Suppose $ M, N $ are subspaces of $ V$. Define $T : M × N → V$ by $T((x, y)) = x + y.$
I proved that $T$ is a linear map. Now I'm trying to show that $ker(T) ⊆ (M ∩ N) × (M ∩ N)$.
Here's my proof so far:
Suppose $(x,y)∈ker(T)$.
So $T(x,y) = x+y=0∈V$. (since $T$ maps to $V$)
And $ker(T)$ is contained in $M × N$, so $(x,y)∈M × N$.
So, $x∈M$ and $y∈N$, and $x+y = 0$, then $x∈N$ and $y∈M$.
Since $x,y∈M$ and $x,y∈M$, we have that $x,y∈(M ∩ N)$.
So we can write $x∈(M ∩ N) and y∈(M ∩ N) $
So, $x∈(M ∩ N) × y∈(M ∩ N) $, by definition of cartesian product.
So, $(x,y)∈(M ∩ N) × (M ∩ N) $.
Thus, $ker(T)$ is contained in $(M ∩ N) × (M ∩ N)$.
I'm also trying this problem: Show that $ker(T) ∼= M ∩ N$.
I need to give an isomorphism. So, I know I need to prove it's injective, surjective and linear. So far I have some thoughts on this one. To prove it's injective I was thinking that I need to show that $ker(T)$ is in $M ∩ N$. For surjective, I'm stuck. For linear, I thought that it was given since $T$ is linear map, but I'm not sure if it's enough to say that.
Your problematic line doesn't even make sense, but if you meant that $(0,0)$ is the only element in the kernel, it's not true in general.
So we have $x\in M,\ y\in N$ and $x+y=0$.
Then $x\in N$ and $y\in M$ as well, can you see why?
For the other question, verify that the projection $\ker T\to M\cap N,\ (x,y)\mapsto x$ is an isomorphism (e.g. find its inverse).