Mixing
Reduced row echelon of Z/2Z
$begingroup$
I've gotten (1 0 1 0 1 0; 0 1 1 0 0 1; 0 0 0 0 0 0), which looks like it is of row reduced form, but would I not also be able to simplify this further to (1 0 0 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0), this is confusing me since I thought rref was unique?
linear-algebra
asked Jan 27 at 20:12
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
add a comment |
$begingroup$
I've gotten (1 0 1 0 1 0; 0 1 1 0 0 1; 0 0 0 0 0 0), which looks like it is of row reduced form, but would I not also be able to simplify this further to (1 0 0 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0), this is confusing me since I thought rref was unique?
linear-algebra
asked Jan 27 at 20:12
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
$begingroup$
The matrix after "I've gotten" is the same as that after "would I not also be able to simplify this further to". Did you mistype something?
$endgroup$
– coffeemath
Jan 27 at 20:17
$begingroup$
Yes, I've corrected it now.
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 20:21
$begingroup$
Could you show in question the steps how you got to the second matrix?
$endgroup$
– coffeemath
Jan 27 at 20:24
add a comment |
$begingroup$
I've gotten (1 0 1 0 1 0; 0 1 1 0 0 1; 0 0 0 0 0 0), which looks like it is of row reduced form, but would I not also be able to simplify this further to (1 0 0 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0), this is confusing me since I thought rref was unique?
linear-algebra
asked Jan 27 at 20:12
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
I've gotten (1 0 1 0 1 0; 0 1 1 0 0 1; 0 0 0 0 0 0), which looks like it is of row reduced form, but would I not also be able to simplify this further to (1 0 0 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0), this is confusing me since I thought rref was unique?
linear-algebra
linear-algebra
asked Jan 27 at 20:12
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Jan 27 at 20:12
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
edited Jan 27 at 20:21
4M4D3U5 M0Z4RT
asked Jan 27 at 20:12
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Jan 27 at 20:12
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Jan 27 at 20:12
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
386
$begingroup$
The matrix after "I've gotten" is the same as that after "would I not also be able to simplify this further to". Did you mistype something?
$endgroup$
– coffeemath
Jan 27 at 20:17
$begingroup$
Yes, I've corrected it now.
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 20:21
$begingroup$
Could you show in question the steps how you got to the second matrix?
$endgroup$
– coffeemath
Jan 27 at 20:24
add a comment |
$begingroup$
The matrix after "I've gotten" is the same as that after "would I not also be able to simplify this further to". Did you mistype something?
$endgroup$
– coffeemath
Jan 27 at 20:17
$begingroup$
Yes, I've corrected it now.
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 20:21
$begingroup$
Could you show in question the steps how you got to the second matrix?
$endgroup$
– coffeemath
Jan 27 at 20:24
$begingroup$
The matrix after "I've gotten" is the same as that after "would I not also be able to simplify this further to". Did you mistype something?
$endgroup$
– coffeemath
Jan 27 at 20:17
$begingroup$
The matrix after "I've gotten" is the same as that after "would I not also be able to simplify this further to". Did you mistype something?
$endgroup$
– coffeemath
Jan 27 at 20:17
$begingroup$
Yes, I've corrected it now.
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 20:21
$begingroup$
Yes, I've corrected it now.
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 20:21
$begingroup$
Could you show in question the steps how you got to the second matrix?
$endgroup$
– coffeemath
Jan 27 at 20:24
$begingroup$
Could you show in question the steps how you got to the second matrix?
$endgroup$
– coffeemath
Jan 27 at 20:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
begin{align}
begin{bmatrix}
0 & 1 & 1 & 0 & 0 & 1 \
1 & 0 & 1 & 0 & 1 & 0 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&& R_1 leftrightarrow R_2
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 1 & 1 & 0 & 0 & 1
end{bmatrix}
&& R_3gets R_3+R_1
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
&& R_3gets R_3+R_2
end{align}
The RREF is unique. The matrix
begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
is not row-equivalent, because the third column is not the sum of the first and second columns like in the matrix we started with. Row-equivalence doesn't change linear relations among columns.
answered Jan 27 at 20:31
egregegreg
185k1486206
$endgroup$
$begingroup$
Could I not just add rows 1 and 2 together and get a different matrix?
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:19
$begingroup$
@4M4D3U5M0Z4RT But it wouldn't be the RREF any longer.
$endgroup$
– egreg
Jan 27 at 21:29
$begingroup$
is that not the same operation you used to, for example do R3 <-- R3+R1, what I meant gives, [1 0 1 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0]
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:32
$begingroup$
@4M4D3U5M0Z4RT If you perform a nontrivial row operation on the RREF, you get a matrix which is not RREF. Anyway, you don't get the matrix you wish to get. It's impossible, because the matrix you'd like to reduce to is not row-equivalent to the given matrix.
$endgroup$
– egreg
Jan 27 at 21:35
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
votes
$begingroup$
begin{align}
begin{bmatrix}
0 & 1 & 1 & 0 & 0 & 1 \
1 & 0 & 1 & 0 & 1 & 0 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&& R_1 leftrightarrow R_2
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 1 & 1 & 0 & 0 & 1
end{bmatrix}
&& R_3gets R_3+R_1
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
&& R_3gets R_3+R_2
end{align}
The RREF is unique. The matrix
begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
is not row-equivalent, because the third column is not the sum of the first and second columns like in the matrix we started with. Row-equivalence doesn't change linear relations among columns.
answered Jan 27 at 20:31
egregegreg
185k1486206
$endgroup$
$begingroup$
Could I not just add rows 1 and 2 together and get a different matrix?
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:19
$begingroup$
@4M4D3U5M0Z4RT But it wouldn't be the RREF any longer.
$endgroup$
– egreg
Jan 27 at 21:29
$begingroup$
is that not the same operation you used to, for example do R3 <-- R3+R1, what I meant gives, [1 0 1 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0]
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:32
$begingroup$
@4M4D3U5M0Z4RT If you perform a nontrivial row operation on the RREF, you get a matrix which is not RREF. Anyway, you don't get the matrix you wish to get. It's impossible, because the matrix you'd like to reduce to is not row-equivalent to the given matrix.
$endgroup$
– egreg
Jan 27 at 21:35
add a comment |
$begingroup$
begin{align}
begin{bmatrix}
0 & 1 & 1 & 0 & 0 & 1 \
1 & 0 & 1 & 0 & 1 & 0 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&& R_1 leftrightarrow R_2
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 1 & 1 & 0 & 0 & 1
end{bmatrix}
&& R_3gets R_3+R_1
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
&& R_3gets R_3+R_2
end{align}
The RREF is unique. The matrix
begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
is not row-equivalent, because the third column is not the sum of the first and second columns like in the matrix we started with. Row-equivalence doesn't change linear relations among columns.
answered Jan 27 at 20:31
egregegreg
185k1486206
$endgroup$
$begingroup$
Could I not just add rows 1 and 2 together and get a different matrix?
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:19
$begingroup$
@4M4D3U5M0Z4RT But it wouldn't be the RREF any longer.
$endgroup$
– egreg
Jan 27 at 21:29
$begingroup$
is that not the same operation you used to, for example do R3 <-- R3+R1, what I meant gives, [1 0 1 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0]
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:32
$begingroup$
@4M4D3U5M0Z4RT If you perform a nontrivial row operation on the RREF, you get a matrix which is not RREF. Anyway, you don't get the matrix you wish to get. It's impossible, because the matrix you'd like to reduce to is not row-equivalent to the given matrix.
$endgroup$
– egreg
Jan 27 at 21:35
add a comment |
$begingroup$
begin{align}
begin{bmatrix}
0 & 1 & 1 & 0 & 0 & 1 \
1 & 0 & 1 & 0 & 1 & 0 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&& R_1 leftrightarrow R_2
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 1 & 1 & 0 & 0 & 1
end{bmatrix}
&& R_3gets R_3+R_1
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
&& R_3gets R_3+R_2
end{align}
The RREF is unique. The matrix
begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
is not row-equivalent, because the third column is not the sum of the first and second columns like in the matrix we started with. Row-equivalence doesn't change linear relations among columns.
answered Jan 27 at 20:31
egregegreg
185k1486206
$endgroup$
begin{align}
begin{bmatrix}
0 & 1 & 1 & 0 & 0 & 1 \
1 & 0 & 1 & 0 & 1 & 0 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
1 & 1 & 0 & 0 & 1 & 1
end{bmatrix}
&& R_1 leftrightarrow R_2
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 1 & 1 & 0 & 0 & 1
end{bmatrix}
&& R_3gets R_3+R_1
\[6px]
&to
begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 \
0 & 1 & 1 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
&& R_3gets R_3+R_2
end{align}
The RREF is unique. The matrix
begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 & 0
end{bmatrix}
is not row-equivalent, because the third column is not the sum of the first and second columns like in the matrix we started with. Row-equivalence doesn't change linear relations among columns.
answered Jan 27 at 20:31
egregegreg
185k1486206
answered Jan 27 at 20:31
egregegreg
185k1486206
answered Jan 27 at 20:31
egregegreg
185k1486206
answered Jan 27 at 20:31
egregegreg
185k1486206
185k1486206
$begingroup$
Could I not just add rows 1 and 2 together and get a different matrix?
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:19
$begingroup$
@4M4D3U5M0Z4RT But it wouldn't be the RREF any longer.
$endgroup$
– egreg
Jan 27 at 21:29
$begingroup$
is that not the same operation you used to, for example do R3 <-- R3+R1, what I meant gives, [1 0 1 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0]
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:32
$begingroup$
@4M4D3U5M0Z4RT If you perform a nontrivial row operation on the RREF, you get a matrix which is not RREF. Anyway, you don't get the matrix you wish to get. It's impossible, because the matrix you'd like to reduce to is not row-equivalent to the given matrix.
$endgroup$
– egreg
Jan 27 at 21:35
add a comment |
$begingroup$
Could I not just add rows 1 and 2 together and get a different matrix?
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:19
$begingroup$
@4M4D3U5M0Z4RT But it wouldn't be the RREF any longer.
$endgroup$
– egreg
Jan 27 at 21:29
$begingroup$
is that not the same operation you used to, for example do R3 <-- R3+R1, what I meant gives, [1 0 1 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0]
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:32
$begingroup$
@4M4D3U5M0Z4RT If you perform a nontrivial row operation on the RREF, you get a matrix which is not RREF. Anyway, you don't get the matrix you wish to get. It's impossible, because the matrix you'd like to reduce to is not row-equivalent to the given matrix.
$endgroup$
– egreg
Jan 27 at 21:35
$begingroup$
Could I not just add rows 1 and 2 together and get a different matrix?
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:19
$begingroup$
Could I not just add rows 1 and 2 together and get a different matrix?
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:19
$begingroup$
@4M4D3U5M0Z4RT But it wouldn't be the RREF any longer.
$endgroup$
– egreg
Jan 27 at 21:29
$begingroup$
@4M4D3U5M0Z4RT But it wouldn't be the RREF any longer.
$endgroup$
– egreg
Jan 27 at 21:29
$begingroup$
is that not the same operation you used to, for example do R3 <-- R3+R1, what I meant gives, [1 0 1 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0]
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:32
$begingroup$
is that not the same operation you used to, for example do R3 <-- R3+R1, what I meant gives, [1 0 1 0 1 0; 0 1 0 0 0 1; 0 0 0 0 0 0]
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 21:32
$begingroup$
@4M4D3U5M0Z4RT If you perform a nontrivial row operation on the RREF, you get a matrix which is not RREF. Anyway, you don't get the matrix you wish to get. It's impossible, because the matrix you'd like to reduce to is not row-equivalent to the given matrix.
$endgroup$
– egreg
Jan 27 at 21:35
$begingroup$
@4M4D3U5M0Z4RT If you perform a nontrivial row operation on the RREF, you get a matrix which is not RREF. Anyway, you don't get the matrix you wish to get. It's impossible, because the matrix you'd like to reduce to is not row-equivalent to the given matrix.
$endgroup$
– egreg
Jan 27 at 21:35
add a comment |
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$begingroup$
The matrix after "I've gotten" is the same as that after "would I not also be able to simplify this further to". Did you mistype something?
$endgroup$
– coffeemath
Jan 27 at 20:17
$begingroup$
Yes, I've corrected it now.
$endgroup$
– 4M4D3U5 M0Z4RT
Jan 27 at 20:21
$begingroup$
Could you show in question the steps how you got to the second matrix?
$endgroup$
– coffeemath
Jan 27 at 20:24