
MHB Master
#1
January 8th, 2019,
02:24
I am reading the book: "Linear Algebra" by Stephen Friedberg, Arnold Insel, and Lawrence Spence ... and am currently focused on Section 2.6: Dual Spaces ... ...
I need help with an aspect of Example 4, Section 2.6 ...
Example 4, Section 2.6 reads as follows: (see below for details of Section 2.6 ...)
Can someone please explain (in detail) how/why $ \displaystyle f_1(2,1) = 1$ ... ?
Help will be appreciated ...
Peter
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To understand the context and notation of the above example it may help MHB readers to have access to the text of Section 2.6 ... so I am providing the same ... as follows ...
Hope that helps ...
Peter

January 8th, 2019 02:24
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MHB Oldtimer
#2
January 8th, 2019,
04:45
Originally Posted by
Peter
Can someone please explain (in detail) how/why $ \displaystyle f_1(2,1) = 1$ ... ?
This comes directly from the definition of the dual basis. In Example 3 of Section 2.6, Friedberg, Insel and Spence say "Note that $\textsf{f}_i(x_j) = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta." In this example, the first element of the given basis is $x_1 = (2,1)$. So $\textsf{f}_1(2,1) = \textsf{f}_1(x_1) = \delta_{11} = 1.$

MHB Master
#3
January 8th, 2019,
07:58
Thread Author
Originally Posted by
Opalg
This comes directly from the definition of the dual basis. In Example 3 of Section 2.6, Friedberg, Insel and Spence say "Note that $\textsf{f}_i(x_j) = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta." In this example, the first element of the given basis is $x_1 = (2,1)$. So $\textsf{f}_1(2,1) = \textsf{f}_1(x_1) = \delta_{11} = 1.$
Thanks Opalg ...
Appreciate your help...
Peter