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 \(n\) 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
4 weeks ago

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Here's my idea of generalization of the formula by induction. Assume we have \(n\) linear homogeneous equations and \(n\) variables: \[\begin{align}a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} &= \sum_{k = 1}^{n}a_{1k}x_{k} = 0, \\ a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} &= \sum_{k = 1}^{n}a_{2k}x_{k} = 0, \\ \vdots ~~~~~~~~~~~~~~~~&\vdots \\ a_{n1}x_{1} + a_{n2}x_{2} + \cdots + a_{nn}x_{n} &= \sum_{k = 1}^{n}a_{nk}x_{k} = 0.\end{align}\] This can be also represented in matrix form as \(A\vec{x} = o\). We can reduce this system to system of \(n-1\) equations by substitution. Respectively: \[x_{n}=-\frac{1}{a_{nn}} \sum_{k = 1}^{n-1}a_{nk}x_{k}.\] Now, plugging in back to the \(j^{\text{th}}\) equation (\(1 \leq j \leq n-1\)), we get: \[\begin{align} a_{nn}\sum_{k = 1}^{n-1}a_{jk}x_{k} - a_{jn}\sum_{k=1}^{n-1}a_{nk}x_{k} = 0 \\ \sum_{k = 1}^{n-1}\left(a_{nn}a_{jk} - a_{jn}a_{nk} \right )x_{k} = 0\end{align}\]

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

Uros Stojkovic - 3 weeks, 4 days 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:

\[\begin{pmatrix} a&b\\ c&d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]

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

\[a x + b y = 0 \\ c x + d y = 0\]

Rearranging gives:

\[ y = -\frac{a}{b} x\]

Plugging into the second equation:

\[c x - d \frac{a}{b} x = 0 \\ c - d \frac{a}{b} = 0 \\ b c - d a = 0 = a d - b c \]

The quantity \((a d - b c)\) 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 weeks, 6 days ago

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Proof which I derived for any \(n \times n\) 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 \(A\vec{x} = 0\) has non-trivial solutions when \(\det(A) = 0.\) Let \(i^{th}\) column entries of \(A\) form vector \(u_{i}\) so that: \[A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}= \begin{bmatrix} | & | & & | \\ u_{1} & u_{2} & \cdots & u_{n} \\ | & | & & | \\ \end{bmatrix}\]

Then, by definition of matrix multiplication, we have:

\[A\vec{x} = x_{1}u_{1} + x_{2}u_{2} + \cdots + x_{n}u_{n} = \theta,\]

where \(x_{i}\) is a scalar and, by conditions, at least one of them is non-zero. The above expression indicates that vectors \(u_{1}, u_{2}, \cdots, u_{n}\) 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 \(u_{i}\) as a linear combination of other \(n-1\) vectors ie. that \(u_{i} \in \mathbb{R}^{k}\) where \(k\leq n-1 \). 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 \(\mathbb{R}^{2}\) case when calculations are pretty straightforward, but what about \(\mathbb{R}^{3}\), \(\mathbb{R}^{4}\), \(\mathbb{R}^{n}\) 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 weeks, 5 days ago

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