# Is this the right way to solve these questions

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

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

i am light years away from calculus 3

HallsofIvy
Homework Helper
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!

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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??

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

HallsofIvy
Homework Helper
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.

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 know that i am wrong in solving it that
way

what is the right way??

HallsofIvy
Homework Helper
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 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.