# Problem regarding rank

• Jul 6th 2009, 04:34 PM
Pinkk
Problem regarding rank
Are there values for $\displaystyle r$ and $\displaystyle s$ for which:

$\displaystyle \left(\begin{array}{ccc}1&0&0\\0&r-2&2\\0&s-1&r+2\\0&0&3\end{array}\right)$

has rank 1 or 2? And if so, find those values.

Well, it's obvious that if $\displaystyle r=2$ and $\displaystyle s=1$. It will have rank 2, and I can see that it can never have rank 1, but I am not sure how to exactly show this. I know in order to have rank 1, each row (or column) must be a scalar of every other column. So, would it be fair to say that $\displaystyle aC_{1}+bC_{2}+cC_{3}+dC_{4}=(0,0,0,0)$ for some non-trivial solution of $\displaystyle a,b,c,d$in order for that to be true. I think using the rows won't work in this situation since the rows can be thought of as vectors in $\displaystyle R^{3}$ and a set of 4 vectors in $\displaystyle R^{3}$ must be linearly dependent. I'm guessing if I can show that the a linear combination of the three column vectors can only equal a column of 0's with only the trivial solution, than that eventually leads to the fact that it is never possible to have any given row (or column) represented as a scalar multiple of all of the other rows or columns.

Any help would be appreciated.
• Jul 6th 2009, 04:46 PM
Pinkk
I just thought of an easier way, but again, I don't know if it suffices. Since $\displaystyle 1 \ne a0$ for any $\displaystyle a \in R$, none of the rows or columns can all be scalar multiples of one another.
• Jul 6th 2009, 09:40 PM
BubbleBrain_103
You've got the right idea. Here is a fleshed out version of what you're getting at.

Let r and s be any real numbers. For the matrix to have rank 1, the three columns vectors must span a 1-dimensional subspace. In this case, any non-zero vector in that subspace would serve as a spanning vector, namely (1,0,0,0) (the first column vector). But obviously the third column vector, no matter what r and s you choose, cannot be a scalar multiple of the first, by what you just said. And that's it.
• Jul 6th 2009, 10:02 PM
Pinkk
Thanks for clarifying. (Clapping)