Exercises:Saul - Algebraic Topology - 1/Exercise 1.5

Exercises

Exercise 1.5

[math]A:\eq\left(\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7& 8 &9\end{array}\right)[/math]
The matrix of [ilmath]A:\mathbb{Z}^3\rightarrow\mathbb{Z}^3[/ilmath]

Let [ilmath]A:\mathbb{Z}^3\rightarrow\mathbb{Z}^3[/ilmath] be the group homomorphism given by the matrix on the right. Calculate its kernel, image and cokernel.

• 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 row-vectors 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]

• Co-kernel, 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...)

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)