How to Understand a Solution to a Linear Algebra Question?

[transgalactic]

i added a link in which i show a question and a solution to it

http://img163.imageshack.us/my.php?image=img8227nu6.jpg

i can't understand the solution that was given

i can't understand how they take a vector of a polinomial (1,0,0) for example and transform it into
another vector
?

 

[HallsofIvy]

You are given a linear transformation T:R³[x]->R³ defined by
$$T(f)= \left[\begin{array}{c}f(0)- f(1) \\ f(0) \\ f(1)\end{array}\right]$$

R³[x] is, of course, the space of all quadratic polynomials in the variable x, a+ bx+ cx². Using basis {1, x, x²} such a polynomial can be written a(1)+ b(x)+ c(x²= [a, b, c]. R³ is the space of all ordered triples of real numbers, (a, b, c). Using basis {(1, 0, 0), (0, 1, 0), (0, 0, 1)} such a triple can be written a(1, 0, 0)+ b(0, 1, 0)+ c(0, 0, 1)= [a, b, c].

That is,T applied to the polynomial corresponding to f= (1, 0, 0)= 1 +0x+ 0x²= 1, is given by, since f(0)= f(1)= 1, T(f)= [f(1)- f(0), f(0), f(1)]= [1-1, 1, 1]= [0, 1, 1].

T applied to the polynomial corresponding to f= (0, 1, 0)= 0+ 1x+ 0x²= x, is given by, since f(0)= 0, f(1)= 1, T(f)= [f(1)- f(0), f(0), f(1)]= [1- 0, 0, 1]= [1, 0, 1].

T applied to the polynomial corresponding to f= (0, 0, 1)= 0+ 0x+ 1x²= x² is given by, since f(0)= 0, f(1)= 1, T(f)= [f(1)- f(0), f(0), f(1)]= [1- 0, 0, 1]= [1, 0, 1].

The matrix of T, in those bases is the matrix having those as columns:
$$\left[\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}\right]$$

 

[transgalactic]

why is f(0)= f(1)= 1 in the first

why is f(0)= 0, f(1)= 1 in the second

why is f(0)= 0, f(1)= 1 in the 3rd

?

it was not given
it was only my conclusion for this solution that the values of these variable are needed
to be like that in order to get to their result

 

[HallsofIvy]

You are applying T to a function of the form f(x)= a+ bx+ cx². In other words, "f" is just the general quadratic, not a specific function. Since we were asked to find a matrix representation of that function in the "standard" basis: e₁= x², e₂= x, e₃= 1, we apply the transformation to each of those functions in turn.

Te₁ is T applied to the function f(x)= x². For that function, f(0)= 0²= 0 and f(1)= 1²= 1.

Te₂ is T applied to the function e₂= x. For that function f(0)= 0 and f(1)= 1.

Te₃ is T applied to the function e₃= 1. For that function, f(0)= 1 and f(1)= 1.

 

[transgalactic]

the bases are the temporary functions??

and we put 0 and one in them each time??

for the base e1=x^2
f(x)=x^2 ==> f(0)=0 f(1)=1

thanks

 

[HallsofIvy]

Are you clear on what a "basis" (not "base") is? It is a collection of vectors so that every vector can be written as a linear combination of them in one way only.

"ex= x^2" is member of the given basis, not a "base" itself. Yes, in this problem, you "put 0 and one in them" because that is what the problem told you to do. The linear transformation was originally defined as:
$$T(f)= \left[\begin{array}{c}f(0)- f(1) \\ f(0) \\ f(1)\end{array}\right]$$

In general, if you are given a linear transformation, T, from a vector space U to a vector space V, you can represent T as a matrix for given bases of U and V (important: the matrix depends on the particular choice of bases, not only on T). A good way to do that is to apply the linear transformation to each of the basis vectors of U in turn, writing the result in terms of basis V. The numbers involved in that linear combination form a column of the matrix.

Take a very abstract, general, example. T goes from vector space U to vector space V. $\{u_1, u_2, u_3\}$ is a basis for U, $\{v_1, v_2, v_4\}$ is a basis for V (so U is 3 dimensional and V is 4 dimensional). $Tu_1$ is, of course, some vector in V and so can be written in terms of $\{v_1, v_2, v_4\}$. Suppose $Tu_1= av_1+ bv_2+ cv_3+ dv_4$. Then the first column of the matrix representing T (in those bases) is a, b, c, d.

The point is that once we have chosen a basis, we represent each vector by the numbers multiplying each basis vector, not writing the basis vectors themselves (which are "understood"). That is, $u_1$ itself is written as just (1, 0, 0, 0) because it is $1*u_1+ 0*u_2+ 0*u_3$ while $Tu_1= av_1+ bv_2+ cv_3+ dv_4$ would be written as (a, b, c, d). Applying T to $u_1$ would be the same as multiplying the matrix representing it by the column vector (1, 0, 0) and the result must be the column (a, b, c, d). Of course
$$\left[\begin{array}{ccc} a & * & * \\ b & * & *\\ c & * & * \\ d & * & * \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ 0 \\\end{array}\right]= \left[\begin{array}{c} a \\ b \\ c \\ d \end{array}\right]$$
where the "*" are other numbers determined by applying T to the other basis vectors.

 

[transgalactic]

i solved it again
in the link i presented the logic of the solution

is the logic of the solution ok??

http://img215.imageshack.us/my.php?image=img8231yd3.jpg