Mixing
Matrices with same row reduced form - Show there's a sequence of row operations
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"Let us assume that Q and W are z x y matrices such that they have the same row reduced form. Prove that there exists a sequence of row operations that takes us from Q to W."
How would one prove this statement? Thank you.
linear-algebra
asked Jan 18 at 5:24
EtezeaEtezea
82
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add a comment |
$begingroup$
"Let us assume that Q and W are z x y matrices such that they have the same row reduced form. Prove that there exists a sequence of row operations that takes us from Q to W."
How would one prove this statement? Thank you.
linear-algebra
asked Jan 18 at 5:24
EtezeaEtezea
82
$endgroup$
$begingroup$
I'm no expert, but a formal version of "we know row operations can take us from Q to rref(Q)=rref(W), and we know row operations can take us from W to rref(W)=rref(Q), so we just do the Q->rref(Q) and then just the steps from W->rref(W) backwards, which works because row operations are linear"?
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– D.R.
Jan 18 at 5:27
$begingroup$
Oh yes that is true? Would that suffice as a proof though?
$endgroup$
– Etezea
Jan 18 at 5:30
add a comment |
$begingroup$
"Let us assume that Q and W are z x y matrices such that they have the same row reduced form. Prove that there exists a sequence of row operations that takes us from Q to W."
How would one prove this statement? Thank you.
linear-algebra
asked Jan 18 at 5:24
EtezeaEtezea
82
$endgroup$
"Let us assume that Q and W are z x y matrices such that they have the same row reduced form. Prove that there exists a sequence of row operations that takes us from Q to W."
How would one prove this statement? Thank you.
linear-algebra
linear-algebra
asked Jan 18 at 5:24
EtezeaEtezea
82
asked Jan 18 at 5:24
EtezeaEtezea
82
asked Jan 18 at 5:24
EtezeaEtezea
82
asked Jan 18 at 5:24
EtezeaEtezea
82
asked Jan 18 at 5:24
EtezeaEtezea
82
82
$begingroup$
I'm no expert, but a formal version of "we know row operations can take us from Q to rref(Q)=rref(W), and we know row operations can take us from W to rref(W)=rref(Q), so we just do the Q->rref(Q) and then just the steps from W->rref(W) backwards, which works because row operations are linear"?
$endgroup$
– D.R.
Jan 18 at 5:27
$begingroup$
Oh yes that is true? Would that suffice as a proof though?
$endgroup$
– Etezea
Jan 18 at 5:30
add a comment |
$begingroup$
I'm no expert, but a formal version of "we know row operations can take us from Q to rref(Q)=rref(W), and we know row operations can take us from W to rref(W)=rref(Q), so we just do the Q->rref(Q) and then just the steps from W->rref(W) backwards, which works because row operations are linear"?
$endgroup$
– D.R.
Jan 18 at 5:27
$begingroup$
Oh yes that is true? Would that suffice as a proof though?
$endgroup$
– Etezea
Jan 18 at 5:30
$begingroup$
I'm no expert, but a formal version of "we know row operations can take us from Q to rref(Q)=rref(W), and we know row operations can take us from W to rref(W)=rref(Q), so we just do the Q->rref(Q) and then just the steps from W->rref(W) backwards, which works because row operations are linear"?
$endgroup$
– D.R.
Jan 18 at 5:27
$begingroup$
I'm no expert, but a formal version of "we know row operations can take us from Q to rref(Q)=rref(W), and we know row operations can take us from W to rref(W)=rref(Q), so we just do the Q->rref(Q) and then just the steps from W->rref(W) backwards, which works because row operations are linear"?
$endgroup$
– D.R.
Jan 18 at 5:27
$begingroup$
Oh yes that is true? Would that suffice as a proof though?
$endgroup$
– Etezea
Jan 18 at 5:30
$begingroup$
Oh yes that is true? Would that suffice as a proof though?
$endgroup$
– Etezea
Jan 18 at 5:30
add a comment |
1 Answer
1
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$begingroup$
The row operations which reduce a matrix to its row reduced form are all reversible, and achievable through premutiplication by an invertible matrix. So if $ R $ is the common row reduced form of $ Q $ and $ W $, we must have:
begin{eqnarray}
R &=& E_1,E_2dots E_r, Q mbox{and}\
R &=& F_1,F_2dots F_s, W ,
end{eqnarray}
where $ E_1, E_2,dots E_r, F_1,F_2,dots F_s $ are matrices by whose premultiplications each of the necessary row operations are effected. It follows that
begin{eqnarray}
W &=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},R \
&=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},E_1,E_2dots E_r, Q .
end{eqnarray}
That is, the sequence of row operations effected by the matrices $ E_r, E_{r-1}, dots E_1, F_1^{-1}, F_2^{-1},dots F_s^{-1} $ will transform $ Q $ into $ W $.
answered Jan 18 at 12:58
lonza leggieralonza leggiera
89117
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The row operations which reduce a matrix to its row reduced form are all reversible, and achievable through premutiplication by an invertible matrix. So if $ R $ is the common row reduced form of $ Q $ and $ W $, we must have:
begin{eqnarray}
R &=& E_1,E_2dots E_r, Q mbox{and}\
R &=& F_1,F_2dots F_s, W ,
end{eqnarray}
where $ E_1, E_2,dots E_r, F_1,F_2,dots F_s $ are matrices by whose premultiplications each of the necessary row operations are effected. It follows that
begin{eqnarray}
W &=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},R \
&=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},E_1,E_2dots E_r, Q .
end{eqnarray}
That is, the sequence of row operations effected by the matrices $ E_r, E_{r-1}, dots E_1, F_1^{-1}, F_2^{-1},dots F_s^{-1} $ will transform $ Q $ into $ W $.
answered Jan 18 at 12:58
lonza leggieralonza leggiera
89117
$endgroup$
add a comment |
$begingroup$
The row operations which reduce a matrix to its row reduced form are all reversible, and achievable through premutiplication by an invertible matrix. So if $ R $ is the common row reduced form of $ Q $ and $ W $, we must have:
begin{eqnarray}
R &=& E_1,E_2dots E_r, Q mbox{and}\
R &=& F_1,F_2dots F_s, W ,
end{eqnarray}
where $ E_1, E_2,dots E_r, F_1,F_2,dots F_s $ are matrices by whose premultiplications each of the necessary row operations are effected. It follows that
begin{eqnarray}
W &=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},R \
&=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},E_1,E_2dots E_r, Q .
end{eqnarray}
That is, the sequence of row operations effected by the matrices $ E_r, E_{r-1}, dots E_1, F_1^{-1}, F_2^{-1},dots F_s^{-1} $ will transform $ Q $ into $ W $.
answered Jan 18 at 12:58
lonza leggieralonza leggiera
89117
$endgroup$
add a comment |
$begingroup$
The row operations which reduce a matrix to its row reduced form are all reversible, and achievable through premutiplication by an invertible matrix. So if $ R $ is the common row reduced form of $ Q $ and $ W $, we must have:
begin{eqnarray}
R &=& E_1,E_2dots E_r, Q mbox{and}\
R &=& F_1,F_2dots F_s, W ,
end{eqnarray}
where $ E_1, E_2,dots E_r, F_1,F_2,dots F_s $ are matrices by whose premultiplications each of the necessary row operations are effected. It follows that
begin{eqnarray}
W &=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},R \
&=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},E_1,E_2dots E_r, Q .
end{eqnarray}
That is, the sequence of row operations effected by the matrices $ E_r, E_{r-1}, dots E_1, F_1^{-1}, F_2^{-1},dots F_s^{-1} $ will transform $ Q $ into $ W $.
answered Jan 18 at 12:58
lonza leggieralonza leggiera
89117
$endgroup$
The row operations which reduce a matrix to its row reduced form are all reversible, and achievable through premutiplication by an invertible matrix. So if $ R $ is the common row reduced form of $ Q $ and $ W $, we must have:
begin{eqnarray}
R &=& E_1,E_2dots E_r, Q mbox{and}\
R &=& F_1,F_2dots F_s, W ,
end{eqnarray}
where $ E_1, E_2,dots E_r, F_1,F_2,dots F_s $ are matrices by whose premultiplications each of the necessary row operations are effected. It follows that
begin{eqnarray}
W &=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},R \
&=& F_s^{-1},F_{s-1}^{-1}dots F_1^{-1},E_1,E_2dots E_r, Q .
end{eqnarray}
That is, the sequence of row operations effected by the matrices $ E_r, E_{r-1}, dots E_1, F_1^{-1}, F_2^{-1},dots F_s^{-1} $ will transform $ Q $ into $ W $.
answered Jan 18 at 12:58
lonza leggieralonza leggiera
89117
answered Jan 18 at 12:58
lonza leggieralonza leggiera
89117
answered Jan 18 at 12:58
lonza leggieralonza leggiera
89117
answered Jan 18 at 12:58
lonza leggieralonza leggiera
89117
89117
add a comment |
add a comment |
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$begingroup$
I'm no expert, but a formal version of "we know row operations can take us from Q to rref(Q)=rref(W), and we know row operations can take us from W to rref(W)=rref(Q), so we just do the Q->rref(Q) and then just the steps from W->rref(W) backwards, which works because row operations are linear"?
$endgroup$
– D.R.
Jan 18 at 5:27
$begingroup$
Oh yes that is true? Would that suffice as a proof though?
$endgroup$
– Etezea
Jan 18 at 5:30