How do I calculate the Basis for Im(T)?

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The basis for the image of a linear transformation T, represented by the matrix T = [[1, 0], [0, 1], [0, 0]], can be calculated by identifying the linearly independent columns of the matrix. The image of T maps R² to R³, resulting in vectors of the form (x, y, 0), where z must equal 0. The basis for Im(T) is the set of vectors {(1, 0, 0), (0, 1, 0)}, which are the columns of the matrix. It is crucial to represent the basis as a set of vectors rather than as a matrix.

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How do I calculate the Basis for Im(T)? I am having troubles finding an example that will best fir here. I know that the I=diagonal matrix with all of all of the i=j entries being 1. Beyond that I am rather confused and don't know where I need to start.
 
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Badger33 said:
How do I calculate the Basis for Im(T)? I am having troubles finding an example that will best fir here. I know that the I=diagonal matrix with all of all of the i=j entries being 1. Beyond that I am rather confused and don't know where I need to start.

This is a confusing question. If you are simply dealing with a situation where T is a matrix, the image is the column space. In other words, it is all vectors that are linearly independent in the columns.

In general, however, the image of a transformation is defined to be the set \left \{ v\in W |T(x)=v, x\in V \right \}where V and W are vector spaces over a field. You need to be more precise as to what exactly you want.
 


So to calculate the basis for the Im(T). I would gather all the Linear Independent parts and put them into a basis?
if I got:
[1] [0]
[0] [1]
[0], [0]

then my basis would be:
[1 0]
[0 1]
[0 0]

Or am I way off on this?
 
Last edited:


From a more fundamental viewpoint, if
T= \begin{bmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}
Then T maps R^2 to R^3. (x, y, z) will be in the image of T if and only if there exist (a, b) such that
\begin{bmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}= \begin{bmatrix}a \\ b \\ 0\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}
which gives the equations a= x, b= y, 0= z. Since a and b could be any numbers so can x and y- but z must be 0. That is, any vector in the image of T must be of the form (x, y, 0)= (x, 0, 0)+ (0, y, 0)= x(1, 0, 0)+ y(0, 1, 0). Yes, the image is two-dimensional and a basis is the set of vectors {(1, 0, 0), (0, 1, 0)}, the columns of the matrix you show. However, it is important to remember that a basis is a set of vectors, not a matrix. Even though you used a matrix to calculate it, you should show the basis as a set of vectors, not as a matrix.
 

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