Expressing the determinant of a sum of two matrices?Is there an operation on matrices such that the...
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Expressing the determinant of a sum of two matrices?
Is there an operation on matrices such that the determinant yields a homomorphism with the additive group of the reals?The Determinant of a Sum of MatricesProve: $v$ is an eigenvector of $A$ with eigenvalue $lambda$ $implies$ $v$ is an eigenvector of $A^{-1}$ with an eigenvalue of $frac{1}{lambda}$determinant of sum of matricesReal matrices whose squares sum up to $O_2$Determinant of unspecified matricesDeterminant of multiplication of two nonsquare matricesFactorization of 2x2 rational matricesDeterminant of the sum of two matrices with positive determinantMatrices with invariant determinantDeterminant of the sum of rank-$1$ matricesDeterminant of sum of squares of two matrices.determinant of sum of matricesFormula for determinant of sum of matricesdeterminant of the sum of two matrices det(A+B)
$begingroup$
Can
$$det(A + B)$$
be expressed in terms of
$$det(A), det(B), n$$
where $A,B$ are $ntimes n$ matrices?
#
I made the edit to allow $n$ to be factored in.
linear-algebra matrices multivariable-calculus determinant
edited Apr 28 '14 at 19:57
Davide Giraudo
127k17154268
asked Feb 12 '14 at 16:04
frogeyedpeasfrogeyedpeas
7,56572053
$endgroup$
add a comment |
$begingroup$
Can
$$det(A + B)$$
be expressed in terms of
$$det(A), det(B), n$$
where $A,B$ are $ntimes n$ matrices?
#
I made the edit to allow $n$ to be factored in.
linear-algebra matrices multivariable-calculus determinant
edited Apr 28 '14 at 19:57
Davide Giraudo
127k17154268
asked Feb 12 '14 at 16:04
frogeyedpeasfrogeyedpeas
7,56572053
$endgroup$
4
$begingroup$
Not in general. Even if $A,B$ are $n times n$ identity matrices, $det(A+B) = 2^n$ while $det(A) = det(B) = 1$, so the connection will depend on $n$ as well...
$endgroup$
– gt6989b
Feb 12 '14 at 16:11
$begingroup$
There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:13
3
$begingroup$
Also, in light of math.stackexchange.com/questions/298454/… , I think such a formula will always depend on more than just $det A, det B$
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:14
$begingroup$
@ABlumenthal I'm having a hard time comprehending your link although it seems to answer my question. Can you explain it to me?
$endgroup$
– frogeyedpeas
Feb 14 '14 at 5:54
$begingroup$
I hope I am not making any mistake but what the link says for this case is that determinant of sum, is sum of determinants of $2^n$ matrices which are constructed by choosing for each column i either ith column of A or ith column of B (all possible choices are $2^n$ if you think about it).
$endgroup$
– kon psych
Apr 12 '16 at 7:26
add a comment |
$begingroup$
Can
$$det(A + B)$$
be expressed in terms of
$$det(A), det(B), n$$
where $A,B$ are $ntimes n$ matrices?
#
I made the edit to allow $n$ to be factored in.
linear-algebra matrices multivariable-calculus determinant
edited Apr 28 '14 at 19:57
Davide Giraudo
127k17154268
asked Feb 12 '14 at 16:04
frogeyedpeasfrogeyedpeas
7,56572053
$endgroup$
Can
$$det(A + B)$$
be expressed in terms of
$$det(A), det(B), n$$
where $A,B$ are $ntimes n$ matrices?
#
I made the edit to allow $n$ to be factored in.
linear-algebra matrices multivariable-calculus determinant
linear-algebra matrices multivariable-calculus determinant
edited Apr 28 '14 at 19:57
Davide Giraudo
127k17154268
asked Feb 12 '14 at 16:04
frogeyedpeasfrogeyedpeas
7,56572053
edited Apr 28 '14 at 19:57
Davide Giraudo
127k17154268
asked Feb 12 '14 at 16:04
frogeyedpeasfrogeyedpeas
7,56572053
edited Apr 28 '14 at 19:57
Davide Giraudo
127k17154268
edited Apr 28 '14 at 19:57
Davide Giraudo
127k17154268
edited Apr 28 '14 at 19:57
Davide Giraudo
127k17154268
127k17154268
asked Feb 12 '14 at 16:04
frogeyedpeasfrogeyedpeas
7,56572053
asked Feb 12 '14 at 16:04
frogeyedpeasfrogeyedpeas
7,56572053
asked Feb 12 '14 at 16:04
frogeyedpeasfrogeyedpeas
7,56572053
7,56572053
4
$begingroup$
Not in general. Even if $A,B$ are $n times n$ identity matrices, $det(A+B) = 2^n$ while $det(A) = det(B) = 1$, so the connection will depend on $n$ as well...
$endgroup$
– gt6989b
Feb 12 '14 at 16:11
$begingroup$
There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:13
3
$begingroup$
Also, in light of math.stackexchange.com/questions/298454/… , I think such a formula will always depend on more than just $det A, det B$
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:14
$begingroup$
@ABlumenthal I'm having a hard time comprehending your link although it seems to answer my question. Can you explain it to me?
$endgroup$
– frogeyedpeas
Feb 14 '14 at 5:54
$begingroup$
I hope I am not making any mistake but what the link says for this case is that determinant of sum, is sum of determinants of $2^n$ matrices which are constructed by choosing for each column i either ith column of A or ith column of B (all possible choices are $2^n$ if you think about it).
$endgroup$
– kon psych
Apr 12 '16 at 7:26
add a comment |
4
$begingroup$
Not in general. Even if $A,B$ are $n times n$ identity matrices, $det(A+B) = 2^n$ while $det(A) = det(B) = 1$, so the connection will depend on $n$ as well...
$endgroup$
– gt6989b
Feb 12 '14 at 16:11
$begingroup$
There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:13
3
$begingroup$
Also, in light of math.stackexchange.com/questions/298454/… , I think such a formula will always depend on more than just $det A, det B$
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:14
$begingroup$
@ABlumenthal I'm having a hard time comprehending your link although it seems to answer my question. Can you explain it to me?
$endgroup$
– frogeyedpeas
Feb 14 '14 at 5:54
$begingroup$
I hope I am not making any mistake but what the link says for this case is that determinant of sum, is sum of determinants of $2^n$ matrices which are constructed by choosing for each column i either ith column of A or ith column of B (all possible choices are $2^n$ if you think about it).
$endgroup$
– kon psych
Apr 12 '16 at 7:26
4
4
$begingroup$
Not in general. Even if $A,B$ are $n times n$ identity matrices, $det(A+B) = 2^n$ while $det(A) = det(B) = 1$, so the connection will depend on $n$ as well...
$endgroup$
– gt6989b
Feb 12 '14 at 16:11
$begingroup$
Not in general. Even if $A,B$ are $n times n$ identity matrices, $det(A+B) = 2^n$ while $det(A) = det(B) = 1$, so the connection will depend on $n$ as well...
$endgroup$
– gt6989b
Feb 12 '14 at 16:11
$begingroup$
There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:13
$begingroup$
There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:13
3
3
$begingroup$
Also, in light of math.stackexchange.com/questions/298454/… , I think such a formula will always depend on more than just $det A, det B$
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:14
$begingroup$
Also, in light of math.stackexchange.com/questions/298454/… , I think such a formula will always depend on more than just $det A, det B$
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:14
$begingroup$
@ABlumenthal I'm having a hard time comprehending your link although it seems to answer my question. Can you explain it to me?
$endgroup$
– frogeyedpeas
Feb 14 '14 at 5:54
$begingroup$
@ABlumenthal I'm having a hard time comprehending your link although it seems to answer my question. Can you explain it to me?
$endgroup$
– frogeyedpeas
Feb 14 '14 at 5:54
$begingroup$
I hope I am not making any mistake but what the link says for this case is that determinant of sum, is sum of determinants of $2^n$ matrices which are constructed by choosing for each column i either ith column of A or ith column of B (all possible choices are $2^n$ if you think about it).
$endgroup$
– kon psych
Apr 12 '16 at 7:26
$begingroup$
I hope I am not making any mistake but what the link says for this case is that determinant of sum, is sum of determinants of $2^n$ matrices which are constructed by choosing for each column i either ith column of A or ith column of B (all possible choices are $2^n$ if you think about it).
$endgroup$
– kon psych
Apr 12 '16 at 7:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When $nge2$, the answer is no. To illustrate, consider
$$
A=I_n,quad B_1=pmatrix{1&1\ 0&0}oplus0,quad B_2=pmatrix{1&1\ 1&1}oplus0.
$$
If $det(A+B)=fleft(det(A),det(B),nright)$ for some function $f$, you should get $det(A+B_1)=f(1,0,n)=det(A+B_2)$. But in fact, $det(A+B_1)=2ne3=det(A+B_2)$ over any field.
answered Apr 29 '14 at 7:21
user1551user1551
73.5k566129
$endgroup$
add a comment |
$begingroup$
When $n=2$, and suppose $A$ has inverse, you can easily show that
$det(A+B)=det A+det B+det A,cdot mathrm{Tr}(A^{-1}B)$.
Let me give a general method to find the determinant of the sum of two matrices $A,B$ with $A$ invertible and symmetric (The following result might also apply to the non-symmetric case. I might verify that later...).
I am a physicist, so I will use the index notation, $A_{ij}$ and $B_{ij}$, with $i,j=1,2,cdots,n$.
Let $A^{ij}$ donate the inverse of $A_{ij}$ such that $A^{il}A_{lj}=delta^i_j=A_{jl}A^{li}$.
We can use $A_{ij}$ to lower the indices, and its inverse to raise.
For example $A^{il}B_{lj}=B^i{}_j$.
Here and in the following, the Einstein summation rule is assumed.
Let $epsilon^{i_1cdots i_n}$ be the totally antisymmetric tensor, with $epsilon^{1cdots n}=1$.
Define a new tensor $tildeepsilon^{i_1cdots i_n}=epsilon^{i_1cdots i_n}/sqrt{|det A|}$.
We can use $A_{ij}$ to lower the indices of $tildeepsilon^{i_1cdots i_n}$, and define
$tildeepsilon_{i_1cdots i_n}=A_{i_1j_1}cdots A_{i_nj_n}tildeepsilon^{j_1cdots j_n}$.
Then there is a useful property:
$$
tildeepsilon_{i_1cdots i_kl_{k+1}cdots l_n}tildeepsilon^{j_1cdots j_kl_{k+1}cdots l_n}=(-1)^sl!(n-l)!delta^{[j_1}_{i_1}cdotsdelta^{j_k]}_{i_k},
$$
where the square brackets $[]$ imply the antisymmetrization of the indices enclosed by them.
$s$ is the number of negative elements of $A_{ij}$ after it has been diagonalized.
So now the determinant of $A+B$ can be obtained in the following way
$$
det(A+B)=frac{1}{n!}epsilon^{i_1cdots i_n}epsilon^{j_1cdots j_n}(A+B)_{i_1j_1}cdots(A+B)_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}tildeepsilon^{i_1cdots i_n}tildeepsilon^{j_1cdots j_n}sum_{k=0}^n C_n^kA_{i_1j_1}cdots A_{i_kj_k}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon^{j_1cdots j_k}{}_{i_{k+1}cdots i_n}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon_{j_1cdots j_ki_{k+1}cdots i_n}B_{i_{k+1}}{}^{j_{k+1}}cdots B_{i_n}{}^{j_n}
$$
$$
=frac{det A}{n!}sum_{k=0}^nC_n^kk!(n-k)!B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det Asum_{k=0}^nB_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det A+det Asum_{k=1}^{n-1}B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}+det B.
$$
This reproduces the result for $n=2$.
An interesting result for physicists is when $n=3$,
begin{split}
det(A+B)=&det A+det Acdottext{Tr}(A^{-1}B)+frac{det A}{2}{[text{Tr}(A^{-1}B)]^2-text{Tr}(BA^{-1}BA^{-1})}\
&+frac{1}{6}{[text{Tr}(BA^{-1})]^3-3text{Tr}(BA^{-1})text{Tr}(BA^{-1}BA^{-1})+2text{Tr}(BA^{-1}BA^{-1}BA^{-1})}\
&+det B.
end{split}
answered Sep 22 '16 at 13:32
Drake MarquisDrake Marquis
640513
$endgroup$
$begingroup$
This, in my opinion, is the better answer, as it not only falsifies the question, but also salvages the false statement with an interesting and relevant identity. (+1)
$endgroup$
– Frpzzd
Sep 22 '18 at 22:09
1
$begingroup$
Anything for $n=3$?
$endgroup$
– Frpzzd
Sep 22 '18 at 22:17
$begingroup$
@Frpzzd Unfortunately, I did not find any interesting result for $n=3$.
$endgroup$
– Drake Marquis
Oct 8 '18 at 1:24
$begingroup$
@Frpzzd Please check my new answer.
$endgroup$
– Drake Marquis
17 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When $nge2$, the answer is no. To illustrate, consider
$$
A=I_n,quad B_1=pmatrix{1&1\ 0&0}oplus0,quad B_2=pmatrix{1&1\ 1&1}oplus0.
$$
If $det(A+B)=fleft(det(A),det(B),nright)$ for some function $f$, you should get $det(A+B_1)=f(1,0,n)=det(A+B_2)$. But in fact, $det(A+B_1)=2ne3=det(A+B_2)$ over any field.
answered Apr 29 '14 at 7:21
user1551user1551
73.5k566129
$endgroup$
add a comment |
$begingroup$
When $nge2$, the answer is no. To illustrate, consider
$$
A=I_n,quad B_1=pmatrix{1&1\ 0&0}oplus0,quad B_2=pmatrix{1&1\ 1&1}oplus0.
$$
If $det(A+B)=fleft(det(A),det(B),nright)$ for some function $f$, you should get $det(A+B_1)=f(1,0,n)=det(A+B_2)$. But in fact, $det(A+B_1)=2ne3=det(A+B_2)$ over any field.
answered Apr 29 '14 at 7:21
user1551user1551
73.5k566129
$endgroup$
add a comment |
$begingroup$
When $nge2$, the answer is no. To illustrate, consider
$$
A=I_n,quad B_1=pmatrix{1&1\ 0&0}oplus0,quad B_2=pmatrix{1&1\ 1&1}oplus0.
$$
If $det(A+B)=fleft(det(A),det(B),nright)$ for some function $f$, you should get $det(A+B_1)=f(1,0,n)=det(A+B_2)$. But in fact, $det(A+B_1)=2ne3=det(A+B_2)$ over any field.
answered Apr 29 '14 at 7:21
user1551user1551
73.5k566129
$endgroup$
When $nge2$, the answer is no. To illustrate, consider
$$
A=I_n,quad B_1=pmatrix{1&1\ 0&0}oplus0,quad B_2=pmatrix{1&1\ 1&1}oplus0.
$$
If $det(A+B)=fleft(det(A),det(B),nright)$ for some function $f$, you should get $det(A+B_1)=f(1,0,n)=det(A+B_2)$. But in fact, $det(A+B_1)=2ne3=det(A+B_2)$ over any field.
answered Apr 29 '14 at 7:21
user1551user1551
73.5k566129
edited Jul 22 '16 at 15:47
answered Apr 29 '14 at 7:21
user1551user1551
73.5k566129
answered Apr 29 '14 at 7:21
user1551user1551
73.5k566129
answered Apr 29 '14 at 7:21
user1551user1551
73.5k566129
73.5k566129
add a comment |
add a comment |
$begingroup$
When $n=2$, and suppose $A$ has inverse, you can easily show that
$det(A+B)=det A+det B+det A,cdot mathrm{Tr}(A^{-1}B)$.
Let me give a general method to find the determinant of the sum of two matrices $A,B$ with $A$ invertible and symmetric (The following result might also apply to the non-symmetric case. I might verify that later...).
I am a physicist, so I will use the index notation, $A_{ij}$ and $B_{ij}$, with $i,j=1,2,cdots,n$.
Let $A^{ij}$ donate the inverse of $A_{ij}$ such that $A^{il}A_{lj}=delta^i_j=A_{jl}A^{li}$.
We can use $A_{ij}$ to lower the indices, and its inverse to raise.
For example $A^{il}B_{lj}=B^i{}_j$.
Here and in the following, the Einstein summation rule is assumed.
Let $epsilon^{i_1cdots i_n}$ be the totally antisymmetric tensor, with $epsilon^{1cdots n}=1$.
Define a new tensor $tildeepsilon^{i_1cdots i_n}=epsilon^{i_1cdots i_n}/sqrt{|det A|}$.
We can use $A_{ij}$ to lower the indices of $tildeepsilon^{i_1cdots i_n}$, and define
$tildeepsilon_{i_1cdots i_n}=A_{i_1j_1}cdots A_{i_nj_n}tildeepsilon^{j_1cdots j_n}$.
Then there is a useful property:
$$
tildeepsilon_{i_1cdots i_kl_{k+1}cdots l_n}tildeepsilon^{j_1cdots j_kl_{k+1}cdots l_n}=(-1)^sl!(n-l)!delta^{[j_1}_{i_1}cdotsdelta^{j_k]}_{i_k},
$$
where the square brackets $[]$ imply the antisymmetrization of the indices enclosed by them.
$s$ is the number of negative elements of $A_{ij}$ after it has been diagonalized.
So now the determinant of $A+B$ can be obtained in the following way
$$
det(A+B)=frac{1}{n!}epsilon^{i_1cdots i_n}epsilon^{j_1cdots j_n}(A+B)_{i_1j_1}cdots(A+B)_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}tildeepsilon^{i_1cdots i_n}tildeepsilon^{j_1cdots j_n}sum_{k=0}^n C_n^kA_{i_1j_1}cdots A_{i_kj_k}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon^{j_1cdots j_k}{}_{i_{k+1}cdots i_n}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon_{j_1cdots j_ki_{k+1}cdots i_n}B_{i_{k+1}}{}^{j_{k+1}}cdots B_{i_n}{}^{j_n}
$$
$$
=frac{det A}{n!}sum_{k=0}^nC_n^kk!(n-k)!B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det Asum_{k=0}^nB_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det A+det Asum_{k=1}^{n-1}B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}+det B.
$$
This reproduces the result for $n=2$.
An interesting result for physicists is when $n=3$,
begin{split}
det(A+B)=&det A+det Acdottext{Tr}(A^{-1}B)+frac{det A}{2}{[text{Tr}(A^{-1}B)]^2-text{Tr}(BA^{-1}BA^{-1})}\
&+frac{1}{6}{[text{Tr}(BA^{-1})]^3-3text{Tr}(BA^{-1})text{Tr}(BA^{-1}BA^{-1})+2text{Tr}(BA^{-1}BA^{-1}BA^{-1})}\
&+det B.
end{split}
answered Sep 22 '16 at 13:32
Drake MarquisDrake Marquis
640513
$endgroup$
$begingroup$
This, in my opinion, is the better answer, as it not only falsifies the question, but also salvages the false statement with an interesting and relevant identity. (+1)
$endgroup$
– Frpzzd
Sep 22 '18 at 22:09
1
$begingroup$
Anything for $n=3$?
$endgroup$
– Frpzzd
Sep 22 '18 at 22:17
$begingroup$
@Frpzzd Unfortunately, I did not find any interesting result for $n=3$.
$endgroup$
– Drake Marquis
Oct 8 '18 at 1:24
$begingroup$
@Frpzzd Please check my new answer.
$endgroup$
– Drake Marquis
17 hours ago
add a comment |
$begingroup$
When $n=2$, and suppose $A$ has inverse, you can easily show that
$det(A+B)=det A+det B+det A,cdot mathrm{Tr}(A^{-1}B)$.
Let me give a general method to find the determinant of the sum of two matrices $A,B$ with $A$ invertible and symmetric (The following result might also apply to the non-symmetric case. I might verify that later...).
I am a physicist, so I will use the index notation, $A_{ij}$ and $B_{ij}$, with $i,j=1,2,cdots,n$.
Let $A^{ij}$ donate the inverse of $A_{ij}$ such that $A^{il}A_{lj}=delta^i_j=A_{jl}A^{li}$.
We can use $A_{ij}$ to lower the indices, and its inverse to raise.
For example $A^{il}B_{lj}=B^i{}_j$.
Here and in the following, the Einstein summation rule is assumed.
Let $epsilon^{i_1cdots i_n}$ be the totally antisymmetric tensor, with $epsilon^{1cdots n}=1$.
Define a new tensor $tildeepsilon^{i_1cdots i_n}=epsilon^{i_1cdots i_n}/sqrt{|det A|}$.
We can use $A_{ij}$ to lower the indices of $tildeepsilon^{i_1cdots i_n}$, and define
$tildeepsilon_{i_1cdots i_n}=A_{i_1j_1}cdots A_{i_nj_n}tildeepsilon^{j_1cdots j_n}$.
Then there is a useful property:
$$
tildeepsilon_{i_1cdots i_kl_{k+1}cdots l_n}tildeepsilon^{j_1cdots j_kl_{k+1}cdots l_n}=(-1)^sl!(n-l)!delta^{[j_1}_{i_1}cdotsdelta^{j_k]}_{i_k},
$$
where the square brackets $[]$ imply the antisymmetrization of the indices enclosed by them.
$s$ is the number of negative elements of $A_{ij}$ after it has been diagonalized.
So now the determinant of $A+B$ can be obtained in the following way
$$
det(A+B)=frac{1}{n!}epsilon^{i_1cdots i_n}epsilon^{j_1cdots j_n}(A+B)_{i_1j_1}cdots(A+B)_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}tildeepsilon^{i_1cdots i_n}tildeepsilon^{j_1cdots j_n}sum_{k=0}^n C_n^kA_{i_1j_1}cdots A_{i_kj_k}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon^{j_1cdots j_k}{}_{i_{k+1}cdots i_n}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon_{j_1cdots j_ki_{k+1}cdots i_n}B_{i_{k+1}}{}^{j_{k+1}}cdots B_{i_n}{}^{j_n}
$$
$$
=frac{det A}{n!}sum_{k=0}^nC_n^kk!(n-k)!B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det Asum_{k=0}^nB_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det A+det Asum_{k=1}^{n-1}B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}+det B.
$$
This reproduces the result for $n=2$.
An interesting result for physicists is when $n=3$,
begin{split}
det(A+B)=&det A+det Acdottext{Tr}(A^{-1}B)+frac{det A}{2}{[text{Tr}(A^{-1}B)]^2-text{Tr}(BA^{-1}BA^{-1})}\
&+frac{1}{6}{[text{Tr}(BA^{-1})]^3-3text{Tr}(BA^{-1})text{Tr}(BA^{-1}BA^{-1})+2text{Tr}(BA^{-1}BA^{-1}BA^{-1})}\
&+det B.
end{split}
answered Sep 22 '16 at 13:32
Drake MarquisDrake Marquis
640513
$endgroup$
$begingroup$
This, in my opinion, is the better answer, as it not only falsifies the question, but also salvages the false statement with an interesting and relevant identity. (+1)
$endgroup$
– Frpzzd
Sep 22 '18 at 22:09
1
$begingroup$
Anything for $n=3$?
$endgroup$
– Frpzzd
Sep 22 '18 at 22:17
$begingroup$
@Frpzzd Unfortunately, I did not find any interesting result for $n=3$.
$endgroup$
– Drake Marquis
Oct 8 '18 at 1:24
$begingroup$
@Frpzzd Please check my new answer.
$endgroup$
– Drake Marquis
17 hours ago
add a comment |
$begingroup$
When $n=2$, and suppose $A$ has inverse, you can easily show that
$det(A+B)=det A+det B+det A,cdot mathrm{Tr}(A^{-1}B)$.
Let me give a general method to find the determinant of the sum of two matrices $A,B$ with $A$ invertible and symmetric (The following result might also apply to the non-symmetric case. I might verify that later...).
I am a physicist, so I will use the index notation, $A_{ij}$ and $B_{ij}$, with $i,j=1,2,cdots,n$.
Let $A^{ij}$ donate the inverse of $A_{ij}$ such that $A^{il}A_{lj}=delta^i_j=A_{jl}A^{li}$.
We can use $A_{ij}$ to lower the indices, and its inverse to raise.
For example $A^{il}B_{lj}=B^i{}_j$.
Here and in the following, the Einstein summation rule is assumed.
Let $epsilon^{i_1cdots i_n}$ be the totally antisymmetric tensor, with $epsilon^{1cdots n}=1$.
Define a new tensor $tildeepsilon^{i_1cdots i_n}=epsilon^{i_1cdots i_n}/sqrt{|det A|}$.
We can use $A_{ij}$ to lower the indices of $tildeepsilon^{i_1cdots i_n}$, and define
$tildeepsilon_{i_1cdots i_n}=A_{i_1j_1}cdots A_{i_nj_n}tildeepsilon^{j_1cdots j_n}$.
Then there is a useful property:
$$
tildeepsilon_{i_1cdots i_kl_{k+1}cdots l_n}tildeepsilon^{j_1cdots j_kl_{k+1}cdots l_n}=(-1)^sl!(n-l)!delta^{[j_1}_{i_1}cdotsdelta^{j_k]}_{i_k},
$$
where the square brackets $[]$ imply the antisymmetrization of the indices enclosed by them.
$s$ is the number of negative elements of $A_{ij}$ after it has been diagonalized.
So now the determinant of $A+B$ can be obtained in the following way
$$
det(A+B)=frac{1}{n!}epsilon^{i_1cdots i_n}epsilon^{j_1cdots j_n}(A+B)_{i_1j_1}cdots(A+B)_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}tildeepsilon^{i_1cdots i_n}tildeepsilon^{j_1cdots j_n}sum_{k=0}^n C_n^kA_{i_1j_1}cdots A_{i_kj_k}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon^{j_1cdots j_k}{}_{i_{k+1}cdots i_n}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon_{j_1cdots j_ki_{k+1}cdots i_n}B_{i_{k+1}}{}^{j_{k+1}}cdots B_{i_n}{}^{j_n}
$$
$$
=frac{det A}{n!}sum_{k=0}^nC_n^kk!(n-k)!B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det Asum_{k=0}^nB_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det A+det Asum_{k=1}^{n-1}B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}+det B.
$$
This reproduces the result for $n=2$.
An interesting result for physicists is when $n=3$,
begin{split}
det(A+B)=&det A+det Acdottext{Tr}(A^{-1}B)+frac{det A}{2}{[text{Tr}(A^{-1}B)]^2-text{Tr}(BA^{-1}BA^{-1})}\
&+frac{1}{6}{[text{Tr}(BA^{-1})]^3-3text{Tr}(BA^{-1})text{Tr}(BA^{-1}BA^{-1})+2text{Tr}(BA^{-1}BA^{-1}BA^{-1})}\
&+det B.
end{split}
answered Sep 22 '16 at 13:32
Drake MarquisDrake Marquis
640513
$endgroup$
When $n=2$, and suppose $A$ has inverse, you can easily show that
$det(A+B)=det A+det B+det A,cdot mathrm{Tr}(A^{-1}B)$.
Let me give a general method to find the determinant of the sum of two matrices $A,B$ with $A$ invertible and symmetric (The following result might also apply to the non-symmetric case. I might verify that later...).
I am a physicist, so I will use the index notation, $A_{ij}$ and $B_{ij}$, with $i,j=1,2,cdots,n$.
Let $A^{ij}$ donate the inverse of $A_{ij}$ such that $A^{il}A_{lj}=delta^i_j=A_{jl}A^{li}$.
We can use $A_{ij}$ to lower the indices, and its inverse to raise.
For example $A^{il}B_{lj}=B^i{}_j$.
Here and in the following, the Einstein summation rule is assumed.
Let $epsilon^{i_1cdots i_n}$ be the totally antisymmetric tensor, with $epsilon^{1cdots n}=1$.
Define a new tensor $tildeepsilon^{i_1cdots i_n}=epsilon^{i_1cdots i_n}/sqrt{|det A|}$.
We can use $A_{ij}$ to lower the indices of $tildeepsilon^{i_1cdots i_n}$, and define
$tildeepsilon_{i_1cdots i_n}=A_{i_1j_1}cdots A_{i_nj_n}tildeepsilon^{j_1cdots j_n}$.
Then there is a useful property:
$$
tildeepsilon_{i_1cdots i_kl_{k+1}cdots l_n}tildeepsilon^{j_1cdots j_kl_{k+1}cdots l_n}=(-1)^sl!(n-l)!delta^{[j_1}_{i_1}cdotsdelta^{j_k]}_{i_k},
$$
where the square brackets $[]$ imply the antisymmetrization of the indices enclosed by them.
$s$ is the number of negative elements of $A_{ij}$ after it has been diagonalized.
So now the determinant of $A+B$ can be obtained in the following way
$$
det(A+B)=frac{1}{n!}epsilon^{i_1cdots i_n}epsilon^{j_1cdots j_n}(A+B)_{i_1j_1}cdots(A+B)_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}tildeepsilon^{i_1cdots i_n}tildeepsilon^{j_1cdots j_n}sum_{k=0}^n C_n^kA_{i_1j_1}cdots A_{i_kj_k}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon^{j_1cdots j_k}{}_{i_{k+1}cdots i_n}B_{i_{k+1}j_{k+1}}cdots B_{i_nj_n}
$$
$$
=frac{(-1)^sdet A}{n!}sum_{k=0}^nC_n^ktildeepsilon^{i_1cdots i_ki_{k+1}cdots i_n}tildeepsilon_{j_1cdots j_ki_{k+1}cdots i_n}B_{i_{k+1}}{}^{j_{k+1}}cdots B_{i_n}{}^{j_n}
$$
$$
=frac{det A}{n!}sum_{k=0}^nC_n^kk!(n-k)!B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det Asum_{k=0}^nB_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}
$$
$$
=det A+det Asum_{k=1}^{n-1}B_{i_{k+1}}{}^{[i_{k+1}}cdots B_{i_n}{}^{i_n]}+det B.
$$
This reproduces the result for $n=2$.
An interesting result for physicists is when $n=3$,
begin{split}
det(A+B)=&det A+det Acdottext{Tr}(A^{-1}B)+frac{det A}{2}{[text{Tr}(A^{-1}B)]^2-text{Tr}(BA^{-1}BA^{-1})}\
&+frac{1}{6}{[text{Tr}(BA^{-1})]^3-3text{Tr}(BA^{-1})text{Tr}(BA^{-1}BA^{-1})+2text{Tr}(BA^{-1}BA^{-1}BA^{-1})}\
&+det B.
end{split}
answered Sep 22 '16 at 13:32
Drake MarquisDrake Marquis
640513
edited yesterday
answered Sep 22 '16 at 13:32
Drake MarquisDrake Marquis
640513
answered Sep 22 '16 at 13:32
Drake MarquisDrake Marquis
640513
answered Sep 22 '16 at 13:32
Drake MarquisDrake Marquis
640513
640513
$begingroup$
This, in my opinion, is the better answer, as it not only falsifies the question, but also salvages the false statement with an interesting and relevant identity. (+1)
$endgroup$
– Frpzzd
Sep 22 '18 at 22:09
1
$begingroup$
Anything for $n=3$?
$endgroup$
– Frpzzd
Sep 22 '18 at 22:17
$begingroup$
@Frpzzd Unfortunately, I did not find any interesting result for $n=3$.
$endgroup$
– Drake Marquis
Oct 8 '18 at 1:24
$begingroup$
@Frpzzd Please check my new answer.
$endgroup$
– Drake Marquis
17 hours ago
add a comment |
$begingroup$
This, in my opinion, is the better answer, as it not only falsifies the question, but also salvages the false statement with an interesting and relevant identity. (+1)
$endgroup$
– Frpzzd
Sep 22 '18 at 22:09
1
$begingroup$
Anything for $n=3$?
$endgroup$
– Frpzzd
Sep 22 '18 at 22:17
$begingroup$
@Frpzzd Unfortunately, I did not find any interesting result for $n=3$.
$endgroup$
– Drake Marquis
Oct 8 '18 at 1:24
$begingroup$
@Frpzzd Please check my new answer.
$endgroup$
– Drake Marquis
17 hours ago
$begingroup$
This, in my opinion, is the better answer, as it not only falsifies the question, but also salvages the false statement with an interesting and relevant identity. (+1)
$endgroup$
– Frpzzd
Sep 22 '18 at 22:09
$begingroup$
This, in my opinion, is the better answer, as it not only falsifies the question, but also salvages the false statement with an interesting and relevant identity. (+1)
$endgroup$
– Frpzzd
Sep 22 '18 at 22:09
1
1
$begingroup$
Anything for $n=3$?
$endgroup$
– Frpzzd
Sep 22 '18 at 22:17
$begingroup$
Anything for $n=3$?
$endgroup$
– Frpzzd
Sep 22 '18 at 22:17
$begingroup$
@Frpzzd Unfortunately, I did not find any interesting result for $n=3$.
$endgroup$
– Drake Marquis
Oct 8 '18 at 1:24
$begingroup$
@Frpzzd Unfortunately, I did not find any interesting result for $n=3$.
$endgroup$
– Drake Marquis
Oct 8 '18 at 1:24
$begingroup$
@Frpzzd Please check my new answer.
$endgroup$
– Drake Marquis
17 hours ago
$begingroup$
@Frpzzd Please check my new answer.
$endgroup$
– Drake Marquis
17 hours ago
add a comment |
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$begingroup$
Not in general. Even if $A,B$ are $n times n$ identity matrices, $det(A+B) = 2^n$ while $det(A) = det(B) = 1$, so the connection will depend on $n$ as well...
$endgroup$
– gt6989b
Feb 12 '14 at 16:11
$begingroup$
There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:13
3
$begingroup$
Also, in light of math.stackexchange.com/questions/298454/… , I think such a formula will always depend on more than just $det A, det B$
$endgroup$
– A Blumenthal
Feb 12 '14 at 16:14
$begingroup$
@ABlumenthal I'm having a hard time comprehending your link although it seems to answer my question. Can you explain it to me?
$endgroup$
– frogeyedpeas
Feb 14 '14 at 5:54
$begingroup$
I hope I am not making any mistake but what the link says for this case is that determinant of sum, is sum of determinants of $2^n$ matrices which are constructed by choosing for each column i either ith column of A or ith column of B (all possible choices are $2^n$ if you think about it).
$endgroup$
– kon psych
Apr 12 '16 at 7:26