If a $3\times{3}$ matrix has a non-trivial solution to $B\vec{x}=\vec{0}$, then it has linearly dependent rows. Looking at Cramer's rule:

$$x_1=\frac{\begin{vmatrix}0 & b_{12} & b_{13} \\0 & b_{22} & b_{23} \\ 0 & b_{32} & b_{33}\end{vmatrix}}{\det(B)}$$

$$\det(B)x_1=\begin{vmatrix}0 & b_{12} & b_{13} \\0 & b_{22} & b_{23} \\ 0 & b_{32} & b_{33}\end{vmatrix}$$

Since $\begin{vmatrix}0 & b_{12} & b_{13} \\0 & b_{22} & b_{23} \\ 0 & b_{32} & b_{33}\end{vmatrix}$ is clearly equal to $0$, then either $x_1$ has to be $0$, or $\det{B}$ does. Applying the same reasoning to $x_2$ and $x_3$, for $\vec{x}$ to be non-trivial, at least one of the components must be non-zero, meaning that $\det{B}$ must be $0$.

So here's where I have doubts on this proof. Cramer's rule is defined only valid if $\det{B}\ne0$. However, the second equation isn't Cramer's rule, but something that I'm postulating may be a valid statement that follows from Cramer's rule. Since it doesn't result in a division by $0$, I suspect it may be valid, but I'm not entirely sure.

Is this proof valid?

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    $\begingroup$ You proved that if $\det(B) \neq 0$, then $Bx=0$ has only a trivial solution. So just use a contraposition argument. $\endgroup$ – Seirios Aug 7 '13 at 4:35
  • $\begingroup$ When you write a supposed mathematical expression of the form $\,x_i=\frac ab\;$ , we're implicitly (and even automatically) assuming $\,b\ne 0\,$, otherwise that is not a mathematical expression. Thus, either you can reach $\,\det(B)\,x_i=...\;$ without first having the division by $\,\det(B)\,$ , or else you can prove that $\,\det(B)\ne 0\,$ upon request...which, of course, you can't, otherwise the argument is, imo, logically non-sound. $\endgroup$ – DonAntonio Aug 7 '13 at 6:39

I think a much easier proof (which avoids the issues you are encountering) runs as follows. If $B$ is a matrix with linearly dependent rows, then we have an elementary row operation matrix such that $AB = B'$, where $B'$ is the same as $B$ except it has a row of zeroes in place of one of the linearly dependent rows. Clearly, $\det(B') = 0$. But by the multiplicativity of determinants, we have $\det(B') = \det(AB) = \det(A)\det(B)$. Since every elementary matrix $A$ has the property $\det(A) = 1$, we have $\det(B) = 0$.

(Of course, if your question is deliberately on this specific method of proof and this answer is not useful to you, I will happily delete it. My apologies if this is the case.)

  • $\begingroup$ My question is specifically about the proof in my question, but this is a good proof. I appreciate your input :) $\endgroup$ – Ataraxia Aug 7 '13 at 4:36
  • $\begingroup$ Fair enough! I will leave it up then. Thanks for taking the time to let me know :) $\endgroup$ – Alex Wertheim Aug 7 '13 at 4:38

Since a comment already answers your question about the validity of your proof, I'll just go ahead and talk about Cramer's Rule. :)

The second version of Cramer's Rule holds over any commutative ring with unity and is a common tool in some parts of the subject. (In fact, weaker conditions can probably be assumed about the base ring, but they're not coming to mind at the moment/wouldn't work with the proof below.)

If you haven't seen a proof of Cramer's rule, it's very short and doesn't change in this more general setting. It also just so happens that I wrote up the general statement/proof years ago as part of some notes, and I've adapted them for this post below.

If you aren't familiar with rings, just think of each $R$ in what follows as the real numbers (or some other field). Throughout what follows, $M_n(R)$ denotes the set of $n\times n$ matrices with entries in $R$. Also, $v^T$ denotes the transpose of $v$ for any $m \times n$ matrix.

The precise statement of Cramer's Rule is exactly what you'd expect:

Proposition: Let $R$ be a commutative ring with unity. Let $A = (C_1 \mid \cdots \mid C_n) \in \text{M}_n(R)$, $x = (x_1,\ldots,x_n)^T$ and $b = (b_1,\ldots,b_n)^T$ in $R^n$ such that $$ Ax = b. $$ Then for all $k \in \{1,\ldots,n\}$ $$ \text{det}(A_k) = \text{det}(C_1,\ldots,C_{k-1},Ax,C_{k+1},\ldots,C_n) = x_k \text{det}(A) $$ where $A_k = (C_1 \mid \cdots \mid C_{k-1} \mid Ax \mid C_{k+1} \mid \cdots \mid C_n)$.

Proof: Let $j \in \{1,\ldots,n\}$. Note the first equality is trivial. For the second, observe that $x = x_1 e_1 + \ldots + x_n e_n = \sum_{i = 1}^n x_ie_i$ where $e_1 = (1,0,\ldots,0),\ldots,e_n=(0,\ldots,0,1)$. Hence $$ Ax = \sum_{i = 1}^n x_i Ae_i = \sum_{i = 1}^n x_i C_i. $$ Thus by the multilinearity of the determinant \begin{align} \det(A_j) &= \det(C_1,\ldots,C_{j-1},Ax,C_{j+1},\ldots,C_n) \\ &= \det(C_1,\ldots,C_{j-1},\sum_{i = 1}^n x_i C_i,C_{j+1},\ldots,C_n) \\ &= \sum_{i=1}^n x_i \det(C_1,\ldots,C_{j-1},C_i,C_{j+1},\ldots,C_n).\qquad\qquad(\star) \end{align} An easy corollary of the fact that $\det$ is an alternating multilinear function is that $\det(M) = 0$ whenever the columns of $M$ are $R$-linearly dependent in $R^n$ (in particular, when $D_i = D_j$ for any $i,j$ with $1 \leq i < j \leq n$). Thus within $(\star)$, only the $j$-th summand $x_j \det(C_1,\ldots,C_{j-1},C_j,C_{j+1},\ldots,C_n)$ is nonzero. Hence \begin{align*} \det(A_j) = x_j \det(C_1,\ldots,C_{j-1},C_j,C_{j+1},\ldots,C_n) = x_j \det(A) \end{align*} as claimed.


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