2
$\begingroup$

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.

|cite|improve this question
$\endgroup$
1
$\begingroup$

$\det A = 0$ 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 $AB = 0$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '12 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 '12 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 '12 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 '12 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 '12 at 16:05

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.