Why is linear Algebra so hard to understand?

# Why is linear Algebra so hard to understand?

because it isn't

@whereismyname

because of determinants.

Grab Linear Algebra Done Right. If that doesn't work for you then you are a certified brainlet.

@whereismyname

[math]\mathbf{\vec{i}\times\vec{j}=\vec{k}}[/math]

It's either [math]\mathrm{\vec{i}\times\vec{j}=\vec{k}}[/math] or [math]\mathbf{i\times j=k}[/math] you fucking mouthbreating drooling inbred m*tt.

It's not. If you're really having trouble understanding it, check out this guy's videos about it. He does a good job explaining the essence of linear algebra with great visuals and clear explanation. Also it's only 2 hours long, divided into 15 short clearly defined segments, so you don't have to binge it all at once.

https://www.youtube.com/watch?v=kjBOesZCoqc&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

It's just lines lol how is that hard?

@whereismyname

Tbh it's the epitome of understandable math. It's the only math we really can do

@whereismyname

if you struggle for algebra don't even dare to touch an analysis book

@Spazyfool

yeah but I didn't remember the command to put it that way. I always use bold anyway, looks classier to me.

Arrows are useful when handwritting since bolding is hard (altough I am switching to blackboard bold)

because you're taught to use it before you understand why you use it. after second year of physics degree it was piss easy.

@Spazyfool

i always struggled to convince myself to drop the dot, but it always felt a little weird without it

think of a 2d/3d example and everything is intuitive

@Harmless_Venom

determinants are bad

self studying from "Linear Algebra Done Right"

Not sure what's most disgusting, a brainlet finding linear algebra hard, or a brainlet posturing as a non-brainlet and regurgitating a meme-book pretending he's read even a single page from it.

omg i cant believe you all took the bait

perhaps you all have to reconsiderer who's the retard here

@farquit

never took linear algebra but how does 2 vectors equal a vector? shouldn't it be a scalar?

@Poker_Star

Depends on how you operate on the vectors.

If you add or subtract two vectors, you get a vector. If you take the dot product of two vectors, you get a scalar. If you take the cross product, you get a vector.

@Spazyfool

@farquit

hats denote unit vectors, boys

[math] \hat{\imath}, \hat{\jmath}, \hat{k}[/math]

I agree it can be tedious when you're first learning it, but it sure as shit isn't hard. Just stick with it and learn all the annoying definitions and try to remember how matrix multiplication works.

It is actually a subset of a much cooler area called group theory. It's only really important because it is the only intuitive algebraic system we have. Most of group theory is trying to represent abstract things as matrices so we can do linear algebra with them and prove shit about abstract algebra.

Pretty neato, but you need to learn linear algebra first so yeah... stick in there. Also, if you're not trolling, Veeky Forums is full people who like to act like they are high functioning autistic geniuses when in reality they only check the first part. Things are almost always difficult the first time you learn them.

Unless you actually are a brainlet. In which case, you should find out soon and should give up because linear algebra isn't actually that difficult.

@BlogWobbles

dot product

cross product

Just call them scalar and vector product, brainlet

@TalkBomber

Yeh, doing it is easy. I just dont understand how stuff like matrix multiplication works in the order it does for example.

@LuckyDusty

Row one times column one for row one, column one

Row two times column one for row two, column one

etc.

@FastChef

That book is great

I can tell you haven't read it because you don't think so

@Booteefool

Can't, got to learn it because the only "trade school" in my city is the community college and they want me to take a math test before letting me weld shit together.

@Stupidasole

Otherwise the result isn't meaningful.

Matrices are linear transformations in disguise, and vice-versa. Linear transformations are functions. When you multiply two matrices, you're composing two linear transformations.

Here's a very hot stack exchange post that goes into detail:

https://math.stackexchange.com/questions/31725/intuition-behind-matrix-multiplication

@LuckyDusty

Thats why you should learn proper linear algebra with a general theory of linear functions and it's representation instead of meme matrix algebra that should be thought at hs.

I've had two separate classes where linear algebra is taught and I still have no fucking idea how to find a basis for a vector space, or the usefulness of eigenvectors, or any of that other crazy shit.

why is linear algebra so hard to understand

@FastChef

Not the above poster, but I have read it and recommend it. Also recommend "Linear algebra done wrong" for another perspective. The Schaums book is good too.

Linear algebra is new stuff, you need to climb up one layer of abstractions at a time. Just like with all math.

LA is key to many fields - statistics, machine learning, quantum mechanics. Deal with it.

I got a C in high school so it wasn’t hard for me

I'm right there with you OP... I'm right there with you....

Does this make intuitive sense?

@5mileys

Also:

[eqn]\begin{align}\left[\begin{array}{c}3 \\ 2 \\ 0\end{array}\right] &= 3\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right] + 2\left[\begin{array}{c}0 \\ 1 \\ 0\end{array}\right] + 0\left[\begin{array}{c}0 \\ 0 \\ 0\end{array}\right] \\

\left[\begin{array}{c}2 & 1 & 2 \\ -1 & 3 & 2 \end{array}\right] \left[\begin{array}{ccc}3 \\ 2 \\ 0\end{array}\right] &= 3\left[\begin{array}{c}2 \\ -1 \end{array}\right] + 2\left[\begin{array}{c}1 \\ 3 \end{array}\right] + 0\left[\begin{array}{c}2 \\ 2 \end{array}\right]

\end{align}

[/eqn]

Matrices are just linear operators expressed in a different fancy way, which gives them some interesting properties. A 2x2 matrix with columns (a,b) (c,d) could be written as a function T:R^2 -> R^2, T(x,y) = (ax + cy,bx + dy), which you can also write as x(a,b) + y(c,d), which is a sum of two linear operators T_1:R^2 -> R^2, T(x,y) = x(a,b) and T_2:R^2 -> R^2, T(x,y) = y(c,d).

So as you can see, a matrix is just a collection of linear operators, where the n:th column representing a linear operator maps the n:th component of a vector on some line.

As for multiplying two or more matrices, it's actually just a function composition. Stick another linear function in T and see what happens.

@farquit

it is just notation

it is a non-issue unless you go full retard and do things like <insert example here>

I found it way easier than calculus.

They should teach it before calculus desu.

@whereismyname

It's hard if you are taught it the wrong way; same goes for any other subject.

Linear Algebra is taught wrongly way too often though, because they teach it by focusing on matrices rather than linear maps.

Rule of thumb for Linear Algebra is "Think with maps, compute with matrices".

@5mileys

Yes. Let f represent the left matrix and g represent the right matrix (given the usual base [math] \{e_i\} [/math] .

Then [math] g(e_i) [/math] is the i-th column of the right matrix.

The i-th column of [math] f \circ g [/math] is [math] f \circ g (e_i) = f(g(e_i)) = f( \text{ i-th column of the right matrix } ) [/math] .

@LuckyDusty

Matrix multiplication is defined this way so that it if you have:

the matrix A of a linear map g under the bases u and v

and

the matrix B of a linear map f under the bases v and w,

then the matrix of f ο g under the bases u and w can be acquired by BA.

That's the only reason it is defined this way.

@TalkBomber

It's misguided to call linear algebra a subset of group theory. They really serve totally different purposes. The fact that a vector space forms an abelian group under addition doesn't mean you're doing group theory. In fact, abelian groups are really closer to being linear algebra than vice versa.

everything becomes easier when you understand that matrices are really just functions if they are on the left, and a group of column vectors that serve as arguments a function (matrix) if they are on the right

@FastChef

Linear Algebra Done right is one of the best textbooks out there in general.

@Dreamworx

its ok but not one of the best textbooks in any way

hoffman&kunze is better for linear algebra for example

it's not hard, it's just your first encounter with real math, calculus, it's little bit simpler because you learn it in HS in almost all countries, and you can give it a physical meaning easily...

am I a brainlet for not understanding all the stupid vocab that goes along with linear algebra? I understand all the concepts, I just don't give a fuck about all the stupid 20+ things that all mean the exact same thing. I know linear algebra is just setting up a basis for a whole range of mathematics, but I really don't want to learn all of the terms. Also I'm taking the class online and I'm a piece of shit, so I do the bare minimum to pass the class.

@farquit

Third.

Also preferable [math] { \boldsymbol{i} \times \boldsymbol{j} = \boldsymbol{k} } [/math]

@likme

if you think there are "20+ things that mean the exact same thing", you don't understand as well as you think, my friend.