The Story about The Determinant

Hi folks,

I always had some difficulties understanding the concept of determinant. Whenever I read about it, and that was usually from Linear Algebra books, it was always described with a lot of "mathematical formality" and abstractness, and although I could follow and understand the definition and properties, the concept generally seemed to me like coming out of the blue.

"The best intuition" I have about determinant is that it measures how much does space get scaled by certain transformation. This intuition helped me a lot to understand some other concepts involving determinant, for example deriving the proof and explaining to myself why non-trivial solutions of Homogeneous System of Linear Equations occur when determinant of a matrix of the corresponding transformation equals 0. But, although I can prove this intuition holds for 2-dimension case, I find no way to generalize it for nn dimensions.

I feel that concept of determinant is really important to understand because it shows up in many branches of mathematics. I would really appreciate if someone could tell me a story about the determinant. I say story because it doesn't need to be highly mathematical and should aim to answer the following two questions:

  • Why did mathematicians introduced the concept of determinant? Why they needed it at the first place?

  • Based on that, how they derived the formula and generalized it for any number of dimensions?

Thanks in advance!

Note by Uros Stojkovic
3 years, 1 month ago

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I don't have a story to tell, but I do have an anecdote. Suppose you have a matrix and a vector, such that the matrix pre-multiplied by the vector yields a vector of zeroes:

(abcd)(xy)=(00)\begin{pmatrix} a&b\\ c&d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

Suppose none of the elements (a,b,c,d,x,y)(a,b,c,d,x,y) are zero, for simplicity. What must the relationships between a,b,c,da,b,c,d be?

ax+by=0cx+dy=0a x + b y = 0 \\ c x + d y = 0

Rearranging gives:

y=abx y = -\frac{a}{b} x

Plugging into the second equation:

cxdabx=0cdab=0bcda=0=adbcc x - d \frac{a}{b} x = 0 \\ c - d \frac{a}{b} = 0 \\ b c - d a = 0 = a d - b c

The quantity (adbc)(a d - b c) is well-known as the two-by-two matrix determinant. The determinant must be zero for the pre-multiplication of a vector by a matrix to yield a vector of zeros. It's almost as though the determinant is the matrix equivalent of a scalar magnitude.

Steven Chase - 3 years, 1 month ago

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Proof which I derived for any n×nn \times n matrix and which I presented in my work.

"I will briefly prove here that non-trivial solutions of any homogeneous system of linear equations written in vector form Ax=0A\vec{x} = 0 has non-trivial solutions when det(A)=0.\det(A) = 0. Let ithi^{th} column entries of AA form vector uiu_{i} so that: A=[a11a12a1na21a22a2nan1an2ann]=[u1u2un]A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}= \begin{bmatrix} | & | & & | \\ u_{1} & u_{2} & \cdots & u_{n} \\ | & | & & | \\ \end{bmatrix}

Then, by definition of matrix multiplication, we have:

Ax=x1u1+x2u2++xnun=θ,A\vec{x} = x_{1}u_{1} + x_{2}u_{2} + \cdots + x_{n}u_{n} = \theta,

where xix_{i} is a scalar and, by conditions, at least one of them is non-zero. The above expression indicates that vectors u1,u2,,unu_{1}, u_{2}, \cdots, u_{n} are linearly dependent. Ultimately, the determinant of a matrix formed by linearly dependent vectors is known to be zero. This comes from the fact that linear dependence implies that we can write some vector uiu_{i} as a linear combination of other n1n-1 vectors ie. that uiRku_{i} \in \mathbb{R}^{k} where kn1k\leq n-1 . Hence, the transformation defined by that matrix would squish the space to a lower dimension which is captured by zero determinant."

This comes from the intuition I mentioned: determinant is the quantitative measure of space transformation defined by a certain matrix.

You wrote about R2\mathbb{R}^{2} case when calculations are pretty straightforward, but what about R3\mathbb{R}^{3}, R4\mathbb{R}^{4}, Rn\mathbb{R}^{n} space? I guess that the most elegant way would be to derive a proof using induction. But I'm wondering if that's actually possible?

Uros Stojkovic - 3 years, 1 month ago

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Here's my idea of generalization of the formula by induction. Assume we have nn linear homogeneous equations and nn variables: a11x1+a12x2++a1nxn=k=1na1kxk=0,a21x1+a22x2++a2nxn=k=1na2kxk=0,                an1x1+an2x2++annxn=k=1nankxk=0.\begin{aligned}a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} &= \sum_{k = 1}^{n}a_{1k}x_{k} = 0, \\ a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} &= \sum_{k = 1}^{n}a_{2k}x_{k} = 0, \\ \vdots ~~~~~~~~~~~~~~~~&\vdots \\ a_{n1}x_{1} + a_{n2}x_{2} + \cdots + a_{nn}x_{n} &= \sum_{k = 1}^{n}a_{nk}x_{k} = 0.\end{aligned} This can be also represented in matrix form as Ax=oA\vec{x} = o. We can reduce this system to system of n1n-1 equations by substitution. Respectively: xn=1annk=1n1ankxk.x_{n}=-\frac{1}{a_{nn}} \sum_{k = 1}^{n-1}a_{nk}x_{k}. Now, plugging in back to the jthj^{\text{th}} equation (1jn11 \leq j \leq n-1), we get: annk=1n1ajkxkajnk=1n1ankxk=0k=1n1(annajkajnank)xk=0\begin{aligned} a_{nn}\sum_{k = 1}^{n-1}a_{jk}x_{k} - a_{jn}\sum_{k=1}^{n-1}a_{nk}x_{k} = 0 \\ \sum_{k = 1}^{n-1}\left(a_{nn}a_{jk} - a_{jn}a_{nk} \right )x_{k} = 0\end{aligned}

We are now left with one less equation and one less variable, but with new transformation matrix BB such that bjk=annajkajnankb_{jk} = a_{nn}a_{jk} - a_{jn}a_{nk}. The idea is to prove general formula for evaluating determinant of n×nn \times n matrix by building upon the basis cases which are simple and intuitive.

Uros Stojkovic - 3 years, 1 month ago

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Read up the history of the determinant; it is actually quite interesting. In any case, I think determinants arose as a consequence of solving a system of linear equations; if you have a system of nn linear equations in the variables x1,x2,,xn x_1, x_2, \dots, x_n and expressed them in terms of y1,y2,,yn y_1, y_2, \dots, y_n then inverting the system naturally gives you this quantity called a determinant. The area/volume/hypervolume story came much later, dating back to Euler and Tartaglia, but formalised by Cayley & Menger. If you want to see how Euler derived his story of volumes in terms of determinants, have a look at Dorrie's book 100 Great Problems in Elementary Mathematics (apologies in advance if I get the name wrong).

A Former Brilliant Member - 2 years, 11 months ago

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Wonderful Youtube channel with intuitive videos. Here is their video on the determinant.

Andrei Li - 2 years, 11 months ago

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Actually, I watched that video and all other from Linear Algebra series. I'm familiar with Grant Sanderson's content and I really enjoy it. However, except for the intuition I mentioned in my note above, the video doesn't offer an answer to my second question, namely, how is the formula derived for nn dimensions.

Uros Stojkovic - 2 years, 11 months ago

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Oh, okay! No problem.

Andrei Li - 2 years, 11 months ago

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