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I know the concept. I know the geometric explanation of the transformed volume. But the determinant still troubles me. The geometric explanation does not help when we are solving a non-vector problem, except that it could be assumed that the most different the columns in a matrix are from each other the greater the determinant, as the cube gets a higher volume when the vectors that form it are far away. That brings to the question: what really does the determinant depend on (ie. how large the numbers in the matrix are or how much the differ from each other)?

In my mind, the answer to this question should lead to an intuitive idea of the determinant different from the geometrical explanation.

Many thanks in advance.

chancar
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    What you are asking is hard to grasp, IMO. I'd like to draw your attention to this [ripened post on determinants](https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant) – Hanno Jan 25 '19 at 23:28
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    Thanks for the input. Basically I'm asking if the determinant depends on how different the columns of a matrix are form each other. – chancar Jan 25 '19 at 23:46
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    In order to answer this, you should explain/define, what " ... how different the columns of a matrix are form [typo?] each other" does signify, so that one can work on and with it. – Hanno Jan 25 '19 at 23:58

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To attempt to answer the question you asked in the comments, the answer is "no". The determinant is, roughly, a measure of how close a square matrix is to being singular, but doesn't really help us to tell how close the columns are to each other.

Take for example, the matrices $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ and } \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ They have the same columns, but their determinants differ by $2$. If order matters here, take instead, $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \text{ and } \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix},$$ which both have determinant $1$, have the same columns, but in a totally different order.

Another example: $$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \text{ and } \begin{pmatrix} 100 & 100 \\ -100 & -100 \end{pmatrix}$$ have the same determinant $0$, since both are singular. The columns, as you can see, are very different. The determinant of their difference is also $0$, in case you were thinking $\det(A - B) = 0$ might yield some measure of similarity between the columns.

It is true, however, that the determinant is continuous. Small changes in a matrix will result in small (possibly null) changes in the determinant. But knowing the determinant of two matrices tells you next to nothing about how close they are.

Theo Bendit
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