Difference between revisions of "Math 260, Spring 2012"
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* If the kernel of a linear transformation is nontrivial, does that imply that the transformation is not injective and therefore the transformation matrix is not invertible? [[User:Patrickmclaren|Patrickmclaren]] 02:11, 6 March 2012 (GMT) |
* If the kernel of a linear transformation is nontrivial, does that imply that the transformation is not injective and therefore the transformation matrix is not invertible? [[User:Patrickmclaren|Patrickmclaren]] 02:11, 6 March 2012 (GMT) |
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: Yes. If the kernel of <math>T</math> is non-trivial, then we have some non-zero <math>\vec{v}</math> with <math>T(\vec{v})=\vec{0}</math>. But also <math>T(\vec{0})=\vec{0}</math>, so <math>T</math> is not one-to-one and hence not invertible. [[User:Steven Glenn Jackson|Steven Glenn Jackson]] 15:27, 6 March 2012 (GMT) |
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Revision as of 15:27, 6 March 2012
Math 260 --- Linear Algebra I
Welcome to the wiki! Editing this page is exactly like editing Wikipedia. You may wish to see their help pages on editing and on typesetting mathematics.
(It isn't as hard as the documentation might make it seem. To see how to typeset the sentence "Consider a linear transformation \(T:R^2\rightarrow R^2\)," click the "edit" link at the top of this page and read the source code that generated it.)
Steven Glenn Jackson 02:42, 1 March 2012 (UTC)
Important Dates
- 02/28/2012 - Exam 1
- 04/17/2012 - Exam 2
Questions
- If the kernel of a linear transformation is nontrivial, does that imply that the transformation is not injective and therefore the transformation matrix is not invertible? Patrickmclaren 02:11, 6 March 2012 (GMT)
- Yes. If the kernel of \(T\) is non-trivial, then we have some non-zero \(\vec{v}\) with \(T(\vec{v})=\vec{0}\). But also \(T(\vec{0})=\vec{0}\), so \(T\) is not one-to-one and hence not invertible. Steven Glenn Jackson 15:27, 6 March 2012 (GMT)
