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consider the determinant $$ \begin{vmatrix} i & j & k \\ 1 & 0 & 2 \\ 0 & 2 & 5 \\ \end{vmatrix} $$ that comes when we calculate the cross product $(i+0j+2k)\times(0i+2j+5k)$.But the matrix contains both scalars and vectors,it is not an ordinary matrix $M_{n,n}(\mathbb F)$ where $\mathbb F$ is a field.Becuase here the second and third rows are real numbers but first one are vectors.Then should we define the meaning of this determinant separately?Because if we calculate it,we have,

$i$$ \begin{vmatrix} 0 & 2 \\ 2 & 5 \\ \end{vmatrix} $$-j$$ \begin{vmatrix} 1 & 2 \\ 0 & 5 \\ \end{vmatrix} $$+k$$ \begin{vmatrix} 1 & 0 \\ 0 & 2 \\ \end{vmatrix} $.Here each multiplication stands for scalar multiplication with a vector not multiplication between scalars.So if we do not define the meaning of determinants for such a matrix which is not HOMOGENEOUS in terms of its entries i.e. all elements not coming from a particular field,then how will it make sense.So how should we define determinant in this case?

Kishalay Sarkar
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    The above “determinant” is not really a determinant so much as a notational tool. – msm Oct 22 '19 at 05:07
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    It's just a mnemonic, it doesn't have to use a formal definition of determinant so long as it has an accepted and meaningful interpretation with the formula it has. The formula looks like the one for the determinant essentially because $({\bf a}\times{\bf b})\cdot{\bf c}=\det({\bf a},{\bf b},{\bf c})$ (which is "by definition" for some, I would argue "morally correct," definitions of the cross product). – anon Oct 22 '19 at 05:08
  • what is the link of $(a\times b).c=det(a,b,c)$ with the morally correct definition? – Kishalay Sarkar Oct 22 '19 at 05:14
  • What if I consider a determinant of order n with the first row of vectors. – Kishalay Sarkar Oct 22 '19 at 05:16
  • Setting $\mathbb{c}$ to be one of the basis vectors yields one of the three corresponding minor determinants within $\det({\bf a},{\bf b},{\bf c})$, but alternating in sign, so these minors must be the components of ${\bf a}\times{\bf b}$. The $n$-dimensional version of this would go the same way: if $\mathbf{v}\cdot\mathbf{x}=\det({\bf a}1,\cdots{\bf a}{n-1},{\bf x})$ then the alternating minor determinants of the matrix $({\bf a}1,\cdots,{\bf a}{n-1})$ will be the components of $\bf v$. – anon Oct 22 '19 at 05:16
  • Ok got it.Btw can anyone answer https://math.stackexchange.com/questions/3401021/a-different-proof-of-abels-theorem-of-power-series-and-of-uniform-convergence – Kishalay Sarkar Oct 22 '19 at 05:18

1 Answers1

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Generally, we don't define determinants in this way. This method for computing the cross product something known as a mnemonic: a pattern used for remembering something complex. It's not a coincidence that it works, of course, but at the same time, it's not a real determinant.

If you wanted to define a more general non-homogeneous determinant, then you'll have to look at the reasons why it's well-defined. Between scalars $\Bbb{R}$ and vectors in $\Bbb{R}^3$, we have a multiplication operation $$a \cdot (x_1, x_2, x_3) = (ax_1, ax_2, ax_3).$$ The scalar multiplication operation is inherently asymmetric, so we can also symmetrise the operation by defining $v \cdot a = a \cdot v$ for all $v \in \Bbb{R}^3$ and $a \in \Bbb{R}$, which forces a kind of commutativity.

This scalar multiplication operation acts sort of associatively with scalar multiplication on $\Bbb{R}$, specifically \begin{align*} b \cdot (a \cdot v) &= (ab) \cdot v \\ b \cdot (v \cdot a) &= (b \cdot v) \cdot a \\ v \cdot (ba) &= (v \cdot b) \cdot a . \end{align*} We also have distributivity laws \begin{align*} a \cdot(u + v) &= (a \cdot u) + (a \cdot v) \\ (a + b) \cdot v &= (a \cdot v) + (b \cdot v). \end{align*} All in all, this builds up a partial ring structure on $\Bbb{R}^3 \cup \Bbb{R}$ where not every element of the set can add or multiply to each other, but the associativity, commutativity, and distributivity all hold. Plus we have some partial additive identities (i.e. $0 \in \Bbb{R}$ and $(0, 0, 0) \in \Bbb{R}^3$ each act as identities for the two parts of the set $\Bbb{R} \cup \Bbb{R}^3$), and associated additive inverses.

It's not a ring, because we don't get closure of addition and multiplication. We cannot multiply two vectors, and we cannot add a scalar to a vector. But, provided we avoid performing these operations, the structure is very much ring-like.

Thus, we can define the determinant of a matrix with elements from $\Bbb{R}^3 \cup \Bbb{R}$, so long as we ensure that no vectors product with vectors, or add to scalars. The given determinant is fine, evidently, but these two determinants are not:

$$\begin{vmatrix} i & j & 1 \\ 1 & 2 & 1 \\ -1 & 0 & 3 \end{vmatrix}, \quad \begin{vmatrix} i & 2 & 0 \\ 3 & j & -1 \\ 0 & 2 & k \end{vmatrix}.$$

Theo Bendit
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  • I have yet not studied RIng theory,so I am having a bit problem in understanding completely,but I think I have partially got what you want to say in this answer... – Kishalay Sarkar Oct 22 '19 at 06:29
  • Can you help me with this question too https://math.stackexchange.com/questions/3401021/a-different-proof-of-abels-theorem-of-power-series-and-of-uniform-convergence – Kishalay Sarkar Oct 22 '19 at 06:31
  • @KishalaySarkar Essentially, a ring is a set with an addition and multiplication operation with various nice properties (the $+$ is often a lot nicer than the $\times$). It's usually enough to define polynomials, and hence determinants of matrices. Unlike a usual ring, not every two elements can add or product with each other, so we just have to be a little careful about whether a given matrix has a determinant or not. – Theo Bendit Oct 22 '19 at 10:35
  • @KishalaySarkar As for your other question, it's a little hard to do while it's closed. You also haven't presented the proof that you don't like, or explained what about it you don't like. If you write those into the question, then I'll vote to reopen it, and see what I can do. – Theo Bendit Oct 22 '19 at 10:37
  • I will do it.I will mention the proof there. – Kishalay Sarkar Oct 22 '19 at 12:44
  • See herehttps://math.stackexchange.com/questions/2715379/proof-abels-uniform-convergence-test It is the proof of Abel's theorem I am talking of. – Kishalay Sarkar Oct 22 '19 at 12:49
  • @KishalaySarkar I voted to reopen. We'll see if four others agree it should be reopened. – Theo Bendit Oct 22 '19 at 12:52
  • We may discuss that problem in chatbox. – Kishalay Sarkar Oct 22 '19 at 12:54