Math Help - linear transformation.

1. linear transformation.

Hi

I attached my question.

I do not have any idea for the question.

Can anyone help?

2. 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!)

3. thank you

Originally Posted by HallsofIvy
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?