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Mathematics 13 Online
OpenStudy (anonymous):

find a basis................

OpenStudy (amistre64):

[1,0], [0,1] :)

OpenStudy (anonymous):

find a basis for subspace W of \[P_{3}\]

OpenStudy (anonymous):

consisting of all vectors of the form

OpenStudy (anonymous):

\[a t^{3}+bt ^{2}+ct+d, where b=3a-5d and c=4a+d\]

OpenStudy (anonymous):

* where b=3a-5d and c=4a+d

OpenStudy (amistre64):

1a+0d 3a-5d 4a+1d 0a+1d looks like the basis is [1,3,4,4],[0,5,1,1]

OpenStudy (anonymous):

should i doing row echolen form

OpenStudy (amistre64):

-5 that is

OpenStudy (amistre64):

depends on what you want your basis to look like; a basis generally is just a set of independant vectors

OpenStudy (anonymous):

why basis {1,3,4,4} not 1,3,4,0 ??

OpenStudy (anonymous):

if i want to do row echelon form..how should i start it?

OpenStudy (amistre64):

becasue im typing on my laptop and hit alot of wrong keys :)

OpenStudy (amistre64):

start at the top and do elementary row operations to get it into some sort of echelon form

OpenStudy (amistre64):

you can even do a grahm schmit if you want to form an orthogonal basis

OpenStudy (anonymous):

can you do a matrix form from this question

OpenStudy (anonymous):

so i can do row echelon form

OpenStudy (amistre64):

http://tutorial.math.lamar.edu/Classes/LinAlg/ChangeOfBasis.aspx look at example 2, does that look like what this question is asking? if so i might have been thinking of something else

OpenStudy (anonymous):

what is difference p2 and p3..does it refer to power??

OpenStudy (amistre64):

at3+bt2+ct+d \[\begin{vmatrix}a&b&c&d\end{vmatrix}~\begin{vmatrix}t^3\\t^2\\t\\ 1\end{vmatrix}\] the subscript referes to the degree of the poly

OpenStudy (amistre64):

\[P_n(x)=a_0+a_1x+a_2x^2+...+a_nx^n\]

OpenStudy (amistre64):

id have to do some more reading to refresh me memories on this .... will have to get back at it in the morning tho.

OpenStudy (amistre64):

http://www.csus.edu/indiv/e/elcek/M35/section4_6.pdf question 32 is exactly like yours, and has a followable solution that could help

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