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 January 12th, 2013, 05:32 AM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Linear Transformation, kernel, image, and bases Hello everyone, I need help with 1 question, I am fairly new to this so please explain your answers in a way that i can understand T is a linear transformation from R3 to R3 that is defined like this: T(x,y,z) = (2x-y+z,x+3z,x-y-2z) 1) Find the Matrix M of the linear transformation by the base [(1,0,0) , (0,1,0) , (0,0,1)] [I think it means I need to find a matrix M so that M*(x,y,z)=T(x,y,z)] 2) Find the matrix N of the linear transformation by the base of [(1,1,0) , (-1,1,0) , (0,0,-1)] 3) Find bases for Ker(T) and Im(T) Please help!
January 15th, 2013, 02:57 PM   #2
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Re: Linear Transformation, kernel, image, and bases

Quote:
 Originally Posted by OriaG Hello everyone, I need help with 1 question, I am fairly new to this so please explain your answers in a way that i can understand T is a linear transformation from R3 to R3 that is defined like this: T(x,y,z) = (2x-y+z,x+3z,x-y-2z) 1) Find the Matrix M of the linear transformation by the base [(1,0,0) , (0,1,0) , (0,0,1)] [I think it means I need to find a matrix M so that M*(x,y,z)=T(x,y,z)] 2) Find the matrix N of the linear transformation by the base of [(1,1,0) , (-1,1,0) , (0,0,-1)] 3) Find bases for Ker(T) and Im(T) Please help!
Since you are given a problem like this, I assume you are taking a Linear Algebra class! So you should at least know the definitions of all these things.

Think about exactly what you are asked. What you want in both 1) and 2) is 3 by 3 matrix so a matrix of the form
$\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$
Now look at what you get if you multiply that matrix by the vectors given:
$\begin{bmatrix}a=&b=&c \\ d=&e=&f \\ g=&h=&i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ d \\ g\end{bmatrix}$
So what does T map (1, 0, 0) to? What must a, d, and g be?

Do the same with (0, 1, 0) and (0, 0, 1)

For 2) do the same kind of thing what is
$\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}$?
What does T map (1, 1, 0) to? That gives you three equations to solve.

For 3) can we presume that you know what "Ker(t)" means? The Kernel of a linear transformation, T, (also called the "null space") is the set of all vectors, v, such that Tv= 0. Here, Tv= T(x,y,z) = (2x-y+z,x+3z,x-y-2z) so v= (x, y, z) is in the kernel if and only if 2x- y+ z= 0, x+ 3z= 0, and x- y- 2z= 0. Solve those three equation. Generally, three equations in three unknown values will have a single solution. Obviously, one solution is x= y= z= 0. If that is not the only solution then these equations are not independent. If two equations are independent and the third dependent on one of those two, you can solve for two of the unknowns in terms of the other two. That is, you might have z= x+ 3y (that is NOT the equation you get here- I just made that up) Then we could write (x, y, z)= (x, y, x+ 3y)= (x, 0, x)+ (0, y, 3y)= x(1, 0, 1)+ y(0, 1, 3) showing that the space is spanned by (1, 0, 1) and (0, 1, 3). Or, if all three equations are dependent- they are all multiples of each other, we can solve for two, say x and y, in terms of the third, say, z. That is, we might get x= 3z, y= -2z (again, those are NOT from this problem- I just made them up) so (x, y, z)= (3z, -2z, z)= z(3, -2, 1) and the space is spanned by the single vector (3, -2, 1).

For 4) the "image of T" (also called the "range") is the set of all vectors v such that Tu= v for some u. Here we would be looking for, say, (u, v, w) such that u= 2x- y+ z, v= x+ 3z, w= x- y- 2z and we want to find relationships between u, v, and w not involving x, y, and z. For example, if we Subtract twice the second equation from the first we eliminate x: u- 2v= -y- 5z. If we subtract the third equation from the second we also eliminate x: y+ 5z= u- w. Finally, add those two equations to eliminate y: 2u- 2v- w= 0. Oh, wow, we have eliminated both y and z and have exactly a single equation relating u, v, and w: 2u- 2v- w= 0. From that equation, w= 2u- 2v so that any such vector is of the form <u, v, w>= <u, w, 2u- 2v>= <u, 0, 2u>+ <0, v, -2v>= u<1, 0, 2>+ v<0, 1, -2>.

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