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The definition of one-to-one was pretty straight forward. If the vectors are lin.indep the transformation would be one-to-one. But could it also be onto?

The definition of onto was a little more abstract. Would a zero-row in reduced echelon form have any effect on this? I just assumed that because it has a couple of free variables it would be onto, but that zero-row set me off a bit. Say if the matrix is 4 by 5. With two free variables. And the zero-row in echelon form.

(Sorry for not posting the given matrix, but that is to specific. And I don't want to get a ban from uni for asking online. The task is determine the onto/one-to-one of to matrices)

I'll check back after class and update the question if more information is desirable.

Update

The definitions in the book is this; Onto: $T: \mathbb R^n \to \mathbb R^m $ is said to be onto $\mathbb R^m $ if each b in $\mathbb R^m $ is the image of at least one x in $\mathbb R^n $

One-to-one: $T: \mathbb R^n \to \mathbb R^m $ is said to be one-to-one $\mathbb R^m $ if each b in $\mathbb R^m $ is the image of at most one x in $\mathbb R^n $

And then, there is another theorem that states that a linear transformation is one-to-one iff the equation T(x) = 0 has only the trivial solution.

That doesn't say anything about onto.

Then there is this bit that confused be about onto: Let $T: \mathbb R^n \to \mathbb R^m $ be a linear transformation and let A be the standard matrix for T. Then: T maps $T: \mathbb R^n$ onto $\mathbb R^m $ iff the columns of A span $\mathbb R^m $.

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What is your definition of an onto matrix? Full (row) rank? –  wildildildlife Mar 11 '11 at 11:00
    
I'll update when I get home :) –  Algific Mar 11 '11 at 11:19
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3 Answers 3

up vote 21 down vote accepted

"One-to-one" and "onto" are properties of functions in general, not just linear transformations.

Definition. Let $f\colon X\to Y$ be a function.

  • $f$ is one-to-one if and only if for every $y\in Y$ there is at most one $x\in X$ such that $f(x)=y$; equivalently, if and only if $f(x_1)=f(x_2)$ implies $x_1=x_2$.
  • $f$ is onto (or onto $Y$, if the codomain is not clear from context) if and only if for every $y\in Y$ there at least one $x\in X$ such that $f(x)=y$.

This definition applies to linear transformations as well, and in particular for linear transformations $T\colon \mathbb{R}^n\to\mathbb{R}^m$, and by extension to matrices, since an $m\times n$ matrix $A$ can be identified with the linear transformation $L_A\colon\mathbb{R}^n\to\mathbb{R}^m$ given by $L_A(\mathbf{x}) = A\mathbf{x}$.

So, the definitions are for any functions. But when our sets $X$ and $Y$ have more structure to them, and the functions are not arbitrary, but special kinds of functions, we can often obtain other ways of characterizing a function as one-to-one or onto which is easier/better/more useful/more conceptual/has interesting applications. This is indeed the case when we have such a rich structure as linear transformations and vector spaces.

One-to-one is probably the easiest; this is because whether a function is one-to-one depends only on its domain, and not on its codomain. By contrast, whether a function is onto depends on both on the domain and the codomain (so, for instance, $f(x)=x^2$ is onto if we think of it as a function $f\colon\mathbb{R}\to[0,\infty)$, but not if we think of it as a function $f\colon\mathbb{R}\to\mathbb{R}$, or $f\colon[2,\infty)\to[0,\infty)$).

Theorem. Let $T\colon\mathbb{R}^n\to\mathbb{R}^m$ be a linear transformation. The following are equivalent:

  1. $T$ is one-to-one.
  2. $T(\mathbf{x})=\mathbf{0}$ has only the trivial solution $\mathbf{x}=\mathbf{0}$.
  3. If $A$ is the standard matrix of $T$, then the columns of $A$ are linearly independent.

Proof. The equivalence of (1) and (2) is basic in linear algebra, so let's deal with that:

(1)$\Rightarrow$(2): If $T$ is one-to-one, then for all $\mathbf{x}$, since $T(\mathbf{0})=\mathbf{0}$ (being linear), then $T(\mathbf{x})=\mathbf{0}=T(\mathbf{0})$ implies $\mathbf{x}=\mathbf{0}$; this proves (2).

(2)$\Rightarrow$(1): Suppose $T(\mathbf{x}_1)=T(\mathbf{x}_2)$. Then $$\mathbf{0} = T(\mathbf{x}_1) - T(\mathbf{x}_2) = T(\mathbf{x}_1-\mathbf{x}_2),$$ since $T$ is linear; because we are assuming (2), $T(\mathbf{x}_1-\mathbf{x}_2)=\mathbf{0}$ implies that $\mathbf{x}_1-\mathbf{x}_2 = \mathbf{0}$, so $\mathbf{x}_1=\mathbf{x}_2$, which proves that $T$ is indeed one-to-one.

The key to the connection with (3) (and eventually to your confusion) is that multiplying a matrix by a vector can be seen as an operation on columns. If $$A=\left(\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{array}\right),$$ then let the columns of $A$, $A_1$, $A_2,\ldots,A_n$ be: $$A_1 = \left(\begin{array}{c}a_{11}\\a_{21}\\ \vdots\\a_{m1}\end{array}\right),\quad A_2 = \left(\begin{array}{c}a_{12}\\a_{22}\\ \vdots\\ a_{m2}\end{array}\right),\quad\ldots,A_n = \left(\begin{array}{c}a_{1n}\\a_{2n}\\ \vdots\\ a_{mn}\end{array}\right).$$ Then we have the following: $$A\left(\begin{array}{c}x_1\\x_2\\ \vdots\\x_n\end{array}\right) = x_1 A_1 + a_2 A_2 + \cdots + a_nA_n.$$ That is, multiplying $A$ by $\mathbf{x}$ gives a linear combination of the columns of $A$. This gives the direct connection we need between conditions (1) and (2), and condition (3).

(2)$\Rightarrow$(3): To show that the columns of $A$ are linearly independent, we need to show that if $\alpha_1A_1 + \cdots + \alpha_nA_n = \mathbf{0}$, then $\alpha_1=\cdots=\alpha_n=0$. So suppose $\alpha_1A_1+\cdots+\alpha_nA_n = \mathbf{0}$. Then $$T(\mathbf{\alpha}) = A\mathbf{\alpha} = \alpha_1A_1+\cdots+\alpha_nA_n = \mathbf{0},\qquad\text{where }\mathbf{\alpha}=\left(\begin{array}{c}\alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n\end{array}\right).$$ Because we are assuming (2), that means that from $T(\mathbf{\alpha}) = \mathbf{0}$ we can conclude taht $\alpha=\mathbf{0}$; therefore, $\alpha_1=\cdots=\alpha_n=0$. This proves that $A_1,\ldots,A_n$ are linearly independent.

(3)$\Rightarrow$(2): Suppose the columns of $A$ are linearly independent, and $$\mathbf{0} = T(\mathbf{x}) = A\mathbf{x}\quad\text{where }\mathbf{x}=\left(\begin{array}{c}a_1\\a_2\\ \vdots\\ a_n\end{array}\right).$$ This means that $a_1A_1 + \cdots a_nA_n = \mathbf{0}$; since the columns of $A$ are assumed to be linearly independent, we conclude that $a_1=\cdots=a_n=0$, so $\mathbf{x}=\mathbf{0}$, proving (2). QED

What about onto? There are two things here. One is a theorem similar to the one above; the other is the Rank-Nullity Theorem.

Theorem. Let $T\colon\mathbb{R}^n\to\mathbb{R}^m$ be a linear transformation. The following are equivalent:

  1. $T$ is onto.
  2. The equation $T(\mathbf{x})=\mathbf{b}$ has solutions for every $\mathbf{b}\in\mathbb{R}^m$.
  3. If $A$ is the standard matrix of $T$, then the columns of $A$ span $\mathbb{R}^m$. That is: every $\mathbf{b}\in\mathbb{R}^m$ is a linear combination of the columns of $A$.

Proof. (1)$\Leftrightarrow$(2) is essentially the definition, only cast in terms of equations for the sake of similarity to the previous theorem.

(2)$\Rightarrow$(3) Let $\mathbf{b}\in\mathbb{R}^m$. Then by (2) there exists an $\mathbf{a}\in\mathbb{R}^n$ such that $T(\mathbf{a})=\mathbf{b}$. We have: $$\mathbf{b} = T(\mathbf{a}) = A\mathbf{a} = A\left(\begin{array}{c}a_1\\a_2\\ \vdots\\a_n\end{array}\right) = a_1A_1 + a_2A_2 + \cdots + a_nA_n.$$ That is, we can express $\mathbf{b}$ as a linear combination of the columns of $A$. Since $\mathbf{b}$ is arbitrary, every vector in $\mathbb{R}^m$ can be expressed as a linear combination of the columns of $A$, so the columns of $A$ span $\mathbb{R}^m$; this proves (3).

(3)$\Rightarrow$(2) Suppose the columns of $A$ span $\mathbb{R}^m$ and let $\mathbf{b}\in\mathbb{R}^m$. We want to show that $T(\mathbf{x})=\mathbf{b}$ has at least one solution.

Since the columns of $A$ span $\mathbb{R}^m$, there exist scalars $\alpha_1,\ldots,\alpha_n$ such that $$\mathbf{b} = \alpha_1 A_1 + \cdots + \alpha_n A_n = A\left(\begin{array}{c}\alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n\end{array}\right) = T(\mathbf{\alpha}).$$ So $\mathbf{\alpha})$, where $$\mathbf{\alpha} = \left(\begin{array}{c}\alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n\end{array}\right),$$ is a solution to $T(\mathbf{x})=\mathbf{b}$. This establishes (2). QED

So: "one-to-one"-ness is related to linear independence; "onto"-ness is related to spanning properties. Note that linear independence is an intrinsic property (it depends only on the set of vectors), whereas spanning is an extrinsic properties (it depends also on the space we are considering; it is contextual). This matches the fact that whether a function is one-to-one or not depends only on the domain, but whether it is onto depends on both the domain and the codomain of the function.

But there is a deep connection between the two. Remember the following:

Definition. Let $A$ be an $m\times n$ matrix. The nullity of $A$, $\mathrm{nullity}(A)$, is the dimension of the kernel of $A$, that is, of the subspace of $\mathbb{R}^n$ given by $$\mathrm{ker}(A) = \Bigl\{ \mathbf{x}\in\mathbb{R}^n\Bigm| A\mathbf{x}=\mathbf{0}\Bigr\}.$$ The rank of $A$, $\mathrm{rank}(A)$ is the dimension of the image of $A$; that is, of the subspace of $\mathbb{R}^m$ given by \begin{align*} \mathrm{Im}(A) &= \Bigl\{ \mathbf{b}\in\mathbb{R}^m\Bigm| A\mathbf{x}=\mathbf{b}\text{ has at least one solution}\Bigr\}\\ &= \Bigl\{ A(\mathbf{x})\Bigm|\mathbf{x}\in\mathbb{R}^n\Bigr\}. \end{align*}

The deep connection between them is given by the Rank-Nullity Theorem:

Rank-Nullity Theorem. Let $A$ be an $m\times n$ matrix. Then $$\mathrm{rank}(A) + \mathrm{nullity}(A) = n.$$

Now we get two more equivalences for one-to-one and onto:

Theorem. Let $T\colon\mathbb{R}^n\to\mathbb{R^m}$ be a linear transformation. The following are equivalent:

  1. $T$ is one-to-one.
  2. *The equation $T(\mathbf{x})=\mathbf{0}$ has only the trivial solution $\mathbf{x}=\mathbf{0}$.
  3. If $A$ is the standard matrix of $T$, then the columns of $A$ are linearly independent.
  4. $\mathrm{ker}(A) = \{\mathbf{0}\}$.
  5. $\mathrm{nullity}(A) = 0$.
  6. $\mathrm{rank}(A) = n$.

Proof. The equivalence of (4) and (5) follows because only the trivial subspace has dimension $0$; the equivalence of (4) and (2) follows by definition of the kernel. The equivalence of (5) and (6) follows from the Rank-Nullity Theorem, since $n = \mathrm{nullity}(A)+\mathrm{rank}(A)$, so $\mathrm{nullity}(A) = 0$ if and only if $\mathrm{rank}(A) = n$. Since we already know (1), (2), and (3) are equivalent, the result follows. QED

Theorem. Let $T\colon\mathbb{R}^n\to\mathbb{R}^m$ be a linear transformation. The following are equivalent:

  1. $T$ is onto.
  2. The equation $T(\mathbf{x})=\mathbf{b}$ has solutions for every $\mathbf{b}\in\mathbb{R}^m$.
  3. If $A$ is the standard matrix of $T$, then the columns of $A$ span $\mathbb{R}^m$. That is: every $\mathbf{b}\in\mathbb{R}^m$ is a linear combination of the columns of $A$.
  4. $\mathrm{Im}(A) = \mathbb{R}^m$.
  5. $\mathrm{rank}(A) = m$.
  6. $\mathrm{nullity}(A) = n-m$.

Proof. We already know that (1), (2), and (3) are equivalent. The equivalence of (4) and (2) follows by the definition of the image. The equivalence of (4) and (5) follows because the only subspace of $\mathbb{R}^m$ that has dimension $m$ is the whole space. Finally, the equivalence of (5) and (6) follows from the rank nullity theorem: since $n = \mathrm{rank}(A)+\mathrm{nullity}(A)$, then $\mathrm{nullity}(A) = n - \mathrm{rank}(A)$. So the rank equals $m$ if and only if the nullity equals $n-m$. QED

So now you have a whole bunch of ways of checking if a matrix is one-to-one, and of checking if a matrix is onto. None of them is "better" than the others: for some matrices, one will be easier to check, for other matrices, it may be a different one which is easy to check. Also, the rank of a matrix is closely related to its row-echelon form, so that might help as well.

Note a few things: generally, "onto" and "one-to-one" are independent of one another. You can have a matrix be onto but not one-to-one; or be one-to-one but not onto; or be both; or be neither. The Rank-Nullity Theorem does place some restrictions: if $A$ is $m\times n$ and $m\lt n$, then the matrix cannot be onto (because $1\leq\mathrm{rank}(A)\leq m$, so if $\mathrm{rank}(A)+\mathrm{nullity}(A) = n$, we must have $\mathrm{nullity}(A)\gt 0$); dually, if $m\gt n$ then $A$ cannot be onto. In particular, the only matrices that can be both one-to-one and onto are square matrices. On the other hand, you can have an $m\times n$ matrix with $m\lt n$ that is onto, or one that is not onto. And you can have $m\times n$ matrices with $m\gt n$ that are one-to-one, and matrices that are not one-to-one.

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Sorry if this is a stupid question, but is it always the case that whether a function is one-to-one depends only on the domain? For instance with the example $f(x) = x^2$ if the codomain were $0$, wouldn't the function still be one-to-one? –  ghshtalt Mar 11 '11 at 18:13
    
@ghshtalt: For $f(x)=x^2$ to have codomain $0$, you would need the domain to be just $0$ (otherwise, you wouldn't have a function). But of course, if the domain of a function is a single point, then the function is one-to-one, regardless of what the codomain (or what the function) is. So in fact, it's because the domain is a single point that such a function would be one-to-one. –  Arturo Magidin Mar 11 '11 at 18:19
    
oh ok, so it wouldn't have been a function in the first place... Thanks for clearing that up for me! And thanks for another great answer of course! –  ghshtalt Mar 11 '11 at 18:22
    
I believe I have an example that is not one-to-one nor onto. The matrix is lin.dep (free variables), and for a random value in the codomain the reduced augmented matrix is inconsistent. Hence there is not one value in the domain that corresponds to that value in the codomain. Right, right? –  Algific Mar 13 '11 at 17:26
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@Algific: Matrices by themselves are nor "linearly independent" or "linearly dependent". Sets of vectors are linearly independent or linearly dependent. If you mean that you have a matrix whose columns are linearly dependent (and somehow relating that to "free variables", yet another concept that is not directly applicable to matrices, but rather to other things to which you can associate matrices), then it is indeed true that any matrix whose columns are linearly dependent gives a linear transformation that is not one-to-one. If not all systems Ax=b have solutions, then not onto. –  Arturo Magidin Mar 13 '11 at 20:28
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A function is onto if its codomain is equal to its image, that is, $f: X \to Y$ is onto if $f(X) = Y$.

Assuming you are working with real matrices, a matrix $A$ is usually seen as mapping vectors from $\mathbb R^n$ to $\mathbb R^m$. You want to show that for all $\vec y \in \mathbb R^m$ there exists a $\vec x \in \mathbb R^n$ such that $A \vec x = \vec y$. In some cases this is not possible depending on $m$ and $n$. How does $m$ and $n$ relate to the size of the matrix?

One way to check that a matrix is onto is to check if you can find vectors $\vec v_1, \vec v_2, \dots, \vec v_m$ such that $A\vec v_i = \vec e_i$ for $i = 1, 2, \dots, m$, where $\vec e_1, \vec e_2, \dots, \vec e_m$ is the standard basis in $\mathbb R^m$ (why?). Can you use the echelon form for this? Do you actually need to get the standard basis or is some other set of vectors ok?

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The question is sort of an open one, it says discuss the concepts of one-to-one and onto with regard to the two following matrices. –  Algific Mar 11 '11 at 12:11
    
Well, to start of, check if they are one-to-one and onto. Does a matrix have to be one-to-one to be onto? Does it have to be onto to be one-to-one? –  Calle Mar 11 '11 at 12:17
    
The problem is that the theorems doesn't mention the other state. Funny enough. –  Algific Mar 11 '11 at 16:06
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let $T(x)=Ax$ be a linear transformation.

$T(x)$ is one-to-one if the columns of $A$ are linearly independent.

$T(x)$ is onto if every row of $A$ has a pivot.

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