Questions regarding linear transformations and their associated matrices Lets say we are given a linear transformation $T: V \to W$ where $V = \mathbb{R^n}$ and $W = \mathbb{R^m}$.
I have learned that $\mathbb{R^n} = R(M) \oplus N(M)$, where R(M) is the row space of the matrix M and N(M) is the null space of the matrix M, and matrix M is the matrix associated with the linear transformation T.
I understand how the input space $\mathbb{R^n}$ contains the null space, as we input some vectors and they are mapped to the zero vector. But I do not understand how the input space contains R(M). What part does R(M) play in the input of T?
Another question. Is the output space $\mathbb{R^m}$ made up of only C(M) or C(M) and $N(M^T)$, where $N(M^T)$ is the left null space of M. I am asking this because I have learned that for a matrix $\in \mathbb{R^{m \times n}}$, the right space $\mathbb{R^n} = R(M) \oplus N(M)$ and the left space $\mathbb{R^m} = C(M) \oplus N(M^T)$. So is it the same with linear transformations? If $\mathbb{R^m} = C(M) \oplus N(M^T)$, then what part does $N(M^T)$ play in the output space or output of T?
My third and final question: A proof I saw that for a linear transformation from $\mathbb{R^n}$ to $\mathbb{R^m}$ to be bijective, m must equal n.
The proof went like this: Since bijective implies surjective, the input space must be atleast as greater than the output space, so n $\geq$ m and then since bijective implies injective, the output space must be atleast as greater than the input space, so m $\geq$ n and so m=n.
I have some questions about this. m and n are the dimensions of here of our vector spaces. The input and output vectors can be infinite, how are we even relating the dimensions to number of inputs/outputs? And how can we even compare inputs from $\mathbb{R^2}$ to outputs in $\mathbb{R^3}$ for example, both contain infinite vectors. I hope someone can clear this for me or provide an alternative proof.
Thank you!
 A: You have picked up on some subtleties that people sometimes gloss over (+1).
When you represent a linear transformation by a matrix, you first pick a basis for both the input and output spaces.
With respect to the basis of the input space, we obtain an inner product:$$
\sum_{i=1}^n a_i\vec{e_i}\cdot \sum_{j=1}^n b_j\vec{e_j}= \sum_{i=1}^n a_ib_i
.$$
You are right that a row of the matrix $M$ is not naturally an element of the input space.  It is in fact an element of the dual of the input space.  That is, any row defines a linear map $\mathbb{R}^n\to \mathbb{R}$.
The inner product allows us to identify $\mathbb{R}^n$ with its dual.  Any vector $\vec{v}\in \mathbb{R}^n$ may be identified with the linear map: $$\vec{u}\mapsto \vec{v}\cdot \vec{u}.$$
If you unpack what this means, all we are doing here is identifying a row vector with its transpose.
The null space is then by definition just the orthogonal complement of the row space, so it makes sense that their direct sum is the whole space.
Regarding the second question: for any transformation $T:{V\to U}$ we have the dual map $T^*\colon U^*\to V^*$.  If $T$ is represented by a matrix $M$ with respect some basis, then $T^*$ will be represented by the matrix $M^T$ with respect to the dual basis.  Thus your second question reduces to the first.
Finally, in the proof that only vector spaces of the same dimension can have bijective linear maps between them, we do not count the elements in a vector space, but rather the number of elements in certain finite sets.
If we take a linearly independent set in $\mathbb{R}^n$ and map it via an injective linear map to $\mathbb{R}^m$, the resulting set will still be linearly independent.  We have a linearly independent set of size $n$ in $\mathbb{R}^n$, so we obtain one in $\mathbb{R}^m$ telling us that $n\leq m$.
Similarly, if we take a spanning set in $\mathbb{R}^n$ and map it via a surjective linear map to $\mathbb{R}^m$, the resulting set will still be spanning.  We have a spanning set of size $n$ in $\mathbb{R}^n$, so we obtain one in $\mathbb{R}^m$ telling us that $m\leq n$.
A: Regarding row and column spaces, the more fundamental identity is that if $U$ is a subspace of a finite-dimensional inner product space $V$, then $$V = U \oplus U^{\perp},$$
where $$U^{\perp} = \{v \in V : (v, u) = 0 \text{ for all } u \in U\}.$$
Suppose $V, W$ are finite-dimensional inner product spaces over $\mathbb{F}$ and $T \colon V \to W$ is linear. Then there is the adjoint of $T$, which is the unique linear map $T^* \colon W \to V$ satisfying $(Tv, w) = (v, T^*w)$ for all $v \in V, w \in W$. Proof of the existence and uniqueness of the adjoint $T^*$ is simple. Let $\{e_1, \dots, e_n\}$ be an orthonormal basis of $V$ and let $\{f_1, \dots, f_m\}$ be an orthonormal basis of $W$. Let $A$ be the matrix representation of $T$ with respect to these bases. Suppose $T^*$ exists. Let $B$ be the matrix representation of $T^*$ with respect to these bases. We have $$b_{ij} = (T^*f_j, e_i) = (f_j, Te_i) = \overline{(Te_i, f_j)} = \overline{a_{ij}}.$$
Thus $B$ is the conjugate transpose of $A$. Conversley, if we define $T^*$ to be the map $T^* \colon W \to V$ with matrix representation $B$, then $T^*$ satisfies $(T^*f_j, e_i) = (f_j, Te_i)$ for all $i, j$, and then by linearity and conjugate linearity of the inner product, $(T^*w, v) = (w, Tv)$ for all $v \in V, w \in W$.
Let $R$ denote the range (not the rowspace). It is easy to show that $R(T)^{\perp} = N(T^*)$. Using $T^{**} = T$, we deduce $R(T^*)^{\perp} = N(T)$. By the fundamental identity, it follows that $V = R(T^*) \oplus N(T)$.
Your results are stated in the case where $M$ is an $m \times n$ matrix. In that case you are interpreting $M$ as the linear map $T_M \colon \mathbb{F}^n \to \mathbb{F}^m$ represented by $M$ with respect to the standard bases on $\mathbb{F}^n$ and $\mathbb{F}^m$ (people commonly write $M$ instead of $T_M$). The adjoint of $M$ is therefore it's conjugate transpose (again identifying $M^*$ with it's matrix representation).
