View source for Exercises:Saul - Algebraic Topology - 1/Exercise 1.5

<noinclude>
__TOC__
==Exercises==
===Exercise 1.5===
</noinclude><div style="float:right;margin:0px;margin-left:0.2em;max-width:25em;">
{| class="wikitable" border="1" style="margin:0px;"
|-
| <span style="font-size:1.2em;"><center>{{MM|A:\eq\left(\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7& 8 &9\end{array}\right)}}</center></span>
|-
! The [[matrix]] of {{M|A:\mathbb{Z}^3\rightarrow\mathbb{Z}^3}}
|}
</div>Let {{M|A:\mathbb{Z}^3\rightarrow\mathbb{Z}^3}} be the [[group homomorphism]] given by the [[matrix]] on the right. Calculate its [[kernel]], [[image]] and [[cokernel]].
* {{Note|Alec's note:}}
** {{M|\text{Image}(A):\eq A(\mathbb{Z}^3):\eq\{y\in\mathbb{Z}^3\ \vert\ \exists x\in\mathbb{Z}^3[A(x)\eq y]\}\eq\{A(x)\ \vert\ x\in\mathbb{Z}^3\} }}
** {{M|\text{Ker}(A):\eq\{x\in\mathbb{Z}^3 \ \vert\ A(x)\eq 0\} }}
** {{XXX|Cokernel is the homology group here. Check that}}
====Solution====
=====Image of {{M|A}}=====
It is easy to see {{Caveat|In linear algebra.... over vector spaces....}} that the span of the images of basis elements spans the image. We will use this here.
* {{M|\text{Im}(A)\eq\text{Span}\left(\left\{A\left(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\right),A\left(\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\right),A\left(\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}\right)\right)}} {{M|\eq\text{Span}\left(\left\{\left(\begin{array}{c}1\\ 4\\ 7\end{array}\right),\left(\begin{array}{c}2\\ 5\\ 8\end{array}\right),\left(\begin{array}{c}3\\ 6\\ 9\end{array}\right)\right\}\right)}}
Let us now reduce this to a basis (and write these as row-vectors transposed from now on)
* Clearly {{M|(1,4,7)^T}} and {{M|(2,5,8)^T}} are "linearly independent"
* To see if the 3rd vector is linearly independent of these two we must 

=====Answers=====
* Image - the first 2 columns of the matrix form a vector. RREFfing {{M|A}} shows that -1 x the first vector + 2 x the second vector = the third vector
* Kernel - RREFfing {{M|A}} with an extra column of zeros shows the kernel has basis {{M|(1,-2,1)}} or something. 
{{Caveat|This is "dumb luck", it could be nastier, there's no.... like all of these are isomorphic to copies of {{M|\mathbb{Z} }} }}
* Co-kernel, best thought of as {{MM|\frac{\langle (1,0,0)^T,(0,1,0)^T,(0,0,1)^T\rangle}{\langle (1,4,7)^T,(2,5,8)^T\rangle } }}
** Clearly we can replace {{M|(1,0,0)^T}} with {{M|(1,4,7)^T}} as {{M|a(1,4,7)-4a(0,1,0)-7a(0,0,1)\eq a(1,0,0)}} 
*** We cannot do the same thing again, as {{M|a(1,4,7)+b(2,5,8)+c(0,0,1)}} can never be {{M|(0,1,0)}} (RREFing we get fractional values. Which is telling...)
======Thoughts======
Let's think about this a bit more
* {{MM|\frac{\langle (1,0,0)^T,(0,1,0)^T,(0,0,1)^T\rangle}{\langle (1,4,7)^T,(2,5,8)^T\rangle } }} is the same as {{MM|\frac{\langle (1,4,7)^T,(0,1,0)^T,(0,0,1)^T\rangle}{\langle (1,4,7)^T,(2,5,8)^T\rangle } }} (As {{M|(1,0,0)}} is in the span of this, and {{M|(1,4,7)}} is clearly in the span of the standard basis vectors)
<div style="clear:both;"></div><noinclude>
==Notes==
<references group="Note"/>
==References==
<references/>
</noinclude>