Could a function be a norm in one basis and not be a norm in the other one? I have to check if expression $||(x,y)||=\sqrt{(x+y)^2+y^2}$ is a norm for vectors from $R^2$. I have trouble with checking triangular inequality, but I have a hunch that is holds.
If we change basis to $u=x+y$ and $v=y$, our norm in that basis turns into $||(u,v)||=\sqrt{u^2+v^2}$, which certainly is a norm, thus previous expression has to be a norm in standart basis. Is that rigorous enough?
 A: Expanding the square of this norm
$$x^2+2xy+2y^2,$$
then writing it under the matrix form :
$$X^TAX=\pmatrix{x & y}\pmatrix{1&1\\1&2}\pmatrix{x\\y},$$
it suffices to test whether matrix $A$ is symmetric definite positive, which the case because its principal minors, $1$ and $1$, are positive (Sylvester's criterion).

Edit: In order to answer to the question in your title, this work has been done with respect to canonical basis of $\mathbb{R}$ but has an absolute character : it will remain a norm with repect to any any other basis, due to the fact that a basis change that I write backwards under the form $X \to X=BX'$ (with invertible $B$) yields the following change:
$$(BX')^TA(BX')= X'^T (B^TAB)X'$$
where $A$ has been changed into $B^TAB$ which is still symmetric definite positive (proof: $B$ being sym. def. positive, there exists $C$ such that $B=C^TC$ where $C$ is called the Cholesky factor of $B$; therefore $B^TAB=B^TC^TCB=(CB)^T(CB)$ ; such a matrix having the form $D^TD$ with $D$ invertible is itself symmetric definite positive).
A: If $\|.\|$ is a norm  and $A$ is invertible then $\|A.\|$ is also a norm. Invertibilty is needed. In this acse $(x,y)^{T} \mapsto (u,v)^{T}$ is invertible and hence $\|(x,y)\|$ is indeed a norm. Invertibility is important because you want $A(x,y)=(0,0)$ implying $(x,y)=(0,0)$.
[$\|A(u+v)\|\leq \|Au\|+\|Av\|$ and $\|A(cu)\|=\|cAu\|=|c|\|Au\|$. By invertibility $\|Au\|=0$ implies $u=0$].
A: Using your change of variables, consider $u_1 = x+y$ and $u_2 = z+w$. And, $||\cdot||$ denotes that norm, while $||\cdot||_2$ denotes the usual $2-$norm. Then, I reckon we can use your reasoning to deduce the following:
\begin{align*}
||(x,y) + (z,w)|| = ||(u_1 - y,y) + (u_2-w,w)|| = ||(u_1 - y + u_2 -w, y+w)|| =\\ \sqrt{(u_1+u_2)^2 + (y+w)^2} = ||(u_1,y) + (u_2,w)||_2 \leq ||(u_1,y)||_2 ||(u_2,w)||_2 =\\ \sqrt{(x+y)^2 + y^2} \sqrt{(z+w)^2 + w^2} = ||(x,y)|| \cdot ||(z,w)||
\end{align*}
where we use the fact that $||\cdot ||_2$ is a norm, hence it satisfies the triangle inequality.
