I am self reading Linear Algebra Done Wrong by Treil. In that, I am currently studying Isomorphism and find it hard to understand what it really means.
An invertible linear transformation $A: V\to W$ is called an isomorphism.
What does this mean really?
Theorem 6.6 Let $A: V\to W$ be an isomorphism and let $\mathbf{v_1}, .. \mathbf{v_n}$ be basis in $V$. Then the system $A\mathbf{v_1}, .., A\mathbf{v_n}$ is basis in $W$.
What does the following mean
$M_{m\times n}^{\mathbb{F}} \cong \mathbb{F}^{mn}$
Does the above line mean this: Suppose I take vector space as $M_{2\times 2}^{\mathbb{R}}$. Its basis are chosen as $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\0&1\end{pmatrix}$. Then above means that it is isomorphic to $\mathbb{R^4}$. Basis in $\mathbb{R}^4$ are of form $\begin{pmatrix}1&0&0&0\end{pmatrix}^T$ etc.
So how can any matrix (isomorphism) $A$ transform $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ to $\begin{pmatrix}1&0&0&0\end{pmatrix}^T$.
I think I am going fundamentally wrong somewhere, but I don't understand, maybe in concept of vector space or isomorphism. Please help me as I am self studying and this course is not available in college to us.
That theorem is saying that if $A$ is a vector space isomorphism from $V$ to $W$ and if $\{v_1,\ldots,v_n\}$ is a basis of $V$, then $\{A(v_1),\ldots,A(v_n)\}$ is a basis of $W$.
And, yes, $M_{2\times2}^{\mathbb R}$ and $\mathbb R^4$ are ismorphic. Just take the map$$\begin{array}{rccc}A\colon&M_{2\times2}^{\mathbb R}&\longrightarrow&\mathbb R^4\\&\begin{bmatrix}a&b\\c&d\end{bmatrix}&\mapsto&(a\ \ b\ \ c\ \ d)^T.\end{array}$$It is a vector space isomorphism.