2
$\begingroup$

If you have the following matrix can $k$ be any number?

\begin{pmatrix} 1 & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & 1 \end{pmatrix}

So this is obviously an assignment question, but I couldn't find a concrete answer anywhere.

I would just liketo double check my reasoning with other people (long distance learning, so no other students to chat too)

I say no, because $k$ cannot be zero. To my understanding, an elementary matrix can only be created using a single row operation on an Identity matrix. I can't think of any operation that would create a row of zeros from an Identity matrix.

Is my assumption correct: $k$ can be any number except for zero.

$\endgroup$
2
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Sidharth Ghoshal
    Jul 30, 2014 at 7:42
  • $\begingroup$ after-re reading your comment about 0, i agree with your logic, for all other values though this indeed is an elementary matrix $\endgroup$
    – Sidharth Ghoshal
    Jul 30, 2014 at 7:43

1 Answer 1

Reset to default
1
$\begingroup$

enter image description here

Like in the definition of Wikipedia you are correct, $k$ cannot be equal $0$. In fact elementary matrices represent the steps that you do with the gaussian algorithm, and you are not allowed to multiply a row/column by $0$ to get the Row Echelon Form because when you multiply a row/column by $0$ "you loose the information" of the row/column and that is obviously not good.

It can happen that you get a row/column with all zeros, but that's because the rank of the matrix isn't equal to its dimension.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .