But that still won't get me the answer of 3x^2 y + y^3 = A. will it? = 2xy + h'(y)$ to $f_{x}(x,y)$ instead of to $f_{y}(x,y)$, which is why it yielded the wrong answer.$)
Oh so I think I'm close... = 2xy$. Partially integrating this wrt $x$ yields $f(x,y) = x^{2}y + h(y)$ (you did $xy^{2}$ instead of $x^2 y$).$)
Oh I missed you saying this.
But I do not believe that the converse need be true.
He was more asking about the other direction, i.e. whether he could conclude from T(something) = 0 that something = 0. This is guaranteed if and only if the linear map T is a one-to-one map. If T is not one-to-one, it'll have a non-trivial kernel.
I see, really nice
 = \mathbf{0}$ on the RHS, then we have $\alpha_{1}T\left(\mathbf{v}_{1}\right) + \cdots + \alpha_{n} T\left(\mathbf{v}_{n}\right) = \mathbf{0}$. So we have expressed $\mathbf{0}$ as a non-trivial linear combination of the elements of $R$ (since not all $\alpha_{j}$ were 0), whence $R$ is a linearly dependent set. This completes the proof.$)
Not really, because to conclude just from T(something) = 0 that something = 0 (which is what you seemed to want to do), we'd have to know T was one-to-one, but T can be any linear map (and the result holds regardless). You'd end up having to do it by contradiction I think (which makes it similar to contrapositive).I see, really nice
Is there a way to do it in a forwards manner though? Because I asked it for a friend and I'm not sure if, in their course, they need to consider the contrapositive
If it's probably too hard then maybe don't worry
If you have two unknowns in an equation let one of them equal to a variable.Finding the basis for the kernel of this matrix (already row reduced):
The answers show (sorry you can barely see):
My understanding to solve this you make the non leading column = lambda, so e.g. x3 = lambda. Then you solve the equations.
But what happens when you have multiple non-leading columns like here? I don't understand what they've done in the answers, and why they have made two things = lamda, and one other = ,u.
I remember doing proof by contradiction. I also remember an interesting discussion on mathstackexchange about proof by contradiction and contrapositive.
It may just be unfair that we know a method some normal maths students wouldn't know but let's not take the risk.
Nah they're definitely allowed. It's only the unfairness factor in that not everyone would think in that direction due to zero exposure to it.