Talk:06-240/Classnotes For Tuesday November 14: Difference between revisions
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Reduced row echelon form - Is there a reason to make column with entry 1 to the form of e<sub>n</sub> (1 at n<sup>th</sup> row, 0 for all other entries)? According to some books, matrix <math>\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}</math> is good enough to show that the rank of the matrix is 3. This is because the first three rows are linearly independent, they can't form linear combination for preceeding rows. Anyone could please explain why we have to reduce it to <math>\begin{pmatrix}1&0&-4&0&-2\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}</math>? Thank you. [[User:Wongpak|Wongpak]] 23:54, 15 November 2006 (EST) |
Reduced row echelon form - Is there a reason to make column with entry 1 to the form of e<sub>n</sub> (1 at n<sup>th</sup> row, 0 for all other entries)? According to some books, matrix <math>\begin{pmatrix}1&3&2&4&2\\0&1&2&3&4\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}</math> is good enough to show that the rank of the matrix is 3. This is because the first three rows are linearly independent, they can't form linear combination for preceeding rows. Anyone could please explain why we have to reduce it to <math>\begin{pmatrix}1&0&-4&0&-2\\0&1&2&0&-2\\0&0&0&1&2\\0&0&0&0&0 \end{pmatrix}</math>? Thank you. [[User:Wongpak|Wongpak]] 23:54, 15 November 2006 (EST) |
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'''Answer.''' For the purpose of figuring out the rank of a matrix, indeed there is no reason to go to "reduced row echelon form", and the slightly easier "row echelon form" (as in your first matrix above) is sufficient. But |
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# "Row echelon form" doesn't enjoy the nicer characterization as "the most you can do with row reduction". |
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# It is possible to prove (though most likely we won't) that the reduced row echelon form of a matrix is unique, while the row echelon form certainly isn't. |
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# We did use the fact that we could get via row operation to the reduced row echelon form in our algorithm for matrix inversion. |
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--[[User:Drorbn|Drorbn]] 09:47, 16 November 2006 (EST) |
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Revision as of 09:47, 16 November 2006
Reduced row echelon form - Is there a reason to make column with entry 1 to the form of en (1 at nth row, 0 for all other entries)? According to some books, matrix is good enough to show that the rank of the matrix is 3. This is because the first three rows are linearly independent, they can't form linear combination for preceeding rows. Anyone could please explain why we have to reduce it to ? Thank you. Wongpak 23:54, 15 November 2006 (EST)
Answer. For the purpose of figuring out the rank of a matrix, indeed there is no reason to go to "reduced row echelon form", and the slightly easier "row echelon form" (as in your first matrix above) is sufficient. But
- "Row echelon form" doesn't enjoy the nicer characterization as "the most you can do with row reduction".
- It is possible to prove (though most likely we won't) that the reduced row echelon form of a matrix is unique, while the row echelon form certainly isn't.
- We did use the fact that we could get via row operation to the reduced row echelon form in our algorithm for matrix inversion.
--Drorbn 09:47, 16 November 2006 (EST)
