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Why is the reduced echelon form of a set of independent vectors, the identity matrix?

Why is a square rref with no zero rows always the identity matrix?Why is reduced echelon form of a matrix always the identity matrixExample of Matrix in Reduced Row Echelon FormMatrices - Understanding row echelon form and reduced echelon formWhy is reduced echelon form of a matrix always the identity matrixDifferent representations of a matrix in reduced row echelon formwhy does the reduced row echelon form have the same null space as the original matrix?Is it okay to determine pivot positions in a matrix in echelon form, not in reduced echelon form?Anybody knows a proof of Uniqueness of the Reduced Echelon Form Theorem?Why is a matrix non-invertible if its row-reduced echelon form matrix is not identity?Different answers for reduced row echelon form (ROW vector and Column Vector cases)Adding identity matrix to A and its reduced echelon form

3

1

$begingroup$

If a matrix has linearly independent rows, then its reduced echelon form is the identity matrix.

I haven't found a concise explanation for this... I have the whole notion in my head but I cannot express this in words. Can someone explain it?

linear-algebra matrices

share|cite|improve this question

edited Oct 27 '14 at 7:04

Joonas Ilmavirta

20.7k94282

asked Oct 17 '14 at 6:13

JOXJOX

7841920

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming this is about row reduction, it is simply not true: if a matrix contains fewer rows than columns, then so will its reduced row echelon form. Also, the title mentions vectors, not rows; in coordinates, vectors are usually associated with columns. So your title does not match the highlighted statement.
  $endgroup$
  – Marc van Leeuwen
  Nov 17 '16 at 17:55
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

If a matrix has linearly independent rows, then its reduced echelon form is the identity matrix.

I haven't found a concise explanation for this... I have the whole notion in my head but I cannot express this in words. Can someone explain it?

linear-algebra matrices

share|cite|improve this question

edited Oct 27 '14 at 7:04

Joonas Ilmavirta

20.7k94282

asked Oct 17 '14 at 6:13

JOXJOX

7841920

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming this is about row reduction, it is simply not true: if a matrix contains fewer rows than columns, then so will its reduced row echelon form. Also, the title mentions vectors, not rows; in coordinates, vectors are usually associated with columns. So your title does not match the highlighted statement.
  $endgroup$
  – Marc van Leeuwen
  Nov 17 '16 at 17:55
  
  

  
  
  
  

add a comment | 

$begingroup$

If a matrix has linearly independent rows, then its reduced echelon form is the identity matrix.

I haven't found a concise explanation for this... I have the whole notion in my head but I cannot express this in words. Can someone explain it?

linear-algebra matrices

share|cite|improve this question

edited Oct 27 '14 at 7:04

Joonas Ilmavirta

20.7k94282

asked Oct 17 '14 at 6:13

JOXJOX

7841920

$endgroup$

share|cite|improve this question

edited Oct 27 '14 at 7:04

Joonas Ilmavirta

20.7k94282

asked Oct 17 '14 at 6:13

JOXJOX

7841920

share|cite|improve this question

edited Oct 27 '14 at 7:04

Joonas Ilmavirta

20.7k94282

asked Oct 17 '14 at 6:13

JOXJOX

7841920

edited Oct 27 '14 at 7:04

Joonas Ilmavirta

20.7k94282

edited Oct 27 '14 at 7:04

Joonas Ilmavirta

20.7k94282

asked Oct 17 '14 at 6:13

JOXJOX

7841920

asked Oct 17 '14 at 6:13

JOXJOX

7841920

3 Answers
3

active

oldest

votes

2

$begingroup$

An $ntimes n$ matrix whose rows (or columns) form a linearly independent set has rank $n$. The elementary row (or column) operations are rank preserving, so the reduced row echelon form of the matrix will have rank $n$.

Now, using the properties of a matrix in reduced row echelon form (i.e., first nonzero term in each row is $1$, and no nonzero terms in any column with a leading $1$) see if you can show that the only $ntimes n$ matrix of rank $n$ in reduced row echelon form is the identity matrix.

share|cite|improve this answer

answered Oct 17 '14 at 7:05

JaredJared

24.2k1047102

$endgroup$

add a comment | 

0

$begingroup$

This might help you:

Suppose we have a matrix $A$ whose columns $v_1,dots,v_k$ are vectors in $Bbb R^n$.

The reduced echelon form of $A$ will have a pivot in every column if and only if the vectors $v_1,dots,v_k$ are linearly independent. The reduced row echelon form of $A$ will have a pivot in every row if and only if the vectors $v_1,dots,v_k$ span $Bbb R^n$.

The only reduced matrix with a pivot in every row and every column is the identity matrix.

share|cite|improve this answer

answered Oct 17 '14 at 14:36

OmnomnomnomOmnomnomnom

129k792185

$endgroup$

add a comment | 

0

$begingroup$

The contrapositive of this is: If the reduced row echelon form is not the identity matrix, then the rows are linearly dependent.

This might be easier to understand. First, if the RREF of a square matrix is not the identity matrix, then one row will not have a leading one. That row must be made all of zeroes. Now track how you got this row of zeroes: if you track back to your original matrix, this row of zeroes is obtained as a linear combinations of rows in the original matrix, and the corresponding linear combination gives you a linear dependency of some rows in the matrix.

As you get a linear dependency, the rows are linearly dependent.

share|cite|improve this answer

answered Oct 9 '16 at 3:16

N. S.N. S.

105k7114210

$endgroup$

add a comment | 

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2

$begingroup$

An $ntimes n$ matrix whose rows (or columns) form a linearly independent set has rank $n$. The elementary row (or column) operations are rank preserving, so the reduced row echelon form of the matrix will have rank $n$.

Now, using the properties of a matrix in reduced row echelon form (i.e., first nonzero term in each row is $1$, and no nonzero terms in any column with a leading $1$) see if you can show that the only $ntimes n$ matrix of rank $n$ in reduced row echelon form is the identity matrix.

share|cite|improve this answer

answered Oct 17 '14 at 7:05

JaredJared

24.2k1047102

$endgroup$

add a comment | 

2

$begingroup$

An $ntimes n$ matrix whose rows (or columns) form a linearly independent set has rank $n$. The elementary row (or column) operations are rank preserving, so the reduced row echelon form of the matrix will have rank $n$.

Now, using the properties of a matrix in reduced row echelon form (i.e., first nonzero term in each row is $1$, and no nonzero terms in any column with a leading $1$) see if you can show that the only $ntimes n$ matrix of rank $n$ in reduced row echelon form is the identity matrix.

share|cite|improve this answer

answered Oct 17 '14 at 7:05

JaredJared

24.2k1047102

$endgroup$

add a comment | 

$begingroup$

An $ntimes n$ matrix whose rows (or columns) form a linearly independent set has rank $n$. The elementary row (or column) operations are rank preserving, so the reduced row echelon form of the matrix will have rank $n$.

Now, using the properties of a matrix in reduced row echelon form (i.e., first nonzero term in each row is $1$, and no nonzero terms in any column with a leading $1$) see if you can show that the only $ntimes n$ matrix of rank $n$ in reduced row echelon form is the identity matrix.

share|cite|improve this answer

answered Oct 17 '14 at 7:05

JaredJared

24.2k1047102

$endgroup$

share|cite|improve this answer

answered Oct 17 '14 at 7:05

JaredJared

24.2k1047102

answered Oct 17 '14 at 7:05

JaredJared

24.2k1047102

answered Oct 17 '14 at 7:05

JaredJared

24.2k1047102

0

$begingroup$

This might help you:

Suppose we have a matrix $A$ whose columns $v_1,dots,v_k$ are vectors in $Bbb R^n$.

The reduced echelon form of $A$ will have a pivot in every column if and only if the vectors $v_1,dots,v_k$ are linearly independent. The reduced row echelon form of $A$ will have a pivot in every row if and only if the vectors $v_1,dots,v_k$ span $Bbb R^n$.

The only reduced matrix with a pivot in every row and every column is the identity matrix.

share|cite|improve this answer

answered Oct 17 '14 at 14:36

OmnomnomnomOmnomnomnom

129k792185

$endgroup$

add a comment | 

0

$begingroup$

This might help you:

Suppose we have a matrix $A$ whose columns $v_1,dots,v_k$ are vectors in $Bbb R^n$.

The reduced echelon form of $A$ will have a pivot in every column if and only if the vectors $v_1,dots,v_k$ are linearly independent. The reduced row echelon form of $A$ will have a pivot in every row if and only if the vectors $v_1,dots,v_k$ span $Bbb R^n$.

The only reduced matrix with a pivot in every row and every column is the identity matrix.

share|cite|improve this answer

answered Oct 17 '14 at 14:36

OmnomnomnomOmnomnomnom

129k792185

$endgroup$

add a comment | 

$begingroup$

This might help you:

Suppose we have a matrix $A$ whose columns $v_1,dots,v_k$ are vectors in $Bbb R^n$.

The reduced echelon form of $A$ will have a pivot in every column if and only if the vectors $v_1,dots,v_k$ are linearly independent. The reduced row echelon form of $A$ will have a pivot in every row if and only if the vectors $v_1,dots,v_k$ span $Bbb R^n$.

The only reduced matrix with a pivot in every row and every column is the identity matrix.

share|cite|improve this answer

answered Oct 17 '14 at 14:36

OmnomnomnomOmnomnomnom

129k792185

$endgroup$

share|cite|improve this answer

answered Oct 17 '14 at 14:36

OmnomnomnomOmnomnomnom

129k792185

answered Oct 17 '14 at 14:36

OmnomnomnomOmnomnomnom

129k792185

answered Oct 17 '14 at 14:36

OmnomnomnomOmnomnomnom

129k792185

0

$begingroup$

The contrapositive of this is: If the reduced row echelon form is not the identity matrix, then the rows are linearly dependent.

This might be easier to understand. First, if the RREF of a square matrix is not the identity matrix, then one row will not have a leading one. That row must be made all of zeroes. Now track how you got this row of zeroes: if you track back to your original matrix, this row of zeroes is obtained as a linear combinations of rows in the original matrix, and the corresponding linear combination gives you a linear dependency of some rows in the matrix.

As you get a linear dependency, the rows are linearly dependent.

share|cite|improve this answer

answered Oct 9 '16 at 3:16

N. S.N. S.

105k7114210

$endgroup$

add a comment | 

0

$begingroup$

The contrapositive of this is: If the reduced row echelon form is not the identity matrix, then the rows are linearly dependent.

This might be easier to understand. First, if the RREF of a square matrix is not the identity matrix, then one row will not have a leading one. That row must be made all of zeroes. Now track how you got this row of zeroes: if you track back to your original matrix, this row of zeroes is obtained as a linear combinations of rows in the original matrix, and the corresponding linear combination gives you a linear dependency of some rows in the matrix.

As you get a linear dependency, the rows are linearly dependent.

share|cite|improve this answer

answered Oct 9 '16 at 3:16

N. S.N. S.

105k7114210

$endgroup$

add a comment | 

$begingroup$

The contrapositive of this is: If the reduced row echelon form is not the identity matrix, then the rows are linearly dependent.

This might be easier to understand. First, if the RREF of a square matrix is not the identity matrix, then one row will not have a leading one. That row must be made all of zeroes. Now track how you got this row of zeroes: if you track back to your original matrix, this row of zeroes is obtained as a linear combinations of rows in the original matrix, and the corresponding linear combination gives you a linear dependency of some rows in the matrix.

As you get a linear dependency, the rows are linearly dependent.

share|cite|improve this answer

answered Oct 9 '16 at 3:16

N. S.N. S.

105k7114210

$endgroup$

share|cite|improve this answer

answered Oct 9 '16 at 3:16

N. S.N. S.

105k7114210

answered Oct 9 '16 at 3:16

N. S.N. S.

105k7114210

answered Oct 9 '16 at 3:16

N. S.N. S.

105k7114210