# Is this the right way to solve these questions

1. Mar 1, 2008

2. Mar 1, 2008

### rocomath

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

3. Mar 1, 2008

### transgalactic

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

i am light years away from calculus 3

4. Mar 1, 2008

### HallsofIvy

Staff Emeritus
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!

Last edited: Mar 1, 2008
5. Mar 1, 2008

### 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??

Last edited: Mar 1, 2008
6. Mar 1, 2008

### HallsofIvy

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

7. Mar 1, 2008

### HallsofIvy

Staff Emeritus
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.

8. Mar 2, 2008

### 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 know that i am wrong in solving it that
way

what is the right way??

9. Mar 2, 2008

### HallsofIvy

Staff Emeritus
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.