Show that every row of matrix $S$ is a linear combination of its bottom row and the row (1 1 1 1 1 1 )

Question

Couldn't solve the following three questions.

$$S=\begin{pmatrix} 36 & 35 & 34 &33&32&31 \\ 25 & 26 & 27&28&29&30 \\ 24&23&22&21&20&19 \\ 13&14&15&16&17&18 \\ 12&11&10&9&8&7 \\ 1&2&3&4&5&6 \\ \end{pmatrix}$$

(a) Show that every row of matrix $S$ is a linear combination of its bottom row and the row${(1 1 1 1 1 1 )}$

(b) Deduce that the rank of $S$ is at most 2

(c) Show that the rank of $S$ can't be 0 or 1, therefore, the rank of $S$ is 2

Answer 1

These hints are probably best read after you've tried what david has to say, so don't read them until you do.

(a) Let $R_n$ be the $n$th row of $S$ as a row vector and let $A$ be the row vector of all ones. What you should do for (a) is take $\pm R_6 + kA$ with the appropriate choice of $k$ to get the row you want.

(b) simply express A in terms of $R_6$ and $R_5$. Then we know from (a) that any row is a linear combination of the two rows, and so we see that $S$ has rank 2. Do you see this?

(c) David already spells it out for you in his answer. I recommend reviewing the definition of rank if you're having trouble.

EDIT: I think $R_5$ is slightly easier to work with than $R_1$, which David used, but it's up to preference. I just like small numbers.

Answer 2

Hints.

(a) What do you get if you take the first row plus the last?

(b) The rank is the minimum number of vectors needed to span the rows.

(c) What does a matrix look like if it has rank $0$? If it has rank $1$?