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Does swapping columns of a matrix cause the rows of the inverse matrix to be swapped?
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This question came up while I was performing some computation on a few matrices on an unrelated problem in computer science.
Let $A$ be an invertible matrix with columns $A_1, dots A_n$. Let $B$ be its inverse, with rows $B_1, dots, B_n$. Now construct a new matrix $hatA$ by swapping two columns $A_i$ and $A_j$. Let $hatB$ be the inverse of $hatA$.
In my specific computations, I noticed that $hatB$ was nothing more than $B$ with the rows $B_i$ and $B_j$ swapped. Is this always true? I doubt it was a coincidence because the numbers were quite random.
Note: I was working with real numbers but I'd be interested to know if the field makes any difference.
linear-algebra inverse
asked 48 mins ago
Pedro A
1,6931723
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up vote
2
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This question came up while I was performing some computation on a few matrices on an unrelated problem in computer science.
Let $A$ be an invertible matrix with columns $A_1, dots A_n$. Let $B$ be its inverse, with rows $B_1, dots, B_n$. Now construct a new matrix $hatA$ by swapping two columns $A_i$ and $A_j$. Let $hatB$ be the inverse of $hatA$.
In my specific computations, I noticed that $hatB$ was nothing more than $B$ with the rows $B_i$ and $B_j$ swapped. Is this always true? I doubt it was a coincidence because the numbers were quite random.
Note: I was working with real numbers but I'd be interested to know if the field makes any difference.
linear-algebra inverse
asked 48 mins ago
Pedro A
1,6931723
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question came up while I was performing some computation on a few matrices on an unrelated problem in computer science.
Let $A$ be an invertible matrix with columns $A_1, dots A_n$. Let $B$ be its inverse, with rows $B_1, dots, B_n$. Now construct a new matrix $hatA$ by swapping two columns $A_i$ and $A_j$. Let $hatB$ be the inverse of $hatA$.
In my specific computations, I noticed that $hatB$ was nothing more than $B$ with the rows $B_i$ and $B_j$ swapped. Is this always true? I doubt it was a coincidence because the numbers were quite random.
Note: I was working with real numbers but I'd be interested to know if the field makes any difference.
linear-algebra inverse
asked 48 mins ago
Pedro A
1,6931723
This question came up while I was performing some computation on a few matrices on an unrelated problem in computer science.
Let $A$ be an invertible matrix with columns $A_1, dots A_n$. Let $B$ be its inverse, with rows $B_1, dots, B_n$. Now construct a new matrix $hatA$ by swapping two columns $A_i$ and $A_j$. Let $hatB$ be the inverse of $hatA$.
In my specific computations, I noticed that $hatB$ was nothing more than $B$ with the rows $B_i$ and $B_j$ swapped. Is this always true? I doubt it was a coincidence because the numbers were quite random.
Note: I was working with real numbers but I'd be interested to know if the field makes any difference.
linear-algebra inverse
linear-algebra inverse
asked 48 mins ago
Pedro A
1,6931723
asked 48 mins ago
Pedro A
1,6931723
asked 48 mins ago
Pedro A
1,6931723
asked 48 mins ago
Pedro A
1,6931723
asked 48 mins ago
Pedro A
1,6931723
1,6931723
add a comment |Â
add a comment |Â
3 Answers
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Yes, this is always true. Note that swapping columns $i$ and $j$ is equivalent to multiplying on the right side by the elementary matrix $T_ij$ which is defined by swapping rows $i$ and $j$ of the identity matrix. You can check that this matrix is the inverse of itself. Also, multiplying by $T_ij$ on the left side is equivalent to swapping rows $i$ and $j$.
So if we call your matrix $A$ then $AT_ij$ is the matrix that you get that swapping columns $i$ and $j$. Then its inverse is $T_ijA^-1$ which is the matrix that you get when you swap rows $i$ and $j$ in $A^-1$.
answered 39 mins ago
Mark
3,044112
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up vote
1
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Your observation is spot on. You can swap columns of a matrix by right-multiplying by a permutation matrix $P$ that is the identity matrix with the corresponding columns swapped: $hat A = AP$. We then have $$hat A^-1 = (AP)^-1 = P^-1A^-1.$$ $P$ is its own inverse, and left-multiplying by $P$ swaps the rows of $A^-1$ that are swapped in $P$.
answered 40 mins ago
amd
27.1k21046
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up vote
1
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Another way to think of it: you are just re-numbering the basis for one of the vector spaces: The domain space for $A$, which is the range space fo $A^-1$.
answered 34 mins ago
GEdgar
59.3k265165
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes, this is always true. Note that swapping columns $i$ and $j$ is equivalent to multiplying on the right side by the elementary matrix $T_ij$ which is defined by swapping rows $i$ and $j$ of the identity matrix. You can check that this matrix is the inverse of itself. Also, multiplying by $T_ij$ on the left side is equivalent to swapping rows $i$ and $j$.
So if we call your matrix $A$ then $AT_ij$ is the matrix that you get that swapping columns $i$ and $j$. Then its inverse is $T_ijA^-1$ which is the matrix that you get when you swap rows $i$ and $j$ in $A^-1$.
answered 39 mins ago
Mark
3,044112
add a comment |Â
up vote
2
down vote
Yes, this is always true. Note that swapping columns $i$ and $j$ is equivalent to multiplying on the right side by the elementary matrix $T_ij$ which is defined by swapping rows $i$ and $j$ of the identity matrix. You can check that this matrix is the inverse of itself. Also, multiplying by $T_ij$ on the left side is equivalent to swapping rows $i$ and $j$.
So if we call your matrix $A$ then $AT_ij$ is the matrix that you get that swapping columns $i$ and $j$. Then its inverse is $T_ijA^-1$ which is the matrix that you get when you swap rows $i$ and $j$ in $A^-1$.
answered 39 mins ago
Mark
3,044112
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes, this is always true. Note that swapping columns $i$ and $j$ is equivalent to multiplying on the right side by the elementary matrix $T_ij$ which is defined by swapping rows $i$ and $j$ of the identity matrix. You can check that this matrix is the inverse of itself. Also, multiplying by $T_ij$ on the left side is equivalent to swapping rows $i$ and $j$.
So if we call your matrix $A$ then $AT_ij$ is the matrix that you get that swapping columns $i$ and $j$. Then its inverse is $T_ijA^-1$ which is the matrix that you get when you swap rows $i$ and $j$ in $A^-1$.
answered 39 mins ago
Mark
3,044112
Yes, this is always true. Note that swapping columns $i$ and $j$ is equivalent to multiplying on the right side by the elementary matrix $T_ij$ which is defined by swapping rows $i$ and $j$ of the identity matrix. You can check that this matrix is the inverse of itself. Also, multiplying by $T_ij$ on the left side is equivalent to swapping rows $i$ and $j$.
So if we call your matrix $A$ then $AT_ij$ is the matrix that you get that swapping columns $i$ and $j$. Then its inverse is $T_ijA^-1$ which is the matrix that you get when you swap rows $i$ and $j$ in $A^-1$.
answered 39 mins ago
Mark
3,044112
answered 39 mins ago
Mark
3,044112
answered 39 mins ago
Mark
3,044112
answered 39 mins ago
Mark
3,044112
3,044112
add a comment |Â
add a comment |Â
up vote
1
down vote
Your observation is spot on. You can swap columns of a matrix by right-multiplying by a permutation matrix $P$ that is the identity matrix with the corresponding columns swapped: $hat A = AP$. We then have $$hat A^-1 = (AP)^-1 = P^-1A^-1.$$ $P$ is its own inverse, and left-multiplying by $P$ swaps the rows of $A^-1$ that are swapped in $P$.
answered 40 mins ago
amd
27.1k21046
add a comment |Â
up vote
1
down vote
Your observation is spot on. You can swap columns of a matrix by right-multiplying by a permutation matrix $P$ that is the identity matrix with the corresponding columns swapped: $hat A = AP$. We then have $$hat A^-1 = (AP)^-1 = P^-1A^-1.$$ $P$ is its own inverse, and left-multiplying by $P$ swaps the rows of $A^-1$ that are swapped in $P$.
answered 40 mins ago
amd
27.1k21046
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your observation is spot on. You can swap columns of a matrix by right-multiplying by a permutation matrix $P$ that is the identity matrix with the corresponding columns swapped: $hat A = AP$. We then have $$hat A^-1 = (AP)^-1 = P^-1A^-1.$$ $P$ is its own inverse, and left-multiplying by $P$ swaps the rows of $A^-1$ that are swapped in $P$.
answered 40 mins ago
amd
27.1k21046
Your observation is spot on. You can swap columns of a matrix by right-multiplying by a permutation matrix $P$ that is the identity matrix with the corresponding columns swapped: $hat A = AP$. We then have $$hat A^-1 = (AP)^-1 = P^-1A^-1.$$ $P$ is its own inverse, and left-multiplying by $P$ swaps the rows of $A^-1$ that are swapped in $P$.
answered 40 mins ago
amd
27.1k21046
answered 40 mins ago
amd
27.1k21046
answered 40 mins ago
amd
27.1k21046
answered 40 mins ago
amd
27.1k21046
27.1k21046
add a comment |Â
add a comment |Â
up vote
1
down vote
Another way to think of it: you are just re-numbering the basis for one of the vector spaces: The domain space for $A$, which is the range space fo $A^-1$.
answered 34 mins ago
GEdgar
59.3k265165
add a comment |Â
up vote
1
down vote
Another way to think of it: you are just re-numbering the basis for one of the vector spaces: The domain space for $A$, which is the range space fo $A^-1$.
answered 34 mins ago
GEdgar
59.3k265165
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another way to think of it: you are just re-numbering the basis for one of the vector spaces: The domain space for $A$, which is the range space fo $A^-1$.
answered 34 mins ago
GEdgar
59.3k265165
Another way to think of it: you are just re-numbering the basis for one of the vector spaces: The domain space for $A$, which is the range space fo $A^-1$.
answered 34 mins ago
GEdgar
59.3k265165
answered 34 mins ago
GEdgar
59.3k265165
answered 34 mins ago
GEdgar
59.3k265165
answered 34 mins ago
GEdgar
59.3k265165
59.3k265165
add a comment |Â
add a comment |Â
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