Question about isomorphisms

Question

Let $A=\operatorname{span}\big(\begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}, \begin{bmatrix} h \\ 0\end{bmatrix}\big)$ and $B=\operatorname{span}\big(\begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1\end{bmatrix}\big)$, then are $A$ and $B$ isomorphic?

I think since $\dim(A)=2$ and $\dim(B)=2$, then $A$ and $B$ are isomorphic. I don't know if this is true for $h=0$ because if we have the linear combination $a_1\begin{bmatrix} 1 \\ 0\end{bmatrix}+a_2\begin{bmatrix} 0 \\ 1\end{bmatrix}+a_3\begin{bmatrix} h \\ 0\end{bmatrix}=\begin{bmatrix} 0 \\ 0\end{bmatrix}$, then $a_3$ is not necessarily zero if $h=0$.

So $A$ and $B$ are isomorphic for $h\neq 0$?

Please give some hint or ideas.

Answer 1

Note that a basis for $B$ is $$\beta_{B}=\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\} \implies \dim(B)=2$$

also you can see that the set $$\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} , \begin{pmatrix} h \\ 0 \end{pmatrix} \right\}$$ is dependent linearly so, the set $\beta_{A}$ is a basis for $A$ where $$\beta_{A}=\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\} \implies \dim(A)=2$$ Therefore, $A$ is isomorphic to $B$ forall $h \in \mathbb{R}$.

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Note: You can remove (delete) linearly dependent vectors from a span, and the set remains the same as before removing the linearly dependent vector. This can be formally proven.

Answer 2

Yes, they are isomorphic since, for any $h$, $\left[\begin{smallmatrix}h\\0\end{smallmatrix}\right]$ is a linear combination of the other two vectors.