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!

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## Comments

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TopNewestHere'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.

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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.

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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?

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