Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This statement is from the book of Winitzki Linear Algebra via Exterior Products. (Section 3.4, page 123) Let $V$ be finite dimensional vector space, $\dim(V)=N$. The determinant of the matrix $v_{ij}$ is a linear function of each of the vectors $\{\mathbf{v}_i \}$. Consider $\det(v_{ij})$ as a linear function of the first vector, $\mathbf{v}_1$; this function is a covector that we may temporarily denote by $\mathbf{f}_1^{*}$. Show that $\mathbf{f}_1^{*}$ can be represented in the dual basis $\{\mathbf{e}_j^{*} \}$ as $$ \mathbf{f}_1^{*}=\sum_{i=1}^N(-1)^{i-1} B_{1i}\mathbf{e}_i^{*}, $$ where the coefficients $B_{1i}$ are minors of the matrix $v_{ij}$, that is, determinants of the matrix $v_{ij}$ from which row 1 and column $i$ have been deleted.

Solution. Consider one of the coefficients, for example $B_{11}\equiv\mathbf{f}_1^{*}(\mathbf{e}_1)$. This coefficient can be determined from the tensor equality $$ \mathbf{e}_1\wedge\mathbf{v}_2\wedge\ldots\wedge\mathbf{v}_N=B_{11}\mathbf{e}_1\wedge\ldots\wedge\mathbf{e}_N.\qquad(1) $$

Edit: the next sentence is

"We could reduce $B_{11}$ to a determinant of an $(n-1)\times(N-1)$ matrix if we could cancel $\mathbf{e}_1$on both sides of equality."

I don't understand the reason of this equality (1). Accepting this equality to be valid, the remainder part of the proof is correct. Please, give some details.

share|cite|improve this question

This is fairly straightforward to check for say a $3\times 3$ matrix, but the details become messy to do this in general. This should give you enough idea of how to do the general case to convince you it is true though:

Suppose $N=3$. Suppose $v_i=\left(\begin{matrix} a_{i1} \\ a_{i2} \\ a_{i3}\end{matrix}\right)$. Then $B_{11}=det(e_1, v_2, v_3)=a_{22}a_{33}-a_{23}a_{32}$, i.e. the determinant of the lower right $(N-1)\times (N-1)$ minor.

Now just expand out in terms of basis vectors and use standard properties of wedges to see the identity.

$v_2=a_{22}e_2+a_{23}e_3$ and $v_3=a_{32}e_2+a_{33}e_3$,

Thus $e_1\wedge v_2 \wedge v_3 = e_1 \wedge (a_{22}e_2+a_{23}e_3)\wedge (a_{32}e_2+a_{33}e_3) \\ = (a_{22}e_1\wedge e_2 +a_{23}e_1\wedge e_3)\wedge (a_{32}e_2+a_{33}e_3)\\ = (a_{22}a_{33}-a_{23}a_{32})e_1\wedge e_2 \wedge e_3 \\ = B_{11} e_1 \wedge e_2 \wedge e_3$

Edit: I misunderstood the question:

Checking $B_{11}$ is the correct minor directly generalizes from what I wrote above too. There seems to be no need to go through that wedge equality (of course using the determinant interpretation it is equivalent). The definition of $f_1^*(e_1)$ is exactly placing $e_1$ in the first column and taking the determinant. So in general $B_{11}=f_1^*(e_1)=det(e_1, v_2, \ldots , v_N)$, thus computing the determinant using the cofactor expansion down the first column, the only non-zero entry is the first one and hence $B_{11}=(-1)^2$(times the correct minor).

share|cite|improve this answer
we want to prove that $B_{11}$ is the minor. Here you used this to prove the wedge equality. – vesszabo Aug 5 '12 at 7:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.