- #1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter transgalactic
- Start date

- #1

- #2

- 1,753

- 1

What course is this? I already hate Calculus 3 :-[

- #3

- 1,395

- 0

this course is lenear algebra 1

i am a rookey

:)

i am light years away from calculus 3

i am a rookey

:)

i am light years away from calculus 3

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

I can see why rocophysics would dislike Calculus III- if he thinks these problems have anything to do with Calculus III!

Once again, you haven't given the full information: I can guess that you mean that R_{4}[x] is the vector space of cubic polynomials over the real numbers and that T is a linear transformation from R_{4}[x] to itself. You are told that T(x)= x^{2}, that T(x^{2}+ x^{3})= x^{3}+ 1, T(x+1)= x+ x^{2}, and that T(x^{2}+ x)= x^{2}+ x^{3}. That will be enough to completely determine T provided the four polynomials, x, x^{2}+ x^{3}, x+ 1, and x^{2}+ x form a basis for R_{4}[x]. That will be true if and only if they are independent since there are 4 polynomials in that set and the vector space if 4 dimensional.

To see that they are independent, look at a(x)+ b(x^{2}+ x^{3})+ c(x+ 1)+ d(x^{2}+ x)= 0. If they are independent that should be true only if a= b= c= d= 0. Go ahead and collect "like powers": bx^{3}+ (b+ d)x^{2}+ (a+ c+ d)x+ c= 0. A polynomial is 0 (for all x) if and only if the coefficients of each power are 0 (basically x^{3}, x^{2}, x, and 1 are independent) so we must have b= 0, b+ d= 0, a+ c+ d= 0, and c= 0. From the first equation, b= 0 and then from the second d= 0. The last equation tells us c= 0 so the last becomes a+ c+ d= a+ 0+ 0= a= 0. Yes, all coefficients must be 0 so these for "vectors" are independent and form a basis for R_{4}[x]. That means we can write ax^{3}+ bx^{2}+ cx+ d as a linear combination:

[tex]ax^3+ bx^2+ cx+ d= \alpha(x)+ \beta(x^2+ x^3)+ \gamma(x+ 1)+ \delta(x^2+ x)[/tex]

Again multiply that out:

[tex]ax^3+ bx^2+ cx+ d= \beta x^3+ (\beta+ \delta)x^2+ (\alpha+ \gamma+ \delta)x+ \gamma[/tex]

Now that gives [itex]\beta= a[/itex], [itex]\beta+ \delta= a+ \delta= b[/itex] so [itex]\delta= b- a[/itex], [itex]\gamma= d[/itex], [itex]\alpha+ \gamma+ \delta= \alpha+ d+ (b-a)= c[/itex] so [itex]\alpha= a- b+ c- d[/itex].

[tex]T(ax^3+ bx^2+ cx+ d)= (a- b+ c-d )T(x)+ (a)T(x^2+ x^3)+ (d)T(x+1)+ (b-a)T(x^2+ x)[/tex]

Since you know what T does to each of those, you can calculate the result directly now.

After I have done all that, I look back and see that you have done basically the same thing except all in terms of matrices!

Once again, you haven't given the full information: I can guess that you mean that R

To see that they are independent, look at a(x)+ b(x

[tex]ax^3+ bx^2+ cx+ d= \alpha(x)+ \beta(x^2+ x^3)+ \gamma(x+ 1)+ \delta(x^2+ x)[/tex]

Again multiply that out:

[tex]ax^3+ bx^2+ cx+ d= \beta x^3+ (\beta+ \delta)x^2+ (\alpha+ \gamma+ \delta)x+ \gamma[/tex]

Now that gives [itex]\beta= a[/itex], [itex]\beta+ \delta= a+ \delta= b[/itex] so [itex]\delta= b- a[/itex], [itex]\gamma= d[/itex], [itex]\alpha+ \gamma+ \delta= \alpha+ d+ (b-a)= c[/itex] so [itex]\alpha= a- b+ c- d[/itex].

[tex]T(ax^3+ bx^2+ cx+ d)= (a- b+ c-d )T(x)+ (a)T(x^2+ x^3)+ (d)T(x+1)+ (b-a)T(x^2+ x)[/tex]

Since you know what T does to each of those, you can calculate the result directly now.

After I have done all that, I look back and see that you have done basically the same thing except all in terms of matrices!

Last edited by a moderator:

- #5

- 1,395

- 0

so you basicly saying that i did correctly this part

i was confused about the finding of "n" questions in the second file

did i do them correctly??

i was confused about the finding of "n" questions in the second file

did i do them correctly??

Last edited:

- #6

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

Yes, I have not gone in detail through your work but it looks like you are reasoning correctly.

- #7

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

- #8

- 1,395

- 0

i got the identity metrix (I)

there is nothing i can do with it

i get that for every value of N

both in the first and the second part

i get the desirable answer

i know that i am wrong in solving it that

way

what is the right way??

- #9

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

In other words, T is invertible!

i got the identity metrix (I)

there is nothing i can do with it

i get that for every value of N

both in the first and the second part

i get the desirable answer

i know that i am wrong in solving it that

way

what is the right way??

You calculated that

[tex]T= \left(\begin{array}{cccc}0 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\end{array}\right)[/tex]

You asked, first, to apply that to 'x

[tex]\left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right)[/tex]

[tex]\left(\begin{array}{cccc}0 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\end{array}\right)\left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right)= \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0\end{array}\right)[/tex]

That's not 0 so do it again:

[tex]\left(\begin{array}{cccc}0 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\end{array}\right)\left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array}\right)= \left(\begin{array}{c} -2 \\ 0 \\ 0 \\ 1\end{array}\right)[/tex]

Again, that's not 0 so do it again:

[tex]\left(\begin{array}{cccc}0 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\end{array}\right)\left(\begin{array}{c} -2 \\ 0 \\ 0 \\ 1 \end{array}\right)= \left(\begin{array}{c} 4 \\ 0 \\ 1 \\ -2\end{array}\right)[/tex]

Keep doing that until you get [0, 1, 0, 0]

To find an n such that T

[tex]T^2= \left(\begin{array}{cccc}0 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\end{array}\right)\left(\begin{array}{cccc}0 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\end{array}\right)= \(\begin{array}{cccc} -2 & 1 & 0 & 0 \\ 0 & -2 & 0 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right)[/tex]

T

Share: