Hello, This might sound like a dumb question but can one learn Linear Algebra before Calculus? Thank you.
Definitely. I did a course on it without any background in calculus as well. What's interesting is that some results in linear algebra would also be used in a course on differential equations.
My LA class had a few integrals and derivatives, but that as just for proofs, which we didnt need to know.
The only time I ever used integrals and derivatives in my Linear Algebra course was to prove some things about Differential Equations. These were just examples though; they were only applications of Linear Algebra. None of the proofs of Linear Algebra theory require any knowledge of Calculus, so except for some examples of applications, you will not need to know how to do derivatives or integrals at all (though some stuff that you might do later in the course may require some knowledge of the properties of the real and complex numbers)
As already stated, linear algebra is completely independent of calculus. Algebra deals with discrete quantities, not continous ones. You can successfully learn linear algebra without any knowledge of calculus. The only problem may arise in applications of linear algebra, such as viewing the integral as a linear map or differential equations. In any case, these are tiny fractions of the whole subject.
yes linear algebra is actually prerequisite for calculus done right. but as courses go in the us, we usually teach calculus as a collection of computational techniques, and then teach linear algebra more abstractly. so we teach linear algebra second because it si thought more difficult to understand "abstract ideas" than computational ones. in an ideal world linear algebra is taught first, then calculus is taught as an application of linear algebra. i.e. correctly done, calculus is the art of using linear algebra to deduce things about non linear functions. e.g. the inverse function theorem in calculus says: if the derivative of a smooth function f at a is an invertible linear map, then locally near a, f is an invertible smooth map.
The problem with that is that "f is an invertible linear map" doesn't give me any geometric intuition. On the other hand, consider the more "calculus" explanation: that the tangent "plane" at a does not contain any of the axes, so in some area around near a, the function must be a bijection. I can actually visualize this, intuitively understand it, and think of how potentially do a proof. Edit: Certainly, I could try picture every linear map as a geometric object, but I don't yet have that mathematical intuition, and it currently inhibits my ability to intuitively understand the abstract idea.