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In my problem, $R$ is a commutative ring with identity. Suppose $A$ is an n by n matrix with entries from R. Let $\operatorname{rk}(A)$ be the largest $m$ such that some $m$ by $m$ submatrix of $A$ has non-zero determinant (For A=0, put rk A=0.)

a) If $\operatorname{rk}(A)< n$ , show there is some non-zero n by $\operatorname{rk}(A)+1$ matrix B such that AB=0.

b) Hence prove that if $z_1, z_2, \cdots z_n$ are linearly independent elements of $R^{\ell}$ then $n\leq \ell.$

In class we have shown that the usual definitions for the determinant and adjugate matrix still work fine over general comm rings and some basic properties hold like the column/row expansion formulas for the determinant, and $\operatorname{adj}(A)A= A\operatorname{adj}(A)=\det(A)I_n.$

Thus, if $\operatorname{rk}(A)=n-1$ then a suitable matrix $B$ is $\operatorname{adj}(A).$ This satisfies $AB=0$ and $B$ is non-zero because its entries are (up to sign) determinants of n-1 by n-1 submatrices of $A$ and since rk A =n-1, at least one of those entries is non-zero. But this example fails if $\operatorname{rk}(A) < n-1$ and I haven't been able to make any progress for hours now. I also have no idea on how to do part b). Any help is greatly appreciated.

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$\det A = 0$ means that that columns (and rows) of $A$ are not linearly independent, which means that $\sum c_n \vec{v_n} = \vec{0}$ where $\vec{v_n}$ are the column vectors of $A$ and $c_n$ are constants, not all zero. In other words, $A \vec{c} = 0$ where $\vec{c}$ is a vector with coordinates $c_n$. Finally copy $\vec{c}$ several times as columns to get a larger matrix $B$ with $AB = 0$.

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  • $\begingroup$ In case you wonder, I am bothered by the fact that my answer doesn't use the rank anywhere. I wrote it as I would for $\mathbb{R}$. Although I don't seem to be making many assumptions. It is possible that you are expected to take some other approach or that I made a mistake. For the same reason I didn't start on b). $\endgroup$
    – Karolis Juodelė
    Sep 16, 2012 at 15:36
  • $\begingroup$ Thank you. It is not in my notes that $\det A=0$ gives that the columns are linearly dependent. How can I prove that for arbitrary commutative rings? Also, can you help me with how part a) helps for part b)? $\endgroup$
    – Katie Dobbs
    Sep 16, 2012 at 15:37
  • $\begingroup$ Ahh ok. Maybe the rank part only comes in specifically useful for b) and isn't needed essentially for a). $\endgroup$
    – Katie Dobbs
    Sep 16, 2012 at 15:38
  • $\begingroup$ This follows from the fact that elementary column operations do not change the determinant and that they can be used to transform the matrix to a representation where all non-zero elements are on the diagonal. Determinant is then the product of diagonal elements and if any one of them is 0 ten that column can be written as linear combination of the other columns. See Gaussian elimination if you don't know how that works. $\endgroup$
    – Karolis Juodelė
    Sep 16, 2012 at 16:02
  • $\begingroup$ For b) you could construct a base of size $l$ and remember that base is the largest linearly independent subset. I don't know how to use a) for it. $\endgroup$
    – Karolis Juodelė
    Sep 16, 2012 at 16:05

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