2
$\begingroup$

I am a bit confused on how to put the matrix \begin{bmatrix}4&2&4\\3&3&4\\2&2&2\end{bmatrix} in $\mathbb{Z}$-Smith Normal Form.

I know this can be done using unimodular elementary row and column operations, but everytime i've tried I end up with the wrong answer.

Any help will be very much appreciated

$\endgroup$
0
$\begingroup$

I perform a series of row and column operations to get $$\begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&1 \end{pmatrix}$$

Is this your right answer ?

$\endgroup$
  • $\begingroup$ I dont actually have the answer, that was what i ended up with once but then I thought it was wrong since 2 does not divide 1 $\endgroup$ – 1234321 Dec 1 '16 at 15:17
  • $\begingroup$ Thats a small point, and actually I ought to have written it as Marc did below. You can always change the order. $\endgroup$ – Rene Schipperus Dec 1 '16 at 16:18
0
$\begingroup$

The Smith normal form of that matrix is $$\pmatrix{1&0&0\\0&2&0\\0&0&2}.$$ That the top left value is going to be $1$ is pretty obvious; just subtract the second row of the original matrix from the first.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.