Determine if a set is a subspace of another one. What does the word fixed mean here?

Question

I am solving the following problem:

For k ∈ R (real number) and A ∈ R^(n x n) fixed, V = R^n and W = {x ∈ R^n | Ax = kx}

Now I apply the rule which is if you have a c ∈ R and u ∈ V then cu ∈ V I take c = -4 and x = (3,3) => cu = (-12, -12) then to see if c*u is ∈ V we have:

If i take n=2, then I take ux = (3,3), A = [2,2 --- 2,2], k = -5 and then I do the A (cu) = k (cu) and I get as a result (-48, -48) = (60, 60) which is not true.

Now my QUESTION is that what does that word "FIXED" mean in the sentence of the problem, does that change anything?

Answer 1

I think you're mixing up which order we take different vectors and numbers here.

Consider $W\subseteq V$ as a subset. We want to prove that it is a subspace. For it to be a subspace, we need three things:

1. $\vec 0\in W$. Clearly $A\vec 0 = k\vec 0$, so $\vec 0$ fulfills the requirement for being an element of $W$.
2. For any $\vec u, \vec v \in W$, we have $\vec u + \vec v\in W$. This we just check:$$A(\vec u + \vec v) = A\vec u + A\vec v = k\vec u+k\vec v = k(\vec u + \vec v)$$so $\vec u + \vec v$ fulfills the requirement of being in $W$ as long as $\vec u$ and $\vec v$ both do.
3. For any $c\in \Bbb R$ and $\vec v\in W$ we have $c\vec v\in W$. Again, we just check: $$ A(c\vec v) = c(A\vec v) = c(k\vec v) = k(c\vec v) $$so $c\vec v$ fulfills the defining requirement of $W$ as long as $\vec v\in W$.

With all three properties checked, this means that $W$ is indeed a subspace of $V$. It is known as "the eigenspace of $A$ corresponding to the eigenvalue $k$".