column-reducing algorithm for finding nullspace of matrix












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I have found a more "automatic" way to compute a basis for the nullspace of a matrix on Wikipedia, http://en.wikipedia.org/wiki/Nullspace , but from that article, I can see no explanation for why it works. For example, I have no idea as to WHY we are trying to append the identity matrix to the bottom of the matrix to be computed. Also, I do not get why we are trying to get columns whose entries in the top (original) matrix are zeroes. I have tried looking elsewhere, but cannot find anything...



EDIT: I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



1.) If we were to transpose our augmented matrix, we would have $[A'|I]$, where $A'$ simply means "the transpose of A".



2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on $[A'|I]$; what you will end up with for the column-augmented matrix is $[B|C]'$ (which, of course, means $[B|C]$ for the transpose. It should be noted that $B$ is the row-reduced form of $A'$ (and the transpose is the column-reduced form of $A$). If $A$ has a nullity (dimension of the nullspace) greater than 0, then that means that $A$ is singular, which means that if we were to try to row-reduce the top/left square submatrix of $A$ (or $A'$), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix.



I think I have answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...










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    0












    $begingroup$


    I have found a more "automatic" way to compute a basis for the nullspace of a matrix on Wikipedia, http://en.wikipedia.org/wiki/Nullspace , but from that article, I can see no explanation for why it works. For example, I have no idea as to WHY we are trying to append the identity matrix to the bottom of the matrix to be computed. Also, I do not get why we are trying to get columns whose entries in the top (original) matrix are zeroes. I have tried looking elsewhere, but cannot find anything...



    EDIT: I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



    1.) If we were to transpose our augmented matrix, we would have $[A'|I]$, where $A'$ simply means "the transpose of A".



    2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on $[A'|I]$; what you will end up with for the column-augmented matrix is $[B|C]'$ (which, of course, means $[B|C]$ for the transpose. It should be noted that $B$ is the row-reduced form of $A'$ (and the transpose is the column-reduced form of $A$). If $A$ has a nullity (dimension of the nullspace) greater than 0, then that means that $A$ is singular, which means that if we were to try to row-reduce the top/left square submatrix of $A$ (or $A'$), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix.



    I think I have answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...










    share|cite|improve this question











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      0












      0








      0


      1



      $begingroup$


      I have found a more "automatic" way to compute a basis for the nullspace of a matrix on Wikipedia, http://en.wikipedia.org/wiki/Nullspace , but from that article, I can see no explanation for why it works. For example, I have no idea as to WHY we are trying to append the identity matrix to the bottom of the matrix to be computed. Also, I do not get why we are trying to get columns whose entries in the top (original) matrix are zeroes. I have tried looking elsewhere, but cannot find anything...



      EDIT: I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



      1.) If we were to transpose our augmented matrix, we would have $[A'|I]$, where $A'$ simply means "the transpose of A".



      2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on $[A'|I]$; what you will end up with for the column-augmented matrix is $[B|C]'$ (which, of course, means $[B|C]$ for the transpose. It should be noted that $B$ is the row-reduced form of $A'$ (and the transpose is the column-reduced form of $A$). If $A$ has a nullity (dimension of the nullspace) greater than 0, then that means that $A$ is singular, which means that if we were to try to row-reduce the top/left square submatrix of $A$ (or $A'$), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix.



      I think I have answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...










      share|cite|improve this question











      $endgroup$




      I have found a more "automatic" way to compute a basis for the nullspace of a matrix on Wikipedia, http://en.wikipedia.org/wiki/Nullspace , but from that article, I can see no explanation for why it works. For example, I have no idea as to WHY we are trying to append the identity matrix to the bottom of the matrix to be computed. Also, I do not get why we are trying to get columns whose entries in the top (original) matrix are zeroes. I have tried looking elsewhere, but cannot find anything...



      EDIT: I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



      1.) If we were to transpose our augmented matrix, we would have $[A'|I]$, where $A'$ simply means "the transpose of A".



      2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on $[A'|I]$; what you will end up with for the column-augmented matrix is $[B|C]'$ (which, of course, means $[B|C]$ for the transpose. It should be noted that $B$ is the row-reduced form of $A'$ (and the transpose is the column-reduced form of $A$). If $A$ has a nullity (dimension of the nullspace) greater than 0, then that means that $A$ is singular, which means that if we were to try to row-reduce the top/left square submatrix of $A$ (or $A'$), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix.



      I think I have answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...







      linear-algebra gaussian-elimination






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      edited Jan 7 '14 at 20:35







      Mike Warren

















      asked Jan 7 '14 at 18:46









      Mike WarrenMike Warren

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          $begingroup$

          I will repost the EDIT here:



          I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



          1.) If we were to transpose our augmented matrix, we would have [A′|I], where A′ simply means "the transpose of A".



          2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on [A′|I]; what you will end up with for the column-augmented matrix is [B|C]′ (which, of course, means [B|C] for the transpose. It should be noted that B is the row-reduced form of A′ (and the transpose is the column-reduced form of A). If A has a nullity (dimension of the nullspace) greater than 0, then that means that A is singular, which means that if we were to try to row-reduce the top/left square submatrix of A (or A′), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix. The record that corresponds to that row of $A'$ (or column of $A$) full of zeroes is in that same row/column of the "record matrix". That explains why this works.



          I think I have correctly answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...






          share|cite|improve this answer









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            $begingroup$

            I will repost the EDIT here:



            I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



            1.) If we were to transpose our augmented matrix, we would have [A′|I], where A′ simply means "the transpose of A".



            2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on [A′|I]; what you will end up with for the column-augmented matrix is [B|C]′ (which, of course, means [B|C] for the transpose. It should be noted that B is the row-reduced form of A′ (and the transpose is the column-reduced form of A). If A has a nullity (dimension of the nullspace) greater than 0, then that means that A is singular, which means that if we were to try to row-reduce the top/left square submatrix of A (or A′), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix. The record that corresponds to that row of $A'$ (or column of $A$) full of zeroes is in that same row/column of the "record matrix". That explains why this works.



            I think I have correctly answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I will repost the EDIT here:



              I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



              1.) If we were to transpose our augmented matrix, we would have [A′|I], where A′ simply means "the transpose of A".



              2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on [A′|I]; what you will end up with for the column-augmented matrix is [B|C]′ (which, of course, means [B|C] for the transpose. It should be noted that B is the row-reduced form of A′ (and the transpose is the column-reduced form of A). If A has a nullity (dimension of the nullspace) greater than 0, then that means that A is singular, which means that if we were to try to row-reduce the top/left square submatrix of A (or A′), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix. The record that corresponds to that row of $A'$ (or column of $A$) full of zeroes is in that same row/column of the "record matrix". That explains why this works.



              I think I have correctly answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...






              share|cite|improve this answer









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                0












                0








                0





                $begingroup$

                I will repost the EDIT here:



                I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



                1.) If we were to transpose our augmented matrix, we would have [A′|I], where A′ simply means "the transpose of A".



                2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on [A′|I]; what you will end up with for the column-augmented matrix is [B|C]′ (which, of course, means [B|C] for the transpose. It should be noted that B is the row-reduced form of A′ (and the transpose is the column-reduced form of A). If A has a nullity (dimension of the nullspace) greater than 0, then that means that A is singular, which means that if we were to try to row-reduce the top/left square submatrix of A (or A′), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix. The record that corresponds to that row of $A'$ (or column of $A$) full of zeroes is in that same row/column of the "record matrix". That explains why this works.



                I think I have correctly answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...






                share|cite|improve this answer









                $endgroup$



                I will repost the EDIT here:



                I think it works for the same reason that the process of row-augmenting a matrix with the identity matrix, and row-reducing it until we have the identity matrix where the original matrix was works. Heck, by that logic, we could think of the process of column-augmenting our matrix and employing the algorithm talked about in the Wikipedia article as such:



                1.) If we were to transpose our augmented matrix, we would have [A′|I], where A′ simply means "the transpose of A".



                2.) Column-reduction on the column-augmented matrix is equivalent to row reduction on [A′|I]; what you will end up with for the column-augmented matrix is [B|C]′ (which, of course, means [B|C] for the transpose. It should be noted that B is the row-reduced form of A′ (and the transpose is the column-reduced form of A). If A has a nullity (dimension of the nullspace) greater than 0, then that means that A is singular, which means that if we were to try to row-reduce the top/left square submatrix of A (or A′), we would get a number of rows/columns full of zeroes equal to its nullity. We are interested in these columns, and namely, a record of the operations we did to that submatrix. The record that corresponds to that row of $A'$ (or column of $A$) full of zeroes is in that same row/column of the "record matrix". That explains why this works.



                I think I have correctly answered my own question, but do not know. So I will leave it here for critique (as my question IS the how/why the algorithm I found works)...







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 '14 at 1:10









                Mike WarrenMike Warren

                153111




                153111






























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