Talk:06-240/Classnotes For Tuesday November 14 - Revision history

Drorbn at 15:00, 16 November 2006

2006-11-16T15:00:44Z

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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.		# 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.		# We did use the fact that we could get via row operation to the reduced row echelon form in our algorithm for matrix inversion.
			# We will use reduced row echelon form on [[06-240/Classnotes For Thursday November 16\|November 16]] once again, to help us solve systems of linear equations.
	--[[User:Drorbn\|Drorbn]] 09:47, 16 November 2006 (EST)		--[[User:Drorbn\|Drorbn]] 10:00, 16 November 2006 (EST)

Drorbn at 14:47, 16 November 2006

2006-11-16T14:47:39Z

← Older revision		Revision as of 10:47, 16 November 2006
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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)

			'''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.
			--[[User:Drorbn\|Drorbn]] 09:47, 16 November 2006 (EST)

Wongpak at 04:55, 16 November 2006

2006-11-16T04:55:26Z

← Older revision		Revision as of 00:55, 16 November 2006
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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 ~~tell~~ 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)

Wongpak at 04:54, 16 November 2006

2006-11-16T04:54:22Z

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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 tell 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)