Exercises:Saul  Algebraic Topology  1/Exercise 1.5
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Exercise 1.5

The matrix of [ilmath]A:\mathbb{Z}^3\rightarrow\mathbb{Z}^3[/ilmath] 

 Alec's note:
 [ilmath]\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\} [/ilmath]
 [ilmath]\text{Ker}(A):\eq\{x\in\mathbb{Z}^3 \ \vert\ A(x)\eq 0\} [/ilmath]
 TODO: Cokernel is the homology group here. Check that
Solution
Image of [ilmath]A[/ilmath]
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.
 [ilmath]\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)[/ilmath] [ilmath]\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)[/ilmath]
Let us now reduce this to a basis (and write these as rowvectors transposed from now on)
 Clearly [ilmath](1,4,7)^T[/ilmath] and [ilmath](2,5,8)^T[/ilmath] 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 [ilmath]A[/ilmath] shows that 1 x the first vector + 2 x the second vector = the third vector
 Kernel  RREFfing [ilmath]A[/ilmath] with an extra column of zeros shows the kernel has basis [ilmath](1,2,1)[/ilmath] or something.
Caveat:This is "dumb luck", it could be nastier, there's no.... like all of these are isomorphic to copies of [ilmath]\mathbb{Z} [/ilmath]
 Cokernel, best thought of as [math]\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 } [/math]
 Clearly we can replace [ilmath](1,0,0)^T[/ilmath] with [ilmath](1,4,7)^T[/ilmath] as [ilmath]a(1,4,7)4a(0,1,0)7a(0,0,1)\eq a(1,0,0)[/ilmath]
 We cannot do the same thing again, as [ilmath]a(1,4,7)+b(2,5,8)+c(0,0,1)[/ilmath] can never be [ilmath](0,1,0)[/ilmath] (RREFing we get fractional values. Which is telling...)
 Clearly we can replace [ilmath](1,0,0)^T[/ilmath] with [ilmath](1,4,7)^T[/ilmath] as [ilmath]a(1,4,7)4a(0,1,0)7a(0,0,1)\eq a(1,0,0)[/ilmath]
Thoughts
Let's think about this a bit more
 [math]\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 } [/math] is the same as [math]\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 } [/math] (As [ilmath](1,0,0)[/ilmath] is in the span of this, and [ilmath](1,4,7)[/ilmath] is clearly in the span of the standard basis vectors)
Notes
References