Question

SMITH NORMAL FORM

Hi ive been trying to understand the process behind smith normal form, but I cannot get it

can someone please thoroughly explain the process in finding the smith normal form of the following matrix

[ 3 6 / 8 12 ]

Nice and simple matrix, I just want to know the process behind it. Something like LMQ=D something?

Also can you show me how to do

[ 3 6 9 / 8 12 16] if possible?

Answer 1

Answer:

Smith normal form of matrix can be computed using following algorithm:

For the matrix of type

a b c d

Smith normal form will look like this matrix,

a1 0 0 a2

1
and $\small a_2$ can be calculated using this equation,

(A) ai di-1(A)

Here,  
d;(A)
is the greatest common divisor of all $\small i\times i$ minor matrix of given matrix.

Base condition is

1= d1(A)

For the given example,

3 6 C = 8 12

Let's calculate  
1
and and $\small a_2$ for C,

a1d1(C) = gcd(3, 6, 8, 12) = 1

There is only one minor of 2 x 2 and that is matrix itself and determinant of matrix is given by,

$\small \begin{vmatrix} 3 &6 \\ 8 & 12 \end{vmatrix} = 3*12 -6*8 = 36-48 = -12$

$\small d_2(C) = gcd(-12) = gcd(12) = 12$

$\small \therefore a_2 = 12/1 = 12$

So, we have got the answer as,

10 E = 0 12

For conversion of C to E, we will only these three operations:

1. interchange two rows or two columns,
2. multiply a row or column by ±1 (which are the invertible elements in Z),
3. add an integer multiple of row to another row (or an integer multiple of a column to another column).

Let's perform these on C,

$\small \begin{bmatrix} 3 &6 \\ 8&12 \end{bmatrix}\rightarrow \begin{bmatrix} 3 &6 \\ -1&-6 \end{bmatrix}\rightarrow \begin{bmatrix} 0 & -12 \\ -1&-6 \end{bmatrix}\rightarrow \begin{bmatrix} 0 & -12 \\ -1& 0 \end{bmatrix}\rightarrow \begin{bmatrix} -1 &0 \\ 0&-12 \end{bmatrix}$

$\small \begin{bmatrix} -1 &0 \\ 0&-12 \end{bmatrix}\rightarrow \begin{bmatrix} 1 &0 \\ 0&-12 \end{bmatrix}\rightarrow \begin{bmatrix} 1 &0 \\ 0&12 \end{bmatrix}$

Now for problem,

$\small F = \begin{bmatrix} 3 &6 &9 \\ 8&12 &16 \end{bmatrix}$

For the above matrix smith normal form will look like,

$\small G = \begin{bmatrix} a_1&0 &0 \\ 0&a_2 &0 \end{bmatrix}$

Now $\small a_1 = gcd(3,6,9,8,12,16) = 1$

There are three minors of 2 x 2. Here,

$\small \begin{bmatrix} 3 &6 \\ 8& 12 \end{bmatrix}\ \begin{bmatrix} 6 &9 \\ 12&16 \end{bmatrix}\ \begin{bmatrix} 3 &9 \\ 8& 16 \end{bmatrix}$

Their determinant are -12, -12, -24 and their gcd is 12,

so $\small a_2 = 12$

Now we can easily transform

3 6 10 0 0 12 0 8 12 16