Confused on whenever or not non-pivot columns constitute directly a basis of KerKernels and reduced row echelon form - explanationRelationship between the column space of a matrix $A$ and its non-free (pivot) columnsPivot columns and basic variablesCan a set of 4 vectors with 3 entries each only span R2 if the third row reduces to all zeros?Basis of column/row space of $A$: using pivot columns of $A$ vs. $textrref(A)$?True or false: The non-pivot columns of a matrix are always linearly dependent.Could non pivot columns form the basis for the column space of a matrix?find basis for kernel and columnsPivot columns and basis for column spaceProve that pivot columns of row reduced form of any matrix forms a basis in column space of that row reduced matrix.Why are pivot columns the basis of A?
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Confused on whenever or not non-pivot columns constitute directly a basis of Ker
Kernels and reduced row echelon form - explanationRelationship between the column space of a matrix $A$ and its non-free (pivot) columnsPivot columns and basic variablesCan a set of 4 vectors with 3 entries each only span R2 if the third row reduces to all zeros?Basis of column/row space of $A$: using pivot columns of $A$ vs. $textrref(A)$?True or false: The non-pivot columns of a matrix are always linearly dependent.Could non pivot columns form the basis for the column space of a matrix?find basis for kernel and columnsPivot columns and basis for column spaceProve that pivot columns of row reduced form of any matrix forms a basis in column space of that row reduced matrix.Why are pivot columns the basis of A?
$begingroup$
I know that you can find the basis of ker by solving Ax = 0. If you can find the basis of the Image by taking the pivot columns can you do the same with ker with non-pivot columns?
linear-algebra
asked Mar 29 at 20:32
Dr.StoneDr.Stone
335
$endgroup$
add a comment |
$begingroup$
I know that you can find the basis of ker by solving Ax = 0. If you can find the basis of the Image by taking the pivot columns can you do the same with ker with non-pivot columns?
linear-algebra
asked Mar 29 at 20:32
Dr.StoneDr.Stone
335
$endgroup$
1
$begingroup$
No, but yes. Paramaterize the solution. The resulting vectors that span are the basis vectors. You’re only really using the non-pivot columns to tell what the dimension of the kernel is. If you’re well practiced, you don’t have to perform this procedure and can just tell what the basis vectors will be.
$endgroup$
– DaveNine
Mar 29 at 20:36
$begingroup$
See math.stackexchange.com/a/1521354/265466. If $A$ is not square, the column vectors have the wrong dimension to be in the kernel, but there’s a mechanical process for generating a kernel basis from the non-pivot columns.
$endgroup$
– amd
Mar 29 at 21:05
add a comment |
$begingroup$
I know that you can find the basis of ker by solving Ax = 0. If you can find the basis of the Image by taking the pivot columns can you do the same with ker with non-pivot columns?
linear-algebra
asked Mar 29 at 20:32
Dr.StoneDr.Stone
335
$endgroup$
I know that you can find the basis of ker by solving Ax = 0. If you can find the basis of the Image by taking the pivot columns can you do the same with ker with non-pivot columns?
linear-algebra
linear-algebra
asked Mar 29 at 20:32
Dr.StoneDr.Stone
335
asked Mar 29 at 20:32
Dr.StoneDr.Stone
335
asked Mar 29 at 20:32
Dr.StoneDr.Stone
335
asked Mar 29 at 20:32
Dr.StoneDr.Stone
335
asked Mar 29 at 20:32
Dr.StoneDr.Stone
335
335
1
$begingroup$
No, but yes. Paramaterize the solution. The resulting vectors that span are the basis vectors. You’re only really using the non-pivot columns to tell what the dimension of the kernel is. If you’re well practiced, you don’t have to perform this procedure and can just tell what the basis vectors will be.
$endgroup$
– DaveNine
Mar 29 at 20:36
$begingroup$
See math.stackexchange.com/a/1521354/265466. If $A$ is not square, the column vectors have the wrong dimension to be in the kernel, but there’s a mechanical process for generating a kernel basis from the non-pivot columns.
$endgroup$
– amd
Mar 29 at 21:05
add a comment |
1
$begingroup$
No, but yes. Paramaterize the solution. The resulting vectors that span are the basis vectors. You’re only really using the non-pivot columns to tell what the dimension of the kernel is. If you’re well practiced, you don’t have to perform this procedure and can just tell what the basis vectors will be.
$endgroup$
– DaveNine
Mar 29 at 20:36
$begingroup$
See math.stackexchange.com/a/1521354/265466. If $A$ is not square, the column vectors have the wrong dimension to be in the kernel, but there’s a mechanical process for generating a kernel basis from the non-pivot columns.
$endgroup$
– amd
Mar 29 at 21:05
1
1
$begingroup$
No, but yes. Paramaterize the solution. The resulting vectors that span are the basis vectors. You’re only really using the non-pivot columns to tell what the dimension of the kernel is. If you’re well practiced, you don’t have to perform this procedure and can just tell what the basis vectors will be.
$endgroup$
– DaveNine
Mar 29 at 20:36
$begingroup$
No, but yes. Paramaterize the solution. The resulting vectors that span are the basis vectors. You’re only really using the non-pivot columns to tell what the dimension of the kernel is. If you’re well practiced, you don’t have to perform this procedure and can just tell what the basis vectors will be.
$endgroup$
– DaveNine
Mar 29 at 20:36
$begingroup$
See math.stackexchange.com/a/1521354/265466. If $A$ is not square, the column vectors have the wrong dimension to be in the kernel, but there’s a mechanical process for generating a kernel basis from the non-pivot columns.
$endgroup$
– amd
Mar 29 at 21:05
$begingroup$
See math.stackexchange.com/a/1521354/265466. If $A$ is not square, the column vectors have the wrong dimension to be in the kernel, but there’s a mechanical process for generating a kernel basis from the non-pivot columns.
$endgroup$
– amd
Mar 29 at 21:05
add a comment |
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$begingroup$
No, but yes. Paramaterize the solution. The resulting vectors that span are the basis vectors. You’re only really using the non-pivot columns to tell what the dimension of the kernel is. If you’re well practiced, you don’t have to perform this procedure and can just tell what the basis vectors will be.
$endgroup$
– DaveNine
Mar 29 at 20:36
$begingroup$
See math.stackexchange.com/a/1521354/265466. If $A$ is not square, the column vectors have the wrong dimension to be in the kernel, but there’s a mechanical process for generating a kernel basis from the non-pivot columns.
$endgroup$
– amd
Mar 29 at 21:05