 at 15:53
Toby PeterkenToby Peterken
1496
asked May 24 '17 at 15:53
Toby PeterkenToby Peterken
1496
1496
$begingroup$
I am too lazy to write an answer, but if you apply Laplace formula to the first row of a matrix with $A$ of dimension $n+1$, you get the sum of $n+1$ element containing the determinants of block matrices with block of dimensions $n$ and $m$, where by hypotesis the product rule applies.
$endgroup$
– enzotib
May 24 '17 at 16:16
$begingroup$
We haven't covered the Laplace formula and so, although I can look it up, I don't think that its what the question wants Edit: Laplace is just the the co-factor expansion I did, i didn't know what it was called. Although I thought of that and can't see how.
$endgroup$
– Toby Peterken
May 24 '17 at 16:18
add a comment |
$begingroup$
I am too lazy to write an answer, but if you apply Laplace formula to the first row of a matrix with $A$ of dimension $n+1$, you get the sum of $n+1$ element containing the determinants of block matrices with block of dimensions $n$ and $m$, where by hypotesis the product rule applies.
$endgroup$
– enzotib
May 24 '17 at 16:16
$begingroup$
We haven't covered the Laplace formula and so, although I can look it up, I don't think that its what the question wants Edit: Laplace is just the the co-factor expansion I did, i didn't know what it was called. Although I thought of that and can't see how.
$endgroup$
– Toby Peterken
May 24 '17 at 16:18
$begingroup$
I am too lazy to write an answer, but if you apply Laplace formula to the first row of a matrix with $A$ of dimension $n+1$, you get the sum of $n+1$ element containing the determinants of block matrices with block of dimensions $n$ and $m$, where by hypotesis the product rule applies.
$endgroup$
– enzotib
May 24 '17 at 16:16
$begingroup$
I am too lazy to write an answer, but if you apply Laplace formula to the first row of a matrix with $A$ of dimension $n+1$, you get the sum of $n+1$ element containing the determinants of block matrices with block of dimensions $n$ and $m$, where by hypotesis the product rule applies.
$endgroup$
– enzotib
May 24 '17 at 16:16
$begingroup$
We haven't covered the Laplace formula and so, although I can look it up, I don't think that its what the question wants Edit: Laplace is just the the co-factor expansion I did, i didn't know what it was called. Although I thought of that and can't see how.
$endgroup$
– Toby Peterken
May 24 '17 at 16:18
$begingroup$
We haven't covered the Laplace formula and so, although I can look it up, I don't think that its what the question wants Edit: Laplace is just the the co-factor expansion I did, i didn't know what it was called. Although I thought of that and can't see how.
$endgroup$
– Toby Peterken
May 24 '17 at 16:18
add a comment |
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$begingroup$
I am too lazy to write an answer, but if you apply Laplace formula to the first row of a matrix with $A$ of dimension $n+1$, you get the sum of $n+1$ element containing the determinants of block matrices with block of dimensions $n$ and $m$, where by hypotesis the product rule applies.
$endgroup$
– enzotib
May 24 '17 at 16:16
$begingroup$
We haven't covered the Laplace formula and so, although I can look it up, I don't think that its what the question wants Edit: Laplace is just the the co-factor expansion I did, i didn't know what it was called. Although I thought of that and can't see how.
$endgroup$
– Toby Peterken
May 24 '17 at 16:18