# Find the structure of $\mathbb Z ^{3} / K$ with $K$ the image of a matrix

I have this matrix: $$A= \begin{pmatrix} 2 & 5 & -1 & 2\\ -2 & -16 & -4 & 4 \\ -2 &-2 &0 &6 \end{pmatrix}$$ If we set K as the Image of this matrix, how do you find a basis of $$K$$ of this form : $$( d_1 w_1 , \cdots , d_s w_s ), s \leq 4$$ such that we have that $$( w_1 , \cdots , w_4 )$$ is a basis of $$\mathbb Z ^{3}$$ and that $$d_i | d_{i+1}$$

I must use the Smith normal form, but I'm blocked by the fact that I can't find a basis of the Image. In the correction of this exercice, their are using a method I'm not understanding.

I would firstly determine a basis of the image and then do the same computation as I usually do.

$$\DeclareMathOperator{\im}{Im}\DeclareMathOperator{sp}{Span}\require{AMScd}$$First, let us understand where all the maps are going in the Smith normal form:

$$\begin{CD} \mathbb{Z}^4 @>A>> \mathbb{Z}^3\\ @APAA @AAQA \\ \mathbb{Z}^4 @>>D> \mathbb{Z}^3 \end{CD}$$

$$P$$ and $$Q$$ are isomorphisms (invertible), $$D$$ is diagonal and $$A = QDP^{-1}$$. The point of $$P$$ and $$Q$$ is that they are a change of basis such that in the new basis, $$A$$ acts diagonally.

We want to compute the image of $$A$$, or equivalently, the image of $$QDP^{-1}$$.

First, I claim that $$\im(A) = \im(QD)$$ and this is because $$P$$ is invertible.

Let $$y \in \im(A)$$. Then $$y = Ax = QDP^{-1}$$ for some $$x$$. So $$y = QD(P^{-1}x)$$ is in the image of $$QD$$. Next, let $$y \in \im(QD)$$. Then $$y = QDx$$ for some $$x$$. Since $$P$$ (and also $$P^{-1}$$) is invertible, there must be some $$x'$$ such that $$x = P^{-1}x'$$ (namely: $$x' = Px$$). Then $$y = QDP^{-1}x' = Ax' \in \im{A}$$.

The general rule here is that if $$A = BC$$ and $$C$$ is invertible, then $$\im(A) = \im(B)$$.

Next, given any matrix, the image of that matrix is the same as the column space.

To demonstrate, let $$B$$ have columns $$v_1, \dots, v_n$$ and let $$x = (x_1,\dots,x_n)$$. Then $$Bx = \begin{pmatrix} v_1 & \cdots & v_n \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} = x_1v_1 + \cdots + x_nv_n \in \sp\{v_1,\dots,v_n\}$$ And conversely, any element $$x_1v_1 + \cdots + x_n v_n \in \sp\{v_1,\dots,v_n\}$$ can be written as $$Bx$$ where $$x = (x_1,\dots,x_n)$$.

So what we have shown is that $$\im(A) = \im(QD) = \sp\{\text{columns of QD}\}$$.

Now the last step is what I said near the beginning: $$P$$ and $$Q$$ represent a change of basis. So the columns of $$Q$$ are a basis for $$\mathbb{Z}^3$$ and the columns of $$P$$ are a basis for $$\mathbb{Z}^4$$. (In fact, the same is true for $$P^{-1}, Q^{-1}$$ as well as $$P^T$$ and $$Q^T$$ or, more generally, any invertible matrix.)

So the columns of $$Q$$ are a basis for $$\mathbb{Z^3}$$ and the (non-zero) columns of $$QD$$ are a basis for $$\im(A)$$. Then it's just a matter of understanding how diagonal matrices act on other matrices. Multiplying by a diagonal matrix on the right multiplies the columns by the corresponding diagonal element. Multiplying by a diagonal matrix on the left multiplies the rows by the corresponding diagonal element.

This is why $$QD$$ is obtained from $$Q$$ by multiplying the columns by $$-1, -2$$, and $$2$$ respectively.

I don't know if you are follow the same course as mine but I had this exact exercice this semester with D. Testerman. Here is the solution :  • Yes I do and actually i dont understand this solution. That is why i m asking for another one facepalm ^~^'' – Marine Galantin Jan 12 at 19:22
• @Marine What parts do you understand or not understand? Do you understand how $D$ is computed? How $Q$ is computed? Why $\operatorname{Im}(f_A) = \operatorname{Im}(f_{QD})$? Why the columns of $Q$ are a basis of $\mathbb{Z}^{3}$? Why the columns of $QD$ are a basis of $\operatorname{Im}(f_A)$? – Trevor Gunn Jan 12 at 19:31
• The only thing I understand is : how you do the smith normal form, and how you fins the matrices P and Q. The rest is meaningless for me... – Marine Galantin Jan 12 at 19:34
• @MarineGalantin Ah sure no problems, good luck for Wednesday! – NotAbelianGroup Jan 12 at 21:15

This took me six elementary column matrices, the 4 by 4 square matrix has determinant $$1.$$ Actually, I combined some steps, so it might be more reasonable to indicate the square matrix as $$R = R_1 R_2R_3R_4R_5R_6R_7 R_8,$$ this is the order when using column operations rather than the more familiar row operations.

$$\left( \begin{array}{rrrr} 2& 5& -1& 2 \\ -2& -16& -4& 4 \\ -2& -2& 0& 6 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 &-3& -10 & -56 \\ 0 &1 & 3 & 17 \\ 1 &-3 & -9 &-53 \\ 0& -1 & -2 & -13 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 0 &0& 0 \\ -6 &-2& 0& 0 \\ -2 &-2& 2 & 0 \\ \end{array} \right)$$

• Thank you for the help. But then? – Marine Galantin Jan 12 at 19:24
• @MarineGalantin The nonzero columns of the 3 by 4 matrix on the right give an integral basis for the image, which is what you said you wanted. – Will Jagy Jan 12 at 19:28
• Okay and how do you obtain it, did you just reduced the matrix? – Marine Galantin Jan 12 at 19:29
• @MarineGalantin experience is best. Can you find an integral matrix $T$ with determinant $1,$ three by three, so that $TAR$ is in Smith form? This $T$ will be the product of two or three elementary integer matrices, for row operations the order will be $T = T_3 T_2 T_1$ – Will Jagy Jan 12 at 19:35
• Im sorry I don't understand exactly what you re saying. Anyways, if that s what you mean, I know that any integral matrix can be written in it s smith form. This afterwards gives me a decomposition $Q^{-1} AP = D$, is it your $TAR$? – Marine Galantin Jan 12 at 19:37