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I'm fuzzy about the relationship of range and the basis of a subspace:

From book, I understand that:

Range is the set of all images of vectors of domain under T. Basis are set of vectors that span the subspace and are also linearly independent.

It seems like as if the basis could be the range(T). Can anyone clarify on that?

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A basis of a vector space $V$ is a collection of vectors which are linearly independent and have span $V$.

The range of a linear transformation $T : V \to W$ is $R(T) = \{T(v) \mid v \in V\}$.

The two concepts, basis and range, are referring to features of different objects, as I have highlighted. Furthermore, a basis is a finite set of vectors, whereas the range of a linear transformation is a subspace which is only finite if it is $\{0\}$.

Example 1: Consider the vector space $\mathbb{R}^3$. A basis for $\mathbb{R}^3$ is $\{(1, 0, 0), (2, 3, 0), (4, 5, 6)\}$.

Example 2 (a): Consider the linear transformation $T_1 : \mathbb{R}^2 \to \mathbb{R}^3$ given by $T_1(x, y) = (x, 0, y)$. The range of the linear transformation $T_1$ is $R(T_1) = \{(x, 0, y) \mid (x, y) \in \mathbb{R}^2\}$.

Example 2 (b): Consider the linear transformation $T_2 : \mathbb{R}^2 \to \mathbb{R}^3$ given by $T_2(x, y) = (x + y, 0, 0)$. The range of the linear transformation $T_2$ is $R(T_2) = \{(x + y, 0, 0) \mid (x, y) \in \mathbb{R}^2\}$.


Added Later: Note that I have altered example 2 (b).

Let me clarify a few things.

1. Suppose you are given a collection of vectors $\{v_1, \dots, v_k\}$ in $\mathbb{R}^n$ and you want to find a basis for $\operatorname{span}\{v_1, \dots, v_k\}$; note, $\{v_1, \dots, v_k\}$ already spans this space, but it may not be linearly independent. To see if the vectors are linearly independent we consider the vector equation

$$c_1v_1 + \dots + c_mv_k = 0.$$

We can rewrite this equation as

$$\left[\begin{array}\ \,\mid & & \,\mid\\ v_1 & \cdots & v_k\\ \,\mid & & \,\mid \end{array}\right]\left[\begin{array}\ c_1\\ \ \vdots\\ c_k \end{array}\right] = 0$$

where the matrix is formed by placing the vectors $v_1, \dots, v_k$ as columns. As you mention in your comment below, by row reducing the matrix and choosing the pivot columns, we obtain a linearly independent set with the same span as $\operatorname{span}\{v_1, \dots, v_m\}$.

2. Now suppose we have a linear transformation $T : V \to W$ and we want to find a basis for the range $R(T)$. By part 1, if we can find a collection of vectors which span $R(T)$, we can extract a basis from it. So how do we find a collection of vectors which span $R(T)$? One usually uses the following result which I encourage you to prove.

Proposition: Let $T : V \to W$ be a linear transformation and suppose $\{v_1, \dots, v_n\}$ is a basis for $V$, then $R(T) = \operatorname{span}\{T(v_1), \dots, T(v_n)\}$.

Note, this does not say that $\{T(v_1), \dots, T(v_n)\}$ is a basis for $R(T)$, only that it spans $R(T)$, but the vectors may not be linearly independent.

Consider the linear transformations in example 2 above. Fix $v_1 = (1, 0)$ and $v_2 = (0, 1)$ as a basis for $\mathbb{R}^2$.

By the Proposition, in example 2 (a) $R(T_1)$ is spanned by $T(v_1) = T_1(1, 0) = (1, 0, 0)$ and $T(v_2) = T_1(0, 1) = (0, 0 , 1)$. In this case, $\{T_1(v_1), T_1(v_2)\}$ is also linearly independent so it forms a basis for $R(T_1)$.

In example 2 (b) we see that $R(T_2)$ is spanned by $T_2(v_1) = T_2(1, 0) = (1, 0, 0)$ and $T_2(v_2) = T_2(0, 1) = (1, 0, 0)$; in this example $\{T_2(v_1), T_2(v_2)\}$ is not linearly independent, so it does not form a basis for $R(T_2)$. In general, we would then go through the procedure in part 1 to find a basis for $R(T_2)$.


In summary, the process of finding a basis from a spanning set is the same for an arbitrary subspace as it is for the range of a linear transformation. The highlighted Proposition seems to be where your confusion stems from. Given a basis for $V$, you get a spanning set for $R(T)$ at which point the process of finding a basis reduces to the method in part 1.

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  • $\begingroup$ ok for example 2 is it correct to say the span of basis of R^2 is the R(T)? $\endgroup$ – electronicsSS Aug 18 '14 at 3:23
  • $\begingroup$ An example in the book makes me confused: Suppose T: R^2 -> R^3 then T([x1,x2]) = [ x1 - 2x2, -3x1 + 6x2, 2x1 - 4x2]. Then, the range(T) = span{[1 , -3, 2]}. (all are Column vectors) $\endgroup$ – electronicsSS Aug 18 '14 at 3:25
  • $\begingroup$ In example 2 there is no basis. If you were to choose a basis for $\mathbb{R}^2$ its span would be $\mathbb{R}^2$ (that's part of the definition of what it means to be a basis), it would not be $R(T)$. I have added a second part to example 2 to give more contrast. $\endgroup$ – Michael Albanese Aug 18 '14 at 3:28
  • $\begingroup$ What confuses me is that the method used to find range and basis is similar. Row reduce the matrix and the columns that have leading variables say are v1 v2. Then the basis are {[v1],[v2]} and the range(T) is the span{v1,v2}. Am I missing something here? Suppose T: R^2 -> R^3 then T([x1,x2]) = [ x1 - 2x2, -3x1 + 6x2, 2x1 - 4x2]. Then, the range(T) = span{[1 , -3, 2]}. (all are Column vectors) $\endgroup$ – electronicsSS Aug 18 '14 at 3:50

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