3

I am trying to understand the geometric interpretation of determinant and I know it gives the area of the parallelepiped formed by the column vectors. For simplicity let's just talk about 2-D case.

I can understand why determinant gets scaled if I scale one column by a scalar and it's also easy to prove geometrically that If I scale one side of a parallelogram by a scalar λ then the area will also get scaled by λ.

What I can't understand is that when I do some elementary column operation (lets say add one column to other) then how the area remains the same? I tried to view it geometrically and I could see that when I added one vector(column) to other vector(non-zero column) then the resultant vector would always be larger. So why doesn't the area of the parallelogram increases if one side gets increased? I could imagine that though one side gets increased but angle between vectors gets decreases but how geometrically you would prove that the perpendicular would always remain same?

For 2-D case I could prove it analytically but I can not view it in my head:

Area of parallelogram = ||v1|| * ||v2 - proj(v2) on v1 ||

Lets say I add 2*v1 to v2 so now

Area of parallelogram = ||v1|| * || v2 + 2*v1 - proj(v2+2*v1) on v1|| = ||v1|| * || v2 + 2*v1 - proj(v2) on v1 - 2*v1|| = ||v1|| * ||v2 - proj(v2) on v1 ||

user_3pij
  • 185
  • Please use mathjax to format the question. – sai-kartik May 15 '20 at 20:16
  • Essentially, if you have a parallelogram, with sides parallel to $v_1$ and $v_2$, with $v_i$, for $i = 1,2$, linearly independent, well if you add to $v_2$ a multiple of $v_1$, then the new parallelogram will still have $v_1$ as one side, and the height of the parallelogram will remain the same. So the area will remain the same. – Malkoun May 15 '20 at 20:18
  • Yeah it's true for some cases but height does change in many cases. – user_3pij May 15 '20 at 21:57

0 Answers0