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Math Help - linear transformation.

  1. #1
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    linear transformation.

    Hi


    I attached my question.


    I do not have any idea for the question.

    Can anyone help?
    Attached Thumbnails Attached Thumbnails linear transformation.-11.jpg  
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  2. #2
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    First, a little theory:
    Let T be a linear transformation from Vector space U to vector space V. It can be written as a matrix by choosing specific bases for U and V. In particular, if \{ u_1, u_2\} is a basis for U and \{ v_1, v_2, v_3\} is a basis for V, then every vector in U can be written as " au_1+ bu_2" for some numbers a and b, and every vector in V can be written as " xv_1+ yv_2+ zv_3" (I have assumed in this example that U has dimension 2 and V has dimension 3 because that is the case in your problem.)

    When we have agreed to use a particular ordered basis, we can drop the basis vectors themselves and and write au_1+ bu_2 as (a, b). Similarly, we can write xv_1+ yv_2+ zv_3 as (x, y, z). That is how we can then write the linear transformation as a matrix: T is applied to members of U which can be represented by two numbers so the matrix for T must have two columns. The result is a vector in V which can be represented by 3 numbers so the matrix for T must have 3 rows. That much tells us it must be of the form \begin{bmatrix}A & B \\ C & D\\ E & F\end{bmatrix}.

    Now, a very nice thing happens! The basis vectors pf U, themselves, u_1 and u_2 are equal to 1u_1+ 0u_2 and 0u_1+ 1u_2 and so are represented by (1, 0) and (0, 1). Look what happens when we multiply our matrix by those:
    \begin{bmatrix}A & B \\ C & D\\ E & F\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}A \\ C\\ E\end{bmatrix}
    and
    \begin{bmatrix}A & B \\ C & D\\ E & F\end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}B \\ D\\ F\end{bmatrix}

    That is, applying the linear transformation to the basis vectors gives us the columns of the matrix in turn!

    If T is a linear transformation from U to V, and we are given bases for U and V, we find the matrix representing T by applying T to each basis vector of U in turn, writing the result in terms of the basis for V. Each of those gives a column for T.

    Now, for this particular problem, T is from R^2 to R^3 so the matrix form for T will have 2 columns and 3 rows just like the example above. We are told to find the standard matrix which means using the standard bases: {(1,0), (0,1)} for R^2 and {(1,0,0), (0,1,0), (0,0,1)} for R^3.

    Now, what are T(1,0) and T(0,1)? Well, we are told that T(2,-1)= T(2(1,0)-1(0,1))= 2T(1,0)-1T(0,1)= (1,0,2) and that T(-3,4)= T(-3(1,0)+ 4(0,1))= -3T(1,0)+ 4T(0,1)= (3,1,-1).

    Writing A= T(1,0) and B= T(0,1) just to simplify the notation, we have 2A- B= (1,0,2) and -3A+ 4B= (3,1,-1). Mutiply the fist equation by 4 to get 8A- 4B= (4,0,8) and add that to -3A+ 4B= (3,1,-1). The "B" cancel leaving 5A= (7,1,7) so A= (7/5,1/5,7/5). Putting that back into 2A- B= (1,0,2) gives (14/5,2/5,14/5)- B= (1,0,2) so B= (14/5,2/5,14/5)- (1,0,2)= (9/5, 2/5,4/5).

    Now we know that T(1,0)= (7/5,1/5,7/5) and T(0,1)= (9/5,2/5, 4/5), the two columns of the matix, so the matrix representation of T is \begin{bmatrix}7/5 & 9/5 \\ 1/5 & 2/5 \\ 7/5 & 4/5\end{bmatrix}

    (Assuming I haven't made any arithmetic mistakes- better check that!)
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  3. #3
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    thank you

    Quote Originally Posted by HallsofIvy View Post
    First, a little theory:
    Let T be a linear transformation from Vector space U to vector space V. It can be written as a matrix by choosing specific bases for U and V. In particular, if \{ u_1, u_2\} is a basis for U and \{ v_1, v_2, v_3\} is a basis for V, then every vector in U can be written as " au_1+ bu_2" for some numbers a and b, and every vector in V can be written as " xv_1+ yv_2+ zv_3" (I have assumed in this example that U has dimension 2 and V has dimension 3 because that is the case in your problem.)

    When we have agreed to use a particular ordered basis, we can drop the basis vectors themselves and and write au_1+ bu_2 as (a, b). Similarly, we can write xv_1+ yv_2+ zv_3 as (x, y, z). That is how we can then write the linear transformation as a matrix: T is applied to members of U which can be represented by two numbers so the matrix for T must have two columns. The result is a vector in V which can be represented by 3 numbers so the matrix for T must have 3 rows. That much tells us it must be of the form \begin{bmatrix}A & B \\ C & D\\ E & F\end{bmatrix}.

    Now, a very nice thing happens! The basis vectors pf U, themselves, u_1 and u_2 are equal to 1u_1+ 0u_2 and 0u_1+ 1u_2 and so are represented by (1, 0) and (0, 1). Look what happens when we multiply our matrix by those:
    \begin{bmatrix}A & B \\ C & D\\ E & F\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}A \\ C\\ E\end{bmatrix}
    and
    \begin{bmatrix}A & B \\ C & D\\ E & F\end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}B \\ D\\ F\end{bmatrix}

    That is, applying the linear transformation to the basis vectors gives us the columns of the matrix in turn!

    If T is a linear transformation from U to V, and we are given bases for U and V, we find the matrix representing T by applying T to each basis vector of U in turn, writing the result in terms of the basis for V. Each of those gives a column for T.

    Now, for this particular problem, T is from R^2 to R^3 so the matrix form for T will have 2 columns and 3 rows just like the example above. We are told to find the standard matrix which means using the standard bases: {(1,0), (0,1)} for R^2 and {(1,0,0), (0,1,0), (0,0,1)} for R^3.

    Now, what are T(1,0) and T(0,1)? Well, we are told that T(2,-1)= T(2(1,0)-1(0,1))= 2T(1,0)-1T(0,1)= (1,0,2) and that T(-3,4)= T(-3(1,0)+ 4(0,1))= -3T(1,0)+ 4T(0,1)= (3,1,-1).

    Writing A= T(1,0) and B= T(0,1) just to simplify the notation, we have 2A- B= (1,0,2) and -3A+ 4B= (3,1,-1). Mutiply the fist equation by 4 to get 8A- 4B= (4,0,8) and add that to -3A+ 4B= (3,1,-1). The "B" cancel leaving 5A= (7,1,7) so A= (7/5,1/5,7/5). Putting that back into 2A- B= (1,0,2) gives (14/5,2/5,14/5)- B= (1,0,2) so B= (14/5,2/5,14/5)- (1,0,2)= (9/5, 2/5,4/5).

    Now we know that T(1,0)= (7/5,1/5,7/5) and T(0,1)= (9/5,2/5, 4/5), the two columns of the matix, so the matrix representation of T is \begin{bmatrix}7/5 & 9/5 \\ 1/5 & 2/5 \\ 7/5 & 4/5\end{bmatrix}

    (Assuming I haven't made any arithmetic mistakes- better check that!)
    wow. it really helps me.
    So is T one to one?
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