# A question in lenear algebra

1. Jan 29, 2008

### 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 cant understand the solution that was given

i cant understand how they take a vector of a polinomial (1,0,0) for example and transform it into
another vector
???

2. Jan 29, 2008

### HallsofIvy

Staff Emeritus
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:
$$\left[\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}\right]$$

3. Jan 29, 2008

### 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

4. Jan 30, 2008

### HallsofIvy

Staff Emeritus
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.

5. Jan 31, 2008

### 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

Last edited: Jan 31, 2008
6. Jan 31, 2008

### HallsofIvy

Staff Emeritus
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.

7. Jan 31, 2008