For whatever reason, my mind wants to organize determinants and other matrix-related things (like bases) geometrically. But I can't wrap my mind around how that would be possible.

Is there actually a geometric way to look at these relationships? Or am I just out of luck?

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    The determinant of $A$ is the signed volume of the parallelopiped spanned by the columns of $A$. – littleO Jun 06 '16 at 03:51
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    Up to sign, the determinant is the volume of the parallelopiped generated by the columns. The sign is flipped whenever you exchange the order of two columns. – Ian Jun 06 '16 at 03:51
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    Here is a thread that will help: http://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant – Bernard W Jun 06 '16 at 03:57

1 Answers1


Yes, there is. The determinant of a matrix $A$ is the (signed) volume of the parallelopiped which is bounded the vectors whose columns are $A$.

Let's first consider the 2D case, and then extend to $n$-d.

$$ A = \begin{bmatrix} v_0 & w_0 \\ v_1 & w_1 \end{bmatrix} \\ |A| = v_0 w_1 - v_1w_0 $$ Think of $\hat v = (v_0, v_1)$ and $\hat w = (w_0, w_1)$. Now, the cross product of $v$ and $w$, which is

$$ |v \times w| = |v| \cdot |w| \sin(\theta) $$

which is the area of the parallelogram with adjacent sides of length $|v|$ and $|w|$, with the sides having an angle of $\theta$ between them.

This immediately gives us some nice geometric understanding of determinant properties:

  1. a determinant is 0 if the volume generated by the vectors is 0. One way for this to happen is to pick two of the same vectors. Hence, there is $0$ volume "between" the two vectors

  2. The Jacobian (if you've studied multivariable calculus) gives you the "local change of volume" which we need to scale the integral correctly. Hence, the Jacobian pops up as an integration factor

  3. A determinant is maximum if all your column vectors are orthogonal to each other - that is, you have an eigenbasis (which diagonalized your matrix)

Siddharth Bhat
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