Is this the right way to solve these questions

[transgalactic]

file1:
http://img212.imageshack.us/my.php?image=img8605it4.jpg

file2:
http://img212.imageshack.us/my.php?image=img8606jc4.jpg

 

[rocomath]

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

 

[transgalactic]

this course is lenear algebra 1
i am a rookey
:)

i am light years away from calculus 3

 

[HallsofIvy]

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 R₄[x] is the vector space of cubic polynomials over the real numbers and that T is a linear transformation from R₄[x] to itself. You are told that T(x)= x², that T(x²+ x³)= x³+ 1, T(x+1)= x+ x², and that T(x²+ x)= x²+ x³. That will be enough to completely determine T provided the four polynomials, x, x²+ x³, x+ 1, and x²+ x form a basis for R₄[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(x²+ x³)+ c(x+ 1)+ d(x²+ x)= 0. If they are independent that should be true only if a= b= c= d= 0. Go ahead and collect "like powers": bx³+ (b+ d)x²+ (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 x³, x², 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 R₄[x]. That means we can write ax³+ bx²+ cx+ d as a linear combination:
ax^3+ bx^2+ cx+ d= \alpha(x)+ \beta(x^2+ x^3)+ \gamma(x+ 1)+ \delta(x^2+ x)
Again multiply that out:
ax^3+ bx^2+ cx+ d= \beta x^3+ (\beta+ \delta)x^2+ (\alpha+ \gamma+ \delta)x+ \gamma
Now that gives \beta= a, \beta+ \delta= a+ \delta= b so \delta= b- a, \gamma= d, \alpha+ \gamma+ \delta= \alpha+ d+ (b-a)= c so \alpha= a- b+ c- d.

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)
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!

 

[transgalactic]

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??

 

[HallsofIvy]

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

 

[HallsofIvy]

As for the second problem, just do it! You already have the matrix for T. For the first part, calculate T²(x²) T(T(x²). Keep multiplying by T until you get x² again. For the second part, multiply to find T², T³, etc. until your multiplication gives you I.

 

[transgalactic]

but after i made some legal operations on the original matrix T
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??

 

[HallsofIvy]

but after i made some legal operations on the original matrix T
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??

In other words, T is invertible! Why are you doing that? I suggested you just go ahead and multiply. It is not important that powers of T can be row reduced to the identity. Any matrix whose determinant is not 0 can be row reduced to the identity matrix. If T can be row reduced to the identity matrix, then so can all of its powers. Remember when I said, "If all you have is a hammer, every problem looks like a nail"? If all you know is "row-reduction", you will apply that to every problem- whether it works or not!

You calculated that
T= \left(\begin{array}{cccc}0 & 1 & 0 & 0 \\ -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\end{array}\right)
You asked, first, to apply that to 'x² which, in the standard basis for P₄ is 0x³+ 1x²+ 0x + 0=
\left(\begin{array}{c} 0 \\ 1 \\ 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)\left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right)= \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0\end{array}\right)
That's not 0 so do it again:
\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)
Again, that's not 0 so do it again:
\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)
Keep doing that until you get [0, 1, 0, 0]^T again.

To find an n such that Tⁿ= I (and so Tⁿx= x for all x), do the matrix multiplication:
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)

T³ will be T times that. Keep multiplying until you get I.