Every nonsingular $mtimes m$ matrix is row equivalent to identity matrix $I_m$ The Next CEO of...
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Every nonsingular $mtimes m$ matrix is row equivalent to identity matrix $I_m$
The Next CEO of Stack OverflowConfusion over the usage of different terms for ref and rref.Why is this matrix not in reduced row echelon form?Issue understanding the difference between reduced row echelon form on a coefficient matrix and on an augmented matrixDifferent representations of a matrix in reduced row echelon formRegarding ranks of row echelon form matricesMatrix rank and number of linearly independent rowsAnybody knows a proof of Uniqueness of the Reduced Echelon Form Theorem?Is a $1 times n$ matrix already in echelon form?Im confused about obtaining the rank of a matrixBasis of Column Space of matrix that is in Row Reduced Echelon form
$begingroup$
Could anyone give an easy to understand proof of this fact or at least tell if the proof below is correct?
We know that every matrix is row equivalent to a reduced row echelon form. An $mtimes m$ square matrix $A$ of rank $m$ is row equivalent to an identity matrix from the definition of reduced row echelon form:
1) all non-zero rows are above rows of all zeros
2) the leading coefficient is strictly to the right of the leading coefficient of the row above
3) the leading coefficient is $1$ and is the only non-zero number in its column
Plus, there's a theorem saying that the rank of any matrix equals the number of non-zero rows in its reduced row echelon form.
Now we know that $A$ has $m$ non-zero rows, then every row contains a leading coefficient $1$ that is shifted to the right with respect to the leading coefficient in the row above. The only way we can arrange $m$ 'ones' in this manner is by placing them on diagonal (we can prove it by noting the the required number of 'shifts to the right' of the leading coefficients when going one row down is $m$ - actually - is it clear enough for a proof?). And we know there is nothing except $1$ in any of those columns - we have a diagonal matrix.
Can you make it shorter without referring to some more complicated terminology?
linear-algebra
asked Apr 23 '15 at 15:24
user4205580user4205580
4611131
$endgroup$
add a comment |
$begingroup$
Could anyone give an easy to understand proof of this fact or at least tell if the proof below is correct?
We know that every matrix is row equivalent to a reduced row echelon form. An $mtimes m$ square matrix $A$ of rank $m$ is row equivalent to an identity matrix from the definition of reduced row echelon form:
1) all non-zero rows are above rows of all zeros
2) the leading coefficient is strictly to the right of the leading coefficient of the row above
3) the leading coefficient is $1$ and is the only non-zero number in its column
Plus, there's a theorem saying that the rank of any matrix equals the number of non-zero rows in its reduced row echelon form.
Now we know that $A$ has $m$ non-zero rows, then every row contains a leading coefficient $1$ that is shifted to the right with respect to the leading coefficient in the row above. The only way we can arrange $m$ 'ones' in this manner is by placing them on diagonal (we can prove it by noting the the required number of 'shifts to the right' of the leading coefficients when going one row down is $m$ - actually - is it clear enough for a proof?). And we know there is nothing except $1$ in any of those columns - we have a diagonal matrix.
Can you make it shorter without referring to some more complicated terminology?
linear-algebra
asked Apr 23 '15 at 15:24
user4205580user4205580
4611131
$endgroup$
$begingroup$
I think it should be "Every invertible (or regular, or non-singular, etc.) matrix is ...." etc.
$endgroup$
– Timbuc
Apr 23 '15 at 15:27
$begingroup$
this wont work with the zero matrix.
$endgroup$
– abel
Apr 23 '15 at 15:34
$begingroup$
I've corrected the title. Do you have any objections?
$endgroup$
– user4205580
Apr 23 '15 at 15:47
add a comment |
$begingroup$
Could anyone give an easy to understand proof of this fact or at least tell if the proof below is correct?
We know that every matrix is row equivalent to a reduced row echelon form. An $mtimes m$ square matrix $A$ of rank $m$ is row equivalent to an identity matrix from the definition of reduced row echelon form:
1) all non-zero rows are above rows of all zeros
2) the leading coefficient is strictly to the right of the leading coefficient of the row above
3) the leading coefficient is $1$ and is the only non-zero number in its column
Plus, there's a theorem saying that the rank of any matrix equals the number of non-zero rows in its reduced row echelon form.
Now we know that $A$ has $m$ non-zero rows, then every row contains a leading coefficient $1$ that is shifted to the right with respect to the leading coefficient in the row above. The only way we can arrange $m$ 'ones' in this manner is by placing them on diagonal (we can prove it by noting the the required number of 'shifts to the right' of the leading coefficients when going one row down is $m$ - actually - is it clear enough for a proof?). And we know there is nothing except $1$ in any of those columns - we have a diagonal matrix.
Can you make it shorter without referring to some more complicated terminology?
linear-algebra
asked Apr 23 '15 at 15:24
user4205580user4205580
4611131
$endgroup$
Could anyone give an easy to understand proof of this fact or at least tell if the proof below is correct?
We know that every matrix is row equivalent to a reduced row echelon form. An $mtimes m$ square matrix $A$ of rank $m$ is row equivalent to an identity matrix from the definition of reduced row echelon form:
1) all non-zero rows are above rows of all zeros
2) the leading coefficient is strictly to the right of the leading coefficient of the row above
3) the leading coefficient is $1$ and is the only non-zero number in its column
Plus, there's a theorem saying that the rank of any matrix equals the number of non-zero rows in its reduced row echelon form.
Now we know that $A$ has $m$ non-zero rows, then every row contains a leading coefficient $1$ that is shifted to the right with respect to the leading coefficient in the row above. The only way we can arrange $m$ 'ones' in this manner is by placing them on diagonal (we can prove it by noting the the required number of 'shifts to the right' of the leading coefficients when going one row down is $m$ - actually - is it clear enough for a proof?). And we know there is nothing except $1$ in any of those columns - we have a diagonal matrix.
Can you make it shorter without referring to some more complicated terminology?
linear-algebra
linear-algebra
asked Apr 23 '15 at 15:24
user4205580user4205580
4611131
asked Apr 23 '15 at 15:24
user4205580user4205580
4611131
edited Apr 23 '15 at 15:47
user4205580
asked Apr 23 '15 at 15:24
user4205580user4205580
4611131
asked Apr 23 '15 at 15:24
user4205580user4205580
4611131
asked Apr 23 '15 at 15:24
user4205580user4205580
4611131
4611131
$begingroup$
I think it should be "Every invertible (or regular, or non-singular, etc.) matrix is ...." etc.
$endgroup$
– Timbuc
Apr 23 '15 at 15:27
$begingroup$
this wont work with the zero matrix.
$endgroup$
– abel
Apr 23 '15 at 15:34
$begingroup$
I've corrected the title. Do you have any objections?
$endgroup$
– user4205580
Apr 23 '15 at 15:47
add a comment |
$begingroup$
I think it should be "Every invertible (or regular, or non-singular, etc.) matrix is ...." etc.
$endgroup$
– Timbuc
Apr 23 '15 at 15:27
$begingroup$
this wont work with the zero matrix.
$endgroup$
– abel
Apr 23 '15 at 15:34
$begingroup$
I've corrected the title. Do you have any objections?
$endgroup$
– user4205580
Apr 23 '15 at 15:47
$begingroup$
I think it should be "Every invertible (or regular, or non-singular, etc.) matrix is ...." etc.
$endgroup$
– Timbuc
Apr 23 '15 at 15:27
$begingroup$
I think it should be "Every invertible (or regular, or non-singular, etc.) matrix is ...." etc.
$endgroup$
– Timbuc
Apr 23 '15 at 15:27
$begingroup$
this wont work with the zero matrix.
$endgroup$
– abel
Apr 23 '15 at 15:34
$begingroup$
this wont work with the zero matrix.
$endgroup$
– abel
Apr 23 '15 at 15:34
$begingroup$
I've corrected the title. Do you have any objections?
$endgroup$
– user4205580
Apr 23 '15 at 15:47
$begingroup$
I've corrected the title. Do you have any objections?
$endgroup$
– user4205580
Apr 23 '15 at 15:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Matrix equivalence can be completely characterized in terms of a factorization: $A sim B$ if and only if there exist nonsingular $P,Q$ of appropriate order so that $A=PBQ$. The matrices $P$ and $Q$ are the products of elementary row and column operation matrices respectively.
Now if $A$ is a nonsingular $m times m$ matrix, we have $I=A^{-1}A=A^{-1}AI$, with $A^{-1}$ and $I$ nonsingular...so $A sim I$.
Edit: Sorry I see you refer to row equivalence specifically. There is a similar characterization: $A stackrel{R}{sim} B$ iff there exists nonsingular $P$ so that $A=PB$ where $P$ is a product of elementary matrices representing elementary row operations. You can fill in the details....
answered Apr 23 '15 at 18:25
Christiaan HattinghChristiaan Hattingh
3,867922
$endgroup$
add a comment |
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$begingroup$
Matrix equivalence can be completely characterized in terms of a factorization: $A sim B$ if and only if there exist nonsingular $P,Q$ of appropriate order so that $A=PBQ$. The matrices $P$ and $Q$ are the products of elementary row and column operation matrices respectively.
Now if $A$ is a nonsingular $m times m$ matrix, we have $I=A^{-1}A=A^{-1}AI$, with $A^{-1}$ and $I$ nonsingular...so $A sim I$.
Edit: Sorry I see you refer to row equivalence specifically. There is a similar characterization: $A stackrel{R}{sim} B$ iff there exists nonsingular $P$ so that $A=PB$ where $P$ is a product of elementary matrices representing elementary row operations. You can fill in the details....
answered Apr 23 '15 at 18:25
Christiaan HattinghChristiaan Hattingh
3,867922
$endgroup$
add a comment |
$begingroup$
Matrix equivalence can be completely characterized in terms of a factorization: $A sim B$ if and only if there exist nonsingular $P,Q$ of appropriate order so that $A=PBQ$. The matrices $P$ and $Q$ are the products of elementary row and column operation matrices respectively.
Now if $A$ is a nonsingular $m times m$ matrix, we have $I=A^{-1}A=A^{-1}AI$, with $A^{-1}$ and $I$ nonsingular...so $A sim I$.
Edit: Sorry I see you refer to row equivalence specifically. There is a similar characterization: $A stackrel{R}{sim} B$ iff there exists nonsingular $P$ so that $A=PB$ where $P$ is a product of elementary matrices representing elementary row operations. You can fill in the details....
answered Apr 23 '15 at 18:25
Christiaan HattinghChristiaan Hattingh
3,867922
$endgroup$
add a comment |
$begingroup$
Matrix equivalence can be completely characterized in terms of a factorization: $A sim B$ if and only if there exist nonsingular $P,Q$ of appropriate order so that $A=PBQ$. The matrices $P$ and $Q$ are the products of elementary row and column operation matrices respectively.
Now if $A$ is a nonsingular $m times m$ matrix, we have $I=A^{-1}A=A^{-1}AI$, with $A^{-1}$ and $I$ nonsingular...so $A sim I$.
Edit: Sorry I see you refer to row equivalence specifically. There is a similar characterization: $A stackrel{R}{sim} B$ iff there exists nonsingular $P$ so that $A=PB$ where $P$ is a product of elementary matrices representing elementary row operations. You can fill in the details....
answered Apr 23 '15 at 18:25
Christiaan HattinghChristiaan Hattingh
3,867922
$endgroup$
Matrix equivalence can be completely characterized in terms of a factorization: $A sim B$ if and only if there exist nonsingular $P,Q$ of appropriate order so that $A=PBQ$. The matrices $P$ and $Q$ are the products of elementary row and column operation matrices respectively.
Now if $A$ is a nonsingular $m times m$ matrix, we have $I=A^{-1}A=A^{-1}AI$, with $A^{-1}$ and $I$ nonsingular...so $A sim I$.
Edit: Sorry I see you refer to row equivalence specifically. There is a similar characterization: $A stackrel{R}{sim} B$ iff there exists nonsingular $P$ so that $A=PB$ where $P$ is a product of elementary matrices representing elementary row operations. You can fill in the details....
answered Apr 23 '15 at 18:25
Christiaan HattinghChristiaan Hattingh
3,867922
edited Apr 23 '15 at 18:35
answered Apr 23 '15 at 18:25
Christiaan HattinghChristiaan Hattingh
3,867922
answered Apr 23 '15 at 18:25
Christiaan HattinghChristiaan Hattingh
3,867922
answered Apr 23 '15 at 18:25
Christiaan HattinghChristiaan Hattingh
3,867922
3,867922
add a comment |
add a comment |
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$begingroup$
I think it should be "Every invertible (or regular, or non-singular, etc.) matrix is ...." etc.
$endgroup$
– Timbuc
Apr 23 '15 at 15:27
$begingroup$
this wont work with the zero matrix.
$endgroup$
– abel
Apr 23 '15 at 15:34
$begingroup$
I've corrected the title. Do you have any objections?
$endgroup$
– user4205580
Apr 23 '15 at 15:47