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Unformatted text preview: or
gives 3
equations with 5 unknowns, so we are free to pick 2 variables (Note not necessarily any two).
The 3 equations are: 7 December 2012 4 Dept. of ECE, Drexel University ECES511: Systems I Choosing
Choosing Fall 20122013 gives: . gives: . These vectors are linearly independent and therefore serve as a basis for the nullspace of A3. d. The rank of is 3, therefore a basis for the column space is: [ ] [ ] [ ] A basis for the column space of is the set of the 2 linearly independent columns of which is: {[ ] [ ]} . The rank of
is 3, therefore there are 3 linearly independent columns. These can serve as a
basis for the column space of .
We can search linearly independent columns from
and quickly see that columns 1,2 and 4.
Thus a valid basis for the column space is: {[ ] [ ] [ ]}. 7 December 2012 5 Dept. of ECE, Drexel University ECES511: Systems I Fall 20122013 3. Linear Transformations.
Consider Example 1 given on page 20 of Chen. This example showed graphically that rotation
about the origin is a linear transformation. As we know, one can re...
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 Spring '14

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