The Inverse of a Linear Transformation - Revision history

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A '''linear transformation''' is an important concept in mathematics because many real world phenomena can be approximated by linear models.

Unlike a linear function, a linear transformation works on vectors as well as numbers.

== Motivations and definitions ==
Say we have the vector <math>\begin{pmatrix} 1 \\ 0 \end{pmatrix}</math> in <math>\mathbb{R}^2</math>, and we rotate it through 90 degrees, to obtain the vector <math>\begin{pmatrix} 0 \\ 1 \end{pmatrix}</math>.

Another example instead of rotating a vector, we stretch it, so a vector <math>\mathbf{v}</math> becomes <math>2\mathbf{v}</math>, for example. <math>\begin{pmatrix} 2 \\ 3 \end{pmatrix}</math> becomes <math>\begin{pmatrix} 4 \\ 6 \end{pmatrix}</math>

Or, if we look at the ''projection'' of one vector onto the ''x'' axis - extracting its ''x'' component - , e.g. from
<math>\begin{pmatrix} 2 \\ 3 \end{pmatrix}</math> we get <math>\begin{pmatrix} 2 \\ 0 \end{pmatrix}</math>

These examples are all an example of a ''mapping'' between two vectors, and are all linear transformations. If the rule transforming the matrix is called <math>T</math>, we often write <math>T\mathbf{v}</math> for the mapping of the vector <math>\mathbf{v}</math> by the rule <math>T</math>. <math>T</math> is often called the transformation.

Note we do not always write brackets like when we write functions. However we ''should'' write brackets, especially when we want to express the mapping of the sum or the product or the combination of many vectors.

== Definitions ==
===Linear Operators===

Suppose one has a field K, and let x be an element of that field. Let O be a function taking values from K where O(x) is an element of a field J. Define O to be a linear form if and only if:
# O(x+y)=O(x)+O(y)
# O(λx)=λO(x)

===Linear Forms===
Suppose one has a vector space V, and let x be an element of that vector space. Let F be a function taking values from V where F(x) is an element of a field K. Define F to be a linear form if and only if:
# F(x+y)=F(x)+F(y)
# F(λx)=λF(x)

===Linear Transformation===
This time, instead of a field, let us consider functions from one vector space into another vector space. Let T be a function taking values from one vector space V where L(V) are elements of another vector space. Define L to be a linear transformation when it:
# ''preserves scalar multiplication'': T(λ'''x''') = λT'''x'''
# ''preserves addition'': T('''x'''+'''y''') = T'''x''' + T'''y'''

Note that not all transformations are linear. Many simple transformations that are in the real world are also non-linear. Their study is more difficult, and will not be done here. For example, the transformation ''S'' (whose input and output are both vectors in '''R'''2) defined by

<math>S\mathbf{x} = S\begin{pmatrix} x \\
y \end{pmatrix}= \begin{pmatrix} xy\\
\cos(y)\end{pmatrix}</math>

We can learn about nonlinear transformations by studying easier, linear ones.

We often ''describe'' a transformation T in the following way
:<math>T : V \rightarrow W</math>

This means that T, whatever transformation it may be, maps vectors in the vector space V to a vector in the vector space W.

The actual transformation ''could'' be written, for instance, as
:<math>T\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} x + y \\ x - y \end{pmatrix}</math>

== Examples and proofs ==
Here are some examples of some linear transformations. At the same time, let's look at how we can prove that a transformation we may find is linear or not.

=== Projection ===
Let us take the projection of vectors in '''R'''2 to vectors on the ''x''-axis. Let's call this transformation T.

We know that T maps vectors from '''R'''2 to '''R'''2, so we can say
: <math>T: \mathbb{R}^2 \rightarrow \mathbb{R}^2</math>

and we can then write the transformation itself as
: <math>T\begin{pmatrix} x_0 \\ x_1 \end{pmatrix} = \begin{pmatrix} x_0 \\ 0 \end{pmatrix}</math>

Clearly this is linear. (''Can you see why, without looking below?'')

Let's go through a proof that the conditions in the definitions are established.

==== Scalar multiplication is preserved ====
We wish to show that for all vectors '''v''' and all scalars λ, T(λ'''v''')=λT('''v''').

Let
: <math>\mathbf{v}=\begin{pmatrix} v_0 \\ v_1 \end{pmatrix}</math>.
Then
:<math>\lambda\mathbf{v}=\begin{pmatrix} \lambda v_0 \\ \lambda v_1 \end{pmatrix}</math>
Now
:<math> T(\lambda\mathbf{v}) = T\begin{pmatrix} \lambda v_0 \\ \lambda v_1\end{pmatrix} = </math>
:<math> \begin{pmatrix} \lambda v_0 \\ 0 \end{pmatrix} </math>
If we work out λT('''v''') and find it is the same vector, we have proved our result.
:<math> \lambda T\mathbf{v}= \lambda \begin{pmatrix} v_0 \\ 0 \end{pmatrix}=</math>
:<math> \begin{pmatrix} \lambda v_0 \\ 0 \end{pmatrix} </math>
This is the same vector as above, so under the transformation T, ''scalar multiplication is preserved''.

==== Addition is preserved ====
We wish to show for all vectors '''x''' and '''y''', T('''x'''+'''y''')=T'''x'''+T'''y'''.

Let
: <math>\mathbf{x}=\begin{pmatrix} x_0 \\ x_1 \end{pmatrix}</math>.
and
: <math>\mathbf{y}=\begin{pmatrix} y_0 \\ y_1 \end{pmatrix}</math>.
Now
: <math>T(\mathbf{x}+\mathbf{y})=T\left(\begin{pmatrix} x_0 \\ x_1 \end{pmatrix}+\begin{pmatrix} y_0 \\ y_1 \end{pmatrix}\right)=</math>
: <math>T\begin{pmatrix} x_0 + y_0 \\ x_1 + y_1 \end{pmatrix} =</math>
: <math>\begin{pmatrix} x_0 + y_0 \\ 0 \end{pmatrix}</math>
Now if we can show T'''x'''+T'''y''' is this vector above, we have proved this result.
Proceed, then,
:<math>T\begin{pmatrix} x_0 \\ x_1 \end{pmatrix} + T\begin{pmatrix} y_0 \\ y_1 \end{pmatrix}=\begin{pmatrix} x_0 \\ 0 \end{pmatrix} + \begin{pmatrix} y_0 \\0 \end{pmatrix}=</math>
: <math>\begin{pmatrix} x_0 + y_0 \\ 0 \end{pmatrix}</math>
So we have that the transformation T ''preserves addition''.

==== Zero vector is preserved ====
Clearly we have
: <math>T\begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} </math>

==== Conclusion ====
We have shown T preserves addition, scalar multiplication and the zero vector. So T must be linear.

== Disproof of linearity ==
When we want to ''disprove'' linearity - that is, to ''prove'' that a transformation is ''not'' linear, we need only find one counter-example.

If we can find just one case in which the transformation does not preserve addition, scalar multiplication, or the zero vector, we can conclude that the transformation is not linear.

For example, consider the transformation
: <math> T\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} x^3 \\ y^2\end{pmatrix}</math>

We suspect it is not linear. To prove it is not linear, take the vector
: <math> \mathbf{v} = \begin{pmatrix} 2 \\ 2 \end{pmatrix} </math>
then
: <math> T(2\mathbf{v}) = \begin{pmatrix} 64 \\ 16 \end{pmatrix}</math>
but
: <math> 2T(\mathbf{v}) = \begin{pmatrix} 16 \\ 8 \end{pmatrix}</math>

so we can immediately say T is not linear because it doesn't preserve scalar multiplication.

=== Problem set ===
Given the above, determine whether the following transformations are in fact linear or not. Write down each transformation in the form T:V -> W, and identify V and W. (Answers follow to even-numbered questions):
# <math>T\begin{pmatrix} v_0 \\ v_1 \end{pmatrix} = \begin{pmatrix} v_0^2 + v_1 \\ v_1 \end{pmatrix}</math>
# <math>T\begin{pmatrix} v_0 \\ v_1 \end{pmatrix} = \begin{pmatrix} 1 \\ v_0 \end{pmatrix}</math>
# <math>T\begin{pmatrix} v_0 \\ v_1 \end{pmatrix} = \mathbf{0}</math>
# <math>T\begin{pmatrix} v_0 \\ v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} v_0 - v_2 \\ v_1 \end{pmatrix}</math>

==== Answers ====
: 2. No. A check whether the zero vector is preserved readily confirms this fact. T : '''R'''2 -> '''R'''2
: 4. Yes. T : '''R'''3 -> '''R'''2.

== Images and kernels ==
We have some fundamental concepts underlying linear transformations, such as the ''kernel'' and the ''image'' of a linear transformation, which are analogous to the ''zeros'' and ''range'' of a function.

=== Kernel ===
The ''kernel'' of a linear transformation T: V -> W is the set of all vectors in V which are mapped to the zero vector in W, ie.,
: <math> \mathrm{ker}\ T = \{v \in V\ |\ T\mathbf{v} = \mathbf{0}\}</math>

Coincidentally because of the matr to the matrix equation A'''x'''='''0'''.

The kernel of a transform T: V->W is always a subspace of V. The dimension of a transform or a matrix is called the ''nullity''..

=== Image ===
The ''image'' of a linear transformation T:V->W is the set of all vectors in W which were mapped from vectors in V. For example with the trivial mapping T:V->W such that T'''x'''='''0''', the image would be '''0'''. (''What would the kernel be?'').

More formally, we say that the image of a transformation T:V->W is the set
: <math> \mathrm{im}\ T = \{w \in W\ |\ w=T\mathbf{v}\ \mathrm{and}\ \mathbf{v}\in V\}</math>

== Isomorphism ==

A linear transformation T:V -> W is an isomorphic transformation if it is:
* one-to-one and onto.
* kernel(T) = {0} and the range(T) = W.
* an inverse of T exists.
* dim(V) = dim(W).

@@ Line 145: / Line 145: @@
 Some functions have a '''two-sided inverse map''', another function that is the inverse of the first, both from the left and from the right. For instance, the map given by <math>\vec{v}\mapsto 2\cdot \vec{v}</math> has the two-sided inverse <math>\vec{v}\mapsto (1/2)\cdot\vec{v}</math>.
-In this subsection we will focus on two-sided inverses. The appendix shows that a function has a two-sided inverse if and only if it is both one-to-one and onto. {{anchor|def:inverse function}}The appendix also shows that if a function <math>f</math> has a two-sided inverse then it is unique, and so it is called "the" inverse, and is denoted <math>f^{-1}</math>. So our purpose in this subsection is, where a linear map <math>h</math> has an inverse, to find the relationship between <math>{\rm Rep}_{B,D}(h)</math> and <math>{\rm Rep}_{D,B}(h^{-1})</math>.
+In this subsection we will focus on two-sided inverses. The appendix shows that a function has a two-sided inverse if and only if it is both one-to-one and onto. The appendix also shows that if a function <math>f</math> has a two-sided inverse then it is unique, and so it is called "the" inverse, and is denoted <math>f^{-1}</math>. So our purpose in this subsection is, where a linear map <math>h</math> has an inverse, to find the relationship between <math>{\rm Rep}_{B,D}(h)</math> and <math>{\rm Rep}_{D,B}(h^{-1})</math>.
 A matrix <math> G </math> is a '''left inverse matrix''' of the matrix <math> H </math> if <math> GH </math> is the identity matrix. It is a '''right inverse matrix''' if <math> HG </math> is the identity. A matrix <math>H</math> with a two-sided inverse is an '''invertible matrix'''. That two-sided inverse is called '''the inverse matrix''' and is denoted <math> H^{-1} </math>.

@@ Line 423: / Line 423: @@
 ==Licensing==
 Content obtained and/or adapted from:
-* [https://en.wikibooks.org/wiki/Linear_Algebra/Inverses] under a CC BY-SA license
+* [https://en.wikibooks.org/wiki/Linear_Algebra/Inverses Inverses, Wikibooks: Linear Algebra] under a CC BY-SA license
-* [https://en.wikibooks.org/wiki/Linear_Algebra/The_Inverse_of_a_Matrix] under a CC BY-SA license
+* [https://en.wikibooks.org/wiki/Linear_Algebra/The_Inverse_of_a_Matrix The Inverse of a Matrix, Wikibooks: Linear Algebra] under a CC BY-SA license

← Older revision		Revision as of 13:26, 3 November 2021
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	This procedure will find the inverse of a general <math>n \! \times \! n</math> matrix.		This procedure will find the inverse of a general <math>n \! \times \! n</math> matrix.
	The <math>2 \! \times \! 2</math> case is handy.		The <math>2 \! \times \! 2</math> case is handy.
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikibooks.org/wiki/Linear_Algebra/Inverses] under a CC BY-SA license
		+	* [https://en.wikibooks.org/wiki/Linear_Algebra/The_Inverse_of_a_Matrix] under a CC BY-SA license

@@ Line 361: / Line 361: @@
 :<math>
 \begin{pmatrix}
-  &1  \\
+  & 1  \\
-  &-1
+  & -1
 \end{pmatrix}^{-1}
+=
@@ Line 402: / Line 402: @@
 \end{array}
 </math>
 ===Example 5===
@@ Line 419: / Line 418: @@
 \end{array}\right)
 </math>
 This procedure will find the inverse of a general <math>n \! \times \! n</math> matrix.
 The <math>2 \! \times \! 2</math> case is handy.

← Older revision	Revision as of 19:03, 29 September 2021
Line 1:	Line 1:
	+	An n-by-n matrix A is the inverse of n-by-n matrix B (and B the inverse of A) if BA = AB = I,
	+	where I is an identity matrix.

	+	The inverse of an n-by-n matrix can be calculated by creating an n-by-2n matrix which has the original matrix on the left and the identity matrix on the right. Row reduce this matrix and the right half will be the inverse. If the matrix does not row reduce completely (i.e., a row is formed with all zeroes as its entries), it does not have an inverse.
	+
	+	== Example ==
	+	Let <math> \mathrm{A} = \begin{bmatrix}
	+	1 & 4 & 4\\
	+	2 & 5 & 8\\
	+	3 & 6 & 9
	+	\end{bmatrix}</math>
	+
	+	We begin by expanding and partitioning A to include the identity matrix, and then proceed to row reduce A until we reach the identity matrix on the left-hand side.
	+
	+	:<math>
	+	\begin{bmatrix}
	+	1 & 4 & 4 & \big\| & 1 & 0 &0\\
	+	2 & 5 & 8 & \big\| & 0 & 1 &0\\
	+	3 & 6 & 9 & \big\| & 0 & 0 &1
	+	\end{bmatrix}
	+
	+	\rightsquigarrow
	+
	+	\begin{bmatrix}
	+	1 & 4 & 4 & \big\| & 1 & 0 &0\\
	+	0 & -3 & 0 & \big\| & -2 & 1 &0\\
	+	0 & -6 & -3 & \big\| & -3 & 0 &1
	+	\end{bmatrix}
	+
	+	\rightsquigarrow
	+
	+	\begin{bmatrix}
	+	1 & 4 & 4 & \big\| & 1 & 0 &0\\
	+	0 & 1 & 0 & \big\| & 2/3 & -1/3 &0\\
	+	0 & -6 & -3 & \big\| & -3 & 0 &1
	+	\end{bmatrix}
	+
	+	\rightsquigarrow
	+
	+	</math>
	+
	+
	+	:<math>
	+
	+	\begin{bmatrix}
	+	1 & 0 & 4 & \big\| & -5/3 & 4/3 &0\\
	+	0 & 1 & 0 & \big\| & 2/3 & -1/3 &0\\
	+	0 & 0 & -3 & \big\| & 1 & -2 &1
	+	\end{bmatrix}
	+
	+	\rightsquigarrow
	+
	+	\begin{bmatrix}
	+	1 & 0 & 4 & \big\| & -5/3 & 4/3 &0\\
	+	0 & 1 & 0 & \big\| & 2/3 & -1/3 &0\\
	+	0 & 0 & 1 & \big\| & -1/3 & 2/3 &-1/3
	+	\end{bmatrix}
	+
	+	\rightsquigarrow
	+
	+	\begin{bmatrix}
	+	1 & 0 & 0 & \big\| & -1/3 & -4/3 &4/3\\
	+	0 & 1 & 0 & \big\| & 2/3 & -1/3 &0\\
	+	0 & 0 & 1 & \big\| & -1/3 & 2/3 &-1/3
	+	\end{bmatrix}
	+
	+
	+	</math>
	+	<br><br>
	+	The matrix <math> \mathrm{B} = \begin{bmatrix}
	+	-1/3 & -4/3 & 4/3\\
	+	2/3 & -1/3 & 0\\
	+	-1/3 & 2/3 & -1/3
	+	\end{bmatrix}
	+	</math> is then the inverse of the original matrix A.

← Older revision		Revision as of 19:00, 29 September 2021
Line 1:		Line 1:
−	~~A '''linear transformation''' is an important concept in mathematics because many real world phenomena can be approximated by linear models.~~

−	~~Unlike a linear function, a linear transformation works on vectors as well as numbers.~~
−
−	~~== Motivations and definitions ==~~
−	Say we have the vector <math>\begin{pmatrix} 1 \\ 0 \end{pmatrix}</math> in <math>\mathbb{R}^2</math>, and we rotate it through 90 degrees, to obtain the vector <math>\begin{pmatrix} 0 \\ 1 \end{pmatrix}</math>.
−
−	Another example instead of rotating a vector, we stretch it, so a vector <math>\mathbf{v}</math> becomes <math>2\mathbf{v}</math>, for example. <math>\begin{pmatrix} 2 \\ 3 \end{pmatrix}</math> becomes <math>\begin{pmatrix} 4 \\ 6 \end{pmatrix}</math>
−
−	~~Or, if we look at the ''projection'' of one vector onto the ''x'' axis - extracting its ''x'' component - , e.g. from~~
−	~~<math>\begin{pmatrix} 2 \\ 3 \end{pmatrix}</math> we get <math>\begin{pmatrix} 2 \\ 0 \end{pmatrix}</math>~~
−
−	These examples are all an example of a ''mapping'' between two vectors, and are all linear transformations. If the rule transforming the matrix is called <math>T</math>, we often write <math>T\mathbf{v}</math> for the mapping of the vector <math>\mathbf{v}</math> by the rule <math>T</math>. <math>T</math> is often called the transformation.
−
−	Note we do not always write brackets like when we write functions. However we ''should'' write brackets, especially when we want to express the mapping of the sum or the product or the combination of many vectors.
−
−	~~== Definitions ==~~
−	~~===Linear Operators===~~
−
−	~~Suppose one has a field K, and let x be an element of that field. Let O be a function taking values from K where O(x) is an element of a field J. Define O to be a linear form if and only if:~~
−	~~# O(x+y)=O(x)+O(y)~~
−	~~# O(λx)=λO(x)~~
−
−	~~===Linear Forms===~~
−	Suppose one has a vector space V, and let x be an element of that vector space. Let F be a function taking values from V where F(x) is an element of a field K. Define F to be a linear form if and only if:
−	~~# F(x+y)=F(x)+F(y)~~
−	~~# F(λx)=λF(x)~~
−
−	~~===Linear Transformation===~~
−	This time, instead of a field, let us consider functions from one vector space into another vector space. Let T be a function taking values from one vector space V where L(V) are elements of another vector space. Define L to be a linear transformation when it:
−	~~# ''preserves scalar multiplication'': T(λ'''x''') = λT'''x'''~~
−	~~# ''preserves addition'': T('''x'''+'''y''') = T'''x''' + T'''y'''~~
−
−	Note that not all transformations are linear. Many simple transformations that are in the real world are also non-linear. Their study is more difficult, and will not be done here. For example, the transformation ''S'' (whose input and output are both vectors in '''R'''<sup>2</sup>) defined by
−
−	~~<math>S\mathbf{x} = S\begin{pmatrix} x \\~~
−	~~y \end{pmatrix}= \begin{pmatrix} xy\\~~
−	~~\cos(y)\end{pmatrix}</math>~~
−
−	~~We can learn about nonlinear transformations by studying easier, linear ones.~~
−
−	~~We often ''describe'' a transformation T in the following way~~
−	~~:<math>T : V \rightarrow W</math>~~
−
−	~~This means that T, whatever transformation it may be, maps vectors in the vector space V to a vector in the vector space W.~~
−
−	~~The actual transformation ''could'' be written, for instance, as~~
−	~~:<math>T\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} x + y \\ x - y \end{pmatrix}</math>~~
−
−	~~== Examples and proofs ==~~
−	~~Here are some examples of some linear transformations. At the same time, let's look at how we can prove that a transformation we may find is linear or not.~~
−
−	~~=== Projection ===~~
−	~~Let us take the projection of vectors in '''R'''<sup>2</sup> to vectors on the ''x''-axis. Let's call this transformation T.~~
−
−	~~We know that T maps vectors from '''R'''<sup>2</sup> to '''R'''<sup>2</sup>, so we can say~~
−	~~: <math>T: \mathbb{R}^2 \rightarrow \mathbb{R}^2</math>~~
−
−	~~and we can then write the transformation itself as~~
−	~~: <math>T\begin{pmatrix} x_0 \\ x_1 \end{pmatrix} = \begin{pmatrix} x_0 \\ 0 \end{pmatrix}</math>~~
−
−	~~Clearly this is linear. (''Can you see why, without looking below?'')~~
−
−	~~Let's go through a proof that the conditions in the definitions are established.~~
−
−	~~==== Scalar multiplication is preserved ====~~
−	~~We wish to show that for all vectors '''v''' and all scalars λ, T(λ'''v''')=λT('''v''').~~
−
−	~~Let~~
−	~~: <math>\mathbf{v}=\begin{pmatrix} v_0 \\ v_1 \end{pmatrix}</math>.~~
−	~~Then~~
−	~~:<math>\lambda\mathbf{v}=\begin{pmatrix} \lambda v_0 \\ \lambda v_1 \end{pmatrix}</math>~~
−	~~Now~~
−	~~:<math> T(\lambda\mathbf{v}) = T\begin{pmatrix} \lambda v_0 \\ \lambda v_1\end{pmatrix} = </math>~~
−	~~:<math> \begin{pmatrix} \lambda v_0 \\ 0 \end{pmatrix} </math>~~
−	~~If we work out λT('''v''') and find it is the same vector, we have proved our result.~~
−	~~:<math> \lambda T\mathbf{v}= \lambda \begin{pmatrix} v_0 \\ 0 \end{pmatrix}=</math>~~
−	~~:<math> \begin{pmatrix} \lambda v_0 \\ 0 \end{pmatrix} </math>~~
−	~~This is the same vector as above, so under the transformation T, ''scalar multiplication is preserved''.~~
−
−	~~==== Addition is preserved ====~~
−	~~We wish to show for all vectors '''x''' and '''y''', T('''x'''+'''y''')=T'''x'''+T'''y'''.~~
−
−	~~Let~~
−	~~: <math>\mathbf{x}=\begin{pmatrix} x_0 \\ x_1 \end{pmatrix}</math>.~~
−	~~and~~
−	~~: <math>\mathbf{y}=\begin{pmatrix} y_0 \\ y_1 \end{pmatrix}</math>.~~
−	~~Now~~
−	~~: <math>T(\mathbf{x}+\mathbf{y})=T\left(\begin{pmatrix} x_0 \\ x_1 \end{pmatrix}+\begin{pmatrix} y_0 \\ y_1 \end{pmatrix}\right)=</math>~~
−	~~: <math>T\begin{pmatrix} x_0 + y_0 \\ x_1 + y_1 \end{pmatrix} =</math>~~
−	~~: <math>\begin{pmatrix} x_0 + y_0 \\ 0 \end{pmatrix}</math>~~
−	~~Now if we can show T'''x'''+T'''y''' is this vector above, we have proved this result.~~
−	~~Proceed, then,~~
−	~~:<math>T\begin{pmatrix} x_0 \\ x_1 \end{pmatrix} + T\begin{pmatrix} y_0 \\ y_1 \end{pmatrix}=\begin{pmatrix} x_0 \\ 0 \end{pmatrix} + \begin{pmatrix} y_0 \\0 \end{pmatrix}=</math>~~
−	~~: <math>\begin{pmatrix} x_0 + y_0 \\ 0 \end{pmatrix}</math>~~
−	~~So we have that the transformation T ''preserves addition''.~~
−
−	~~==== Zero vector is preserved ====~~
−	~~Clearly we have~~
−	~~: <math>T\begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} </math>~~
−
−	~~==== Conclusion ====~~
−	~~We have shown T preserves addition, scalar multiplication and the zero vector. So T must be linear.~~
−
−	~~== Disproof of linearity ==~~
−	~~When we want to ''disprove'' linearity - that is, to ''prove'' that a transformation is ''not'' linear, we need only find one counter-example.~~
−
−	~~If we can find just one case in which the transformation does not preserve addition, scalar multiplication, or the zero vector, we can conclude that the transformation is not linear.~~
−
−	~~For example, consider the transformation~~
−	~~: <math> T\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} x^3 \\ y^2\end{pmatrix}</math>~~
−
−	~~We suspect it is not linear. To prove it is not linear, take the vector~~
−	~~: <math> \mathbf{v} = \begin{pmatrix} 2 \\ 2 \end{pmatrix} </math>~~
−	~~then~~
−	~~: <math> T(2\mathbf{v}) = \begin{pmatrix} 64 \\ 16 \end{pmatrix}</math>~~
−	~~but~~
−	~~: <math> 2T(\mathbf{v}) = \begin{pmatrix} 16 \\ 8 \end{pmatrix}</math>~~
−
−	~~so we can immediately say T is not linear because it doesn't preserve scalar multiplication.~~
−
−	~~=== Problem set ===~~
−	Given the above, determine whether the following transformations are in fact linear or not. Write down each transformation in the form T:V -> W, and identify V and W. (Answers follow to even-numbered questions):
−	~~# <math>T\begin{pmatrix} v_0 \\ v_1 \end{pmatrix} = \begin{pmatrix} v_0^2 + v_1 \\ v_1 \end{pmatrix}</math>~~
−	~~# <math>T\begin{pmatrix} v_0 \\ v_1 \end{pmatrix} = \begin{pmatrix} 1 \\ v_0 \end{pmatrix}</math>~~
−	~~# <math>T\begin{pmatrix} v_0 \\ v_1 \end{pmatrix} = \mathbf{0}</math>~~
−	~~# <math>T\begin{pmatrix} v_0 \\ v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} v_0 - v_2 \\ v_1 \end{pmatrix}</math>~~
−
−	~~==== Answers ====~~
−	~~: 2. No. A check whether the zero vector is preserved readily confirms this fact. T : '''R'''<sup>2</sup> -> '''R'''<sup>2</sup>~~
−	~~: 4. Yes. T : '''R'''<sup>3</sup> -> '''R'''<sup>2</sup>.~~
−
−	~~== Images and kernels ==~~
−	We have some fundamental concepts underlying linear transformations, such as the ''kernel'' and the ''image'' of a linear transformation, which are analogous to the ''zeros'' and ''range'' of a function.
−
−	~~=== Kernel ===~~
−	~~The ''kernel'' of a linear transformation T: V -> W is the set of all vectors in V which are mapped to the zero vector in W, ie.,~~
−	~~: <math> \mathrm{ker}\ T = \{v \in V\ \|\ T\mathbf{v} = \mathbf{0}\}</math>~~
−
−	~~Coincidentally because of the matr to the matrix equation A'''x'''='''0'''.~~
−
−	~~The kernel of a transform T: V->W is always a subspace of V. The dimension of a transform or a matrix is called the ''nullity''..~~
−
−	~~=== Image ===~~
−	The ''image'' of a linear transformation T:V->W is the set of all vectors in W which were mapped from vectors in V. For example with the trivial mapping T:V->W such that T'''x'''='''0''', the image would be '''0'''. (''What would the kernel be?'').
−
−	~~More formally, we say that the image of a transformation T:V->W is the set~~
−	~~: <math> \mathrm{im}\ T = \{w \in W\ \|\ w=T\mathbf{v}\ \mathrm{and}\ \mathbf{v}\in V\}</math>~~
−
−	~~== Isomorphism ==~~
−
−	~~A linear transformation T:V -> W is an isomorphic transformation if it is:~~
−	* one-to-one and onto.
−	* kernel(T) = {0} and the range(T) = W.
−	* an inverse of T exists.
−	* dim(V) = dim(W).