Linear transformation by a non-invertible matrix According to this book I'm reading, a linear transformation of a vector can be considered as the linear transformation of the basis vectors of the vector. Consider a linear transformation represented by multiplying a non-invertible matrix A shown below to (1, 1), whose basis vectors are (1, 0) and (0, 1).
\begin{bmatrix}a&b\\c&d\end{bmatrix}
By this transformation,  the original vector (1, 1) would be transformed to (a+b, c+d), and the basis vectors would be transformed to (a, c) and (b, d), which are parallel since det A = ad - bc = 0.

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*To me, it seems like the above transformation is a mapping from 2D to 1D: T: R2 → R1, because the basis vectors become parallel. However, the resultant vector (a+b, c+d) is still a 2D vector. Which way of thinking is correct and why? Also, why is the other way of thinking wrong? If the transformation is T: R2 → R1, how can the resultant 2D vector be explained?


*My book says that such transformation by multiplying a non-invertible matrix is  NOT a bijective mapping. However, isn't the output of the transformation still a distinct 2D point (a+b, c+d) so there should be a transformation that transforms (a+b, c+d) to (1, 1)? Why is it not bijective? Is it because such transformation from (a+b, c+d) to (1, 1) cannot be done by A-1 but by other transformation? Is there any way based on the definition of bijection to understand the non-bijective nature of the transformation (without dealing with the non-existence of A-1)?


*If the transformation is T: R2 → R1, since det A = 0, is it true that linear transformation can lower dimensions but cannot increase dimensions? If it is true, is it just because A-1 doesn't exist when A is a linear transformation that lowers the dimension? Or are there any other mathematical ways to prove this nature of linear transformation?
Since I'm a beginner in linear algebra, there might be wrong uses of language and my thoughts on which my questions are based might be totally wrong. Sorry if there are any of them.
Thank you.
 A: First, a side point: vectors do not have "basis vectors". Instead, vector spaces have basis vectors. A basis of a vector space is a set of vectors such that every vector in the vector space can be written uniquely as a linear combination of the vectors in the basis. For the vector space $\Bbb R^2$, the most obvious basis is the set $\{(1,0), (0,1)\}$. Any vector $(u,v) \in \Bbb R^2$ can be written as a linear combination of $(1,0)$ and $(0,1)$ in only one way:
$$(u,v) = u\cdot (1,0) + v\cdot (0,1)$$
But this is only one of many possible bases for $\Bbb R^2$. For example $\{(1,1), (1,-1)\}$ is also a basis:
$$(u,v) = \frac{u+v}2\cdot(1,1) + \frac{u-v}2\cdot(1,-1)$$
In fact, if $v_1, v_2 \in \Bbb R^2$ are two non-parallel (and non-zero) vectors, then $\{v_1, v_2\}$ is another basis for $\Bbb R^2$. A basis vector is just a vector in some basis. But every non-zero vector in a vector space is a member of some basis, so any vector other than $0$ can be called a "basis vector". But the term is meaningless without specifying the basis of which it is a part.

Now to your question.

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*In general if $F : A \to B$ is a map (linear or otherwise) from a set $A$ to a set $B$, then $A$ is called the "domain" of $F$, and $B$ is called the "codomain". $F$ is expected to be defined for every element of $A$, but there is no requirement that $F$ need take on every value in $B$. The set $\{F(a)\mid a\in A\}$ of all values in $B$ that are taken on by $F$ is called the image of $F$. So $T : \Bbb R^2 \to \Bbb R^2$ has $\Bbb R^2$ as codomain. Its values are 2-dimensional vectors. But the image of $T$ is smaller. Since $T$ is a linear map, its image will be a vector space itself, a sub-space of $\Bbb R^2$. The dimension of a vector space is the number of vectors in a basis for it (all bases will have the same number of vectors). In fact, the image of $T$ is the line $\{(at,ct)\mid t \in \Bbb R\}$, which has $\{(a,c)\}$ as a basis, and is thus $1$-dimensional.

*No, your book does not call the linear map of a non-invertible matrix a "bijection". Just the opposite! Non-invertible matrices are exactly the linear maps that are NOT bijections. Being a bijection is exactly the condition that allows a function to be invertible. Invertible matrices are bijective linear maps. Non-invertible matrices are non-bijective linear maps. Either you have confused what the book was saying, or it was an accidental mistatement.

*For any linear map $ T : V \to W$ between two vector spaces $V$ and $W$, the highest dimension the image of $T$ can have is the dimension of $V$. This is because if $\{v_1,\dots,v_n\}$ is a basis for the vector space $V$, then for any $x \in V, x$ can be written as a linear combination of the basis vectors:
$$x = x_1v_1 + x_2v_2 + \dots x_nv_n$$
for some real numbers $x_1, \dots, x_n$. But then by the linearity of $T$,
$$T(x) = x_1T(v_1) + x_2T(v_2) + \dots x_nT(v_n)$$
so every vector in the image of $T$ can be expressed as a linear combination of the $n$ vectors $T(v_1), \dots, T(v_n)$. This means the dimension of the image of $T$ is at most $n$, the dimension of $V$ (the dimension could be lower than $n$ - some of the $T(v_i)$ might be expressible themselves as a combination of the others, which would allow $T(x)$ to also be expressed as a linear combination of the others as well).

