Let w - T- w in {mathbb R}^3 that is not in the image of T(x1- x2-.docx
Let w = T. w in {mathbb R}^3 that is not in the image of T(x1, x2, x3 )= (x1- x2, x2- x3, x3-x1). Find a vector T: {mathbb R}^3 rightarrow {mathbb R}^3 be the linear transformation defined by Solution Note how the sum of the three coordinates of the transformation is 0 - x 1 - x 2 + x 2 - x 3 + x 3 - x 1 = 0. Thus, it is easy to see that the range of this transformation is x 1 + x 2 + x 3 = 0 (the fact that the first two vectors [1 -1 0] and [0 1 -1] are linearly independent assures that the solution will be at least 2 dimensional). Pick any values of x 1 + x 2 + x 3 not equal to 0, and it will not be in the transformation. I select [1 0 0] T We can actually verify this. If (x 1 - x 2 , x 2 - x 3 , x 3 - x 1 ) does include (1, 0, 0) then x 2 - x 3 = 0, or x 2 = x 3 Then, x 3 - x 1 , so, substituting  x 2 = x 3 , x 2 - x 1 = 0, or x 1 - x 2 = 0 Yet, the first coordinate, x 1 - x 2 , = 1, so it can not equal 0. This is a contradiction. Thus, it is not in the range. .
Let w - T- w in {mathbb R}^3 that is not in the image of T(x1- x2-.docx
Let w = T. w in {mathbb R}^3 that is not in the image of T(x1, x2, x3 )= (x1- x2, x2- x3, x3-x1). Find a vector T: {mathbb R}^3 rightarrow {mathbb R}^3 be the linear transformation defined by Solution Note how the sum of the three coordinates of the transformation is 0 - x 1 - x 2 + x 2 - x 3 + x 3 - x 1 = 0. Thus, it is easy to see that the range of this transformation is x 1 + x 2 + x 3 = 0 (the fact that the first two vectors [1 -1 0] and [0 1 -1] are linearly independent assures that the solution will be at least 2 dimensional). Pick any values of x 1 + x 2 + x 3 not equal to 0, and it will not be in the transformation. I select [1 0 0] T We can actually verify this. If (x 1 - x 2 , x 2 - x 3 , x 3 - x 1 ) does include (1, 0, 0) then x 2 - x 3 = 0, or x 2 = x 3 Then, x 3 - x 1 , so, substituting  x 2 = x 3 , x 2 - x 1 = 0, or x 1 - x 2 = 0 Yet, the first coordinate, x 1 - x 2 , = 1, so it can not equal 0. This is a contradiction. Thus, it is not in the range. .
![Let w = T. w in {mathbb R}^3 that is not in the image of T(x1, x2, x3 )= (x1- x2, x2- x3, x3-x1).
Find a vector T: {mathbb R}^3 rightarrow {mathbb R}^3 be the linear transformation defined by
Solution
Note how the sum of the three coordinates of the transformation is 0 -
x 1 - x 2 + x 2 - x 3 + x 3 - x 1 = 0.
Thus, it is easy to see that the range of this transformation is x 1 + x 2 + x 3 = 0 (the fact that the
first two vectors [1 -1 0] and [0 1 -1] are linearly independent assures that the solution will be at
least 2 dimensional).
Pick any values of x 1 + x 2 + x 3 not equal to 0, and it will not be in the transformation.
I select [1 0 0] T
We can actually verify this. If (x 1 - x 2 , x 2 - x 3 , x 3 - x 1 ) does include (1, 0, 0) then x 2 - x 3 = 0,
or x 2 = x 3
Then, x 3 - x 1 , so, substituting  x 2 = x 3 , x 2 - x 1 = 0, or x 1 - x 2 = 0 Yet, the first coordinate, x
1 - x 2 , = 1, so it can not equal 0. This is a contradiction. Thus, it is not in the range.](https://image.slidesharecdn.com/letw-t-winmathbbr3thatisnotintheimageoftx1-x2-230204030507-b2437494/75/let-w-t-w-in-mathbb-r3-that-is-not-in-the-image-of-tx1-x2docx-1-2048.jpg?cb=1675481390)
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Let w - T- w in {mathbb R}^3 that is not in the image of T(x1- x2-.docx
- 1. Let w = T. w in {mathbb R}^3 that is not in the image of T(x1, x2, x3 )= (x1- x2, x2- x3, x3-x1). Find a vector T: {mathbb R}^3 rightarrow {mathbb R}^3 be the linear transformation defined by Solution Note how the sum of the three coordinates of the transformation is 0 - x 1 - x 2 + x 2 - x 3 + x 3 - x 1 = 0. Thus, it is easy to see that the range of this transformation is x 1 + x 2 + x 3 = 0 (the fact that the first two vectors [1 -1 0] and [0 1 -1] are linearly independent assures that the solution will be at least 2 dimensional). Pick any values of x 1 + x 2 + x 3 not equal to 0, and it will not be in the transformation. I select [1 0 0] T We can actually verify this. If (x 1 - x 2 , x 2 - x 3 , x 3 - x 1 ) does include (1, 0, 0) then x 2 - x 3 = 0, or x 2 = x 3 Then, x 3 - x 1 , so, substituting  x 2 = x 3 , x 2 - x 1 = 0, or x 1 - x 2 = 0 Yet, the first coordinate, x 1 - x 2 , = 1, so it can not equal 0. This is a contradiction. Thus, it is not in the range.