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Stuck on a proof about rotating a matrix
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Given a matrix $A in M^n×n(F)$, let $A^rho$ denote the matrix obtained from $A$ by ‘rotating’ it $90^circ$ clockwise.
For example,
$$left[beginarraycc 1 & 2\ 3 & 4 endarray right]^rho = left[beginarraycc 3 & 1\ 4 & 2 endarray right].$$
Find (with proof) an expression for $A^rho$ in terms of $A$.
Through analyzing $2times 2, 3times 3, 4times 4$, and $5times 5$ cases, it seems that the rotation is equivalent to tranposing A and then swapping the first column with the $n^th$ column, the second column with the $(n-1)^th$ column, and so on. Performing these column operations is equivalent to left multiplcation by a permuation matrix.
So I've concluded that $A^rho$ is equivalent to $A^TP$ where $P$ is the the "mirror image" of $I_n$.
For example in the 5x5 case, $P= left[
beginarrayccccc
0 & 0 & 0 & 0 & 1\
0 & 0 &0 &1 &0 \
0 &0 &1 &0 &0\
0 &1 &0 &0 &0\
1 &0 &0 &0 &0
endarray
right]$
I'm struggling with how to prove this is the case in general, or how to express $P$ for any $n$. Any help is appreciated!
linear-algebra matrices transpose
edited 12 mins ago
Gaby Boy Analysis
32011
asked 3 hours ago
kreagle
1314
add a comment |Â
up vote
2
down vote
favorite
Given a matrix $A in M^n×n(F)$, let $A^rho$ denote the matrix obtained from $A$ by ‘rotating’ it $90^circ$ clockwise.
For example,
$$left[beginarraycc 1 & 2\ 3 & 4 endarray right]^rho = left[beginarraycc 3 & 1\ 4 & 2 endarray right].$$
Find (with proof) an expression for $A^rho$ in terms of $A$.
Through analyzing $2times 2, 3times 3, 4times 4$, and $5times 5$ cases, it seems that the rotation is equivalent to tranposing A and then swapping the first column with the $n^th$ column, the second column with the $(n-1)^th$ column, and so on. Performing these column operations is equivalent to left multiplcation by a permuation matrix.
So I've concluded that $A^rho$ is equivalent to $A^TP$ where $P$ is the the "mirror image" of $I_n$.
For example in the 5x5 case, $P= left[
beginarrayccccc
0 & 0 & 0 & 0 & 1\
0 & 0 &0 &1 &0 \
0 &0 &1 &0 &0\
0 &1 &0 &0 &0\
1 &0 &0 &0 &0
endarray
right]$
I'm struggling with how to prove this is the case in general, or how to express $P$ for any $n$. Any help is appreciated!
linear-algebra matrices transpose
edited 12 mins ago
Gaby Boy Analysis
32011
asked 3 hours ago
kreagle
1314
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given a matrix $A in M^n×n(F)$, let $A^rho$ denote the matrix obtained from $A$ by ‘rotating’ it $90^circ$ clockwise.
For example,
$$left[beginarraycc 1 & 2\ 3 & 4 endarray right]^rho = left[beginarraycc 3 & 1\ 4 & 2 endarray right].$$
Find (with proof) an expression for $A^rho$ in terms of $A$.
Through analyzing $2times 2, 3times 3, 4times 4$, and $5times 5$ cases, it seems that the rotation is equivalent to tranposing A and then swapping the first column with the $n^th$ column, the second column with the $(n-1)^th$ column, and so on. Performing these column operations is equivalent to left multiplcation by a permuation matrix.
So I've concluded that $A^rho$ is equivalent to $A^TP$ where $P$ is the the "mirror image" of $I_n$.
For example in the 5x5 case, $P= left[
beginarrayccccc
0 & 0 & 0 & 0 & 1\
0 & 0 &0 &1 &0 \
0 &0 &1 &0 &0\
0 &1 &0 &0 &0\
1 &0 &0 &0 &0
endarray
right]$
I'm struggling with how to prove this is the case in general, or how to express $P$ for any $n$. Any help is appreciated!
linear-algebra matrices transpose
edited 12 mins ago
Gaby Boy Analysis
32011
asked 3 hours ago
kreagle
1314
Given a matrix $A in M^n×n(F)$, let $A^rho$ denote the matrix obtained from $A$ by ‘rotating’ it $90^circ$ clockwise.
For example,
$$left[beginarraycc 1 & 2\ 3 & 4 endarray right]^rho = left[beginarraycc 3 & 1\ 4 & 2 endarray right].$$
Find (with proof) an expression for $A^rho$ in terms of $A$.
Through analyzing $2times 2, 3times 3, 4times 4$, and $5times 5$ cases, it seems that the rotation is equivalent to tranposing A and then swapping the first column with the $n^th$ column, the second column with the $(n-1)^th$ column, and so on. Performing these column operations is equivalent to left multiplcation by a permuation matrix.
So I've concluded that $A^rho$ is equivalent to $A^TP$ where $P$ is the the "mirror image" of $I_n$.
For example in the 5x5 case, $P= left[
beginarrayccccc
0 & 0 & 0 & 0 & 1\
0 & 0 &0 &1 &0 \
0 &0 &1 &0 &0\
0 &1 &0 &0 &0\
1 &0 &0 &0 &0
endarray
right]$
I'm struggling with how to prove this is the case in general, or how to express $P$ for any $n$. Any help is appreciated!
linear-algebra matrices transpose
linear-algebra matrices transpose
edited 12 mins ago
Gaby Boy Analysis
32011
asked 3 hours ago
kreagle
1314
edited 12 mins ago
Gaby Boy Analysis
32011
asked 3 hours ago
kreagle
1314
edited 12 mins ago
Gaby Boy Analysis
32011
edited 12 mins ago
Gaby Boy Analysis
32011
edited 12 mins ago
Gaby Boy Analysis
32011
32011
asked 3 hours ago
kreagle
1314
asked 3 hours ago
kreagle
1314
asked 3 hours ago
kreagle
1314
1314
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
A good way to prove this is to take $A_ij$ and track where it should end up after a rotation, and where it does end up after your expression.
To get you started: $A_i,j = A^rho_j , n-i $
Now show that $A_i,j = (A^TP)_j, n-i$. This will show that they are equivalent matrices, since all their entries are the same. If you found a good expression, it should work. As for defining $P$, your current definition should be good enough.
answered 3 hours ago
Anthony Ter
963
add a comment |Â
up vote
3
down vote
Anthony Ter's suggestion of verifying $(A^rm T P)_j, n-i+1$ actually equals $A_i,j$ is key (I'm assuming indexes go from $1$ to $n$), but in order to do so, it's helpful to rely on a way of defining $P$ entrywise. Consider the Kronecker delta:
$$delta_i,j :=begincases 1,& textif i=j;\ 0,& textif ineq j.endcases$$
Then you can define $I_n$ entrywise as $(I_n)_i,j :=delta_i,j$; and similarly, $(P)_i,j :=delta_i,n-j+1$. Then, you can compute $(A^rm T P)_j, n-i+1$ directly by using the matrix-product entrywise definition and the matrix transpose property: $(A^rm T)_i,j := A_j,i$.
beginalign (A^p)_j, n-i+1 &=(A^rm T P)_j, n-i+1 =sum_k=1^n (A^rm T)_j,k P_k, n-i+1\
&=sum_k=1^n A_k,j delta_k, n-(n-i+1)+1 =sum_k=1^n A_k,j delta_k,i\
&=A_1,j delta_1,i +A_2,j delta_2,i +A_3,j delta_3,i +ldots +A_n,j delta_n,i.endalign
Note that inside the summation, $k$ is variable, whereas $i$ and $j$ are fixed. For $delta_k,i$ not to be zero (i.e., one) then $k$ shall equal $i$. Thus, $(A^p)_j, n-i+1 =A_i,j delta_i,i =A_i,j$, as desired.
answered 20 mins ago
Rócherz
2,2562518
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A good way to prove this is to take $A_ij$ and track where it should end up after a rotation, and where it does end up after your expression.
To get you started: $A_i,j = A^rho_j , n-i $
Now show that $A_i,j = (A^TP)_j, n-i$. This will show that they are equivalent matrices, since all their entries are the same. If you found a good expression, it should work. As for defining $P$, your current definition should be good enough.
answered 3 hours ago
Anthony Ter
963
add a comment |Â
up vote
2
down vote
accepted
A good way to prove this is to take $A_ij$ and track where it should end up after a rotation, and where it does end up after your expression.
To get you started: $A_i,j = A^rho_j , n-i $
Now show that $A_i,j = (A^TP)_j, n-i$. This will show that they are equivalent matrices, since all their entries are the same. If you found a good expression, it should work. As for defining $P$, your current definition should be good enough.
answered 3 hours ago
Anthony Ter
963
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A good way to prove this is to take $A_ij$ and track where it should end up after a rotation, and where it does end up after your expression.
To get you started: $A_i,j = A^rho_j , n-i $
Now show that $A_i,j = (A^TP)_j, n-i$. This will show that they are equivalent matrices, since all their entries are the same. If you found a good expression, it should work. As for defining $P$, your current definition should be good enough.
answered 3 hours ago
Anthony Ter
963
A good way to prove this is to take $A_ij$ and track where it should end up after a rotation, and where it does end up after your expression.
To get you started: $A_i,j = A^rho_j , n-i $
Now show that $A_i,j = (A^TP)_j, n-i$. This will show that they are equivalent matrices, since all their entries are the same. If you found a good expression, it should work. As for defining $P$, your current definition should be good enough.
answered 3 hours ago
Anthony Ter
963
answered 3 hours ago
Anthony Ter
963
answered 3 hours ago
Anthony Ter
963
answered 3 hours ago
Anthony Ter
963
963
add a comment |Â
add a comment |Â
up vote
3
down vote
Anthony Ter's suggestion of verifying $(A^rm T P)_j, n-i+1$ actually equals $A_i,j$ is key (I'm assuming indexes go from $1$ to $n$), but in order to do so, it's helpful to rely on a way of defining $P$ entrywise. Consider the Kronecker delta:
$$delta_i,j :=begincases 1,& textif i=j;\ 0,& textif ineq j.endcases$$
Then you can define $I_n$ entrywise as $(I_n)_i,j :=delta_i,j$; and similarly, $(P)_i,j :=delta_i,n-j+1$. Then, you can compute $(A^rm T P)_j, n-i+1$ directly by using the matrix-product entrywise definition and the matrix transpose property: $(A^rm T)_i,j := A_j,i$.
beginalign (A^p)_j, n-i+1 &=(A^rm T P)_j, n-i+1 =sum_k=1^n (A^rm T)_j,k P_k, n-i+1\
&=sum_k=1^n A_k,j delta_k, n-(n-i+1)+1 =sum_k=1^n A_k,j delta_k,i\
&=A_1,j delta_1,i +A_2,j delta_2,i +A_3,j delta_3,i +ldots +A_n,j delta_n,i.endalign
Note that inside the summation, $k$ is variable, whereas $i$ and $j$ are fixed. For $delta_k,i$ not to be zero (i.e., one) then $k$ shall equal $i$. Thus, $(A^p)_j, n-i+1 =A_i,j delta_i,i =A_i,j$, as desired.
answered 20 mins ago
Rócherz
2,2562518
add a comment |Â
up vote
3
down vote
Anthony Ter's suggestion of verifying $(A^rm T P)_j, n-i+1$ actually equals $A_i,j$ is key (I'm assuming indexes go from $1$ to $n$), but in order to do so, it's helpful to rely on a way of defining $P$ entrywise. Consider the Kronecker delta:
$$delta_i,j :=begincases 1,& textif i=j;\ 0,& textif ineq j.endcases$$
Then you can define $I_n$ entrywise as $(I_n)_i,j :=delta_i,j$; and similarly, $(P)_i,j :=delta_i,n-j+1$. Then, you can compute $(A^rm T P)_j, n-i+1$ directly by using the matrix-product entrywise definition and the matrix transpose property: $(A^rm T)_i,j := A_j,i$.
beginalign (A^p)_j, n-i+1 &=(A^rm T P)_j, n-i+1 =sum_k=1^n (A^rm T)_j,k P_k, n-i+1\
&=sum_k=1^n A_k,j delta_k, n-(n-i+1)+1 =sum_k=1^n A_k,j delta_k,i\
&=A_1,j delta_1,i +A_2,j delta_2,i +A_3,j delta_3,i +ldots +A_n,j delta_n,i.endalign
Note that inside the summation, $k$ is variable, whereas $i$ and $j$ are fixed. For $delta_k,i$ not to be zero (i.e., one) then $k$ shall equal $i$. Thus, $(A^p)_j, n-i+1 =A_i,j delta_i,i =A_i,j$, as desired.
answered 20 mins ago
Rócherz
2,2562518
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Anthony Ter's suggestion of verifying $(A^rm T P)_j, n-i+1$ actually equals $A_i,j$ is key (I'm assuming indexes go from $1$ to $n$), but in order to do so, it's helpful to rely on a way of defining $P$ entrywise. Consider the Kronecker delta:
$$delta_i,j :=begincases 1,& textif i=j;\ 0,& textif ineq j.endcases$$
Then you can define $I_n$ entrywise as $(I_n)_i,j :=delta_i,j$; and similarly, $(P)_i,j :=delta_i,n-j+1$. Then, you can compute $(A^rm T P)_j, n-i+1$ directly by using the matrix-product entrywise definition and the matrix transpose property: $(A^rm T)_i,j := A_j,i$.
beginalign (A^p)_j, n-i+1 &=(A^rm T P)_j, n-i+1 =sum_k=1^n (A^rm T)_j,k P_k, n-i+1\
&=sum_k=1^n A_k,j delta_k, n-(n-i+1)+1 =sum_k=1^n A_k,j delta_k,i\
&=A_1,j delta_1,i +A_2,j delta_2,i +A_3,j delta_3,i +ldots +A_n,j delta_n,i.endalign
Note that inside the summation, $k$ is variable, whereas $i$ and $j$ are fixed. For $delta_k,i$ not to be zero (i.e., one) then $k$ shall equal $i$. Thus, $(A^p)_j, n-i+1 =A_i,j delta_i,i =A_i,j$, as desired.
answered 20 mins ago
Rócherz
2,2562518
Anthony Ter's suggestion of verifying $(A^rm T P)_j, n-i+1$ actually equals $A_i,j$ is key (I'm assuming indexes go from $1$ to $n$), but in order to do so, it's helpful to rely on a way of defining $P$ entrywise. Consider the Kronecker delta:
$$delta_i,j :=begincases 1,& textif i=j;\ 0,& textif ineq j.endcases$$
Then you can define $I_n$ entrywise as $(I_n)_i,j :=delta_i,j$; and similarly, $(P)_i,j :=delta_i,n-j+1$. Then, you can compute $(A^rm T P)_j, n-i+1$ directly by using the matrix-product entrywise definition and the matrix transpose property: $(A^rm T)_i,j := A_j,i$.
beginalign (A^p)_j, n-i+1 &=(A^rm T P)_j, n-i+1 =sum_k=1^n (A^rm T)_j,k P_k, n-i+1\
&=sum_k=1^n A_k,j delta_k, n-(n-i+1)+1 =sum_k=1^n A_k,j delta_k,i\
&=A_1,j delta_1,i +A_2,j delta_2,i +A_3,j delta_3,i +ldots +A_n,j delta_n,i.endalign
Note that inside the summation, $k$ is variable, whereas $i$ and $j$ are fixed. For $delta_k,i$ not to be zero (i.e., one) then $k$ shall equal $i$. Thus, $(A^p)_j, n-i+1 =A_i,j delta_i,i =A_i,j$, as desired.
answered 20 mins ago
Rócherz
2,2562518
answered 20 mins ago
Rócherz
2,2562518
answered 20 mins ago
Rócherz
2,2562518
answered 20 mins ago
Rócherz
2,2562518
2,2562518
add a comment |Â
add a comment |Â
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