Vtgyjfy

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 :

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.

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