Are linear transformations homomorphism
Linear maps
Corollary 1 from "Freedom": the coordination map Φ_{B.}. Is B. = {b_{1}, ..., b_{n}} a basis of V, we get one (and only one) linear map

 The existence of Φ_{B.} follows from the "freedom" of the standard basis of K^{n}. That Φ_{B.} is an isomorphism follows from criterion 2 for isomorphisms.
Corollary 1 '. Every vector space of dimension n is isomorphic to K.^{n}. 
Corollary 1 ". All vector spaces of dimension n are isomorphic to one another. 
 If dim V = dim W = n, then there are isomorphisms Φ_{V} : K^{n} → V and Φ_{W.} : K^{n} → W, i.e. the isomorphism Φ_{W.}Φ_{V}^{1} : V → W
Linear maps and matrices
If you limit yourself to finitely generated vector spaces, there is a clear correspondence between linear mappings and matrices  this is thematized in the fields highlighted in light blue:There is a kind of dictionary of how terms in abstract linear algebra correspond to terms in matrix theory. The fields highlighted in light blue show this.
Every (m × n) matrix A with coefficients in K yields a linear map f_{A.} : K^{n} → K^{m}, by
Are A, B (m × n) matrices with coefficients in K, and is f_{A.} = f_{B.}, then A = B. Conclusion 2 from "Freedom": The reverse is also true! For every linear mapping f: K^{n} → K^{m} there is an (m × n) matrix A = M (f) with coefficients in K such that f = f_{A.}.  
Reformulation: The assignment that a matrix A in M (m × n, K) is mapped to f_{A.} is a bijection between
 
Note rule for determining the matrix A = M (f). The linear mappings f: K^{n} → K^{m} the (m × n) matrix A = M (f) is assigned, with

 Proof: Here, too, we are dealing with a consequence of "freedom"; again, "freedom" becomes the standard basis of K^{n} related.
Composition. Invertibility of matrices. Is A in M (m n, K), B in M (n m, K) with AB = I_{m} and BA = I_{n}, then m = n. 
 The invertibility statement follows from the fact that isomorphic vector spaces have the same dimension: K^{m} and K^{n} are isomorphic only for m = n. (From AB = I and BA = I it follows that f_{A.} and f_{B.} are mutually inverse isomorphisms.)
Core, image, rank

The interplay of Corollaries 1 and 2:Let V, W be vector spaces, let dim V = n, dim W = m. Let V a basis of V, W. a basis of V. Then we get a bijection between the linear mappings f: V → W and the (n × m) matrices with coefficients in K.
Is f_{A.} = Φ_{W.}^{1} f Φ_{V}, we write A = M^{V}_{W.}(f) and say that A is the performing Matrix of f with respect to the bases V and W. is. Is f: K^{n} → K^{m} linear and we denote it with E. each the standard base of K^{n} as well as from K^{m}, so is M^{E.}_{E.}(f) = M (f). 
Is V = {v_{1}, ..., v_{n}} a basis of K^{n} (understood as a space of column vectors), then let M (V) the matrix whose jth column is just v_{j} is.
Then: this is the representing matrix of Φ_{V}.
Now let A be an (m × n) matrix with coefficients in K. Consider the map f_{A.}. Base change in K^{n} and K^{m} delivers:
f_{B.}  
K^{n}  →  K^{m}  
↓ Φ_{V}  ↓ Φ_{W.}  
K^{n}  →  K^{m}  
f_{A.} 
Evaluation of the proof of the dimensional formula:
Sentence. Let V, W be finitegenerated vector spaces, let f: V → W be a linear mapping. Then there are bases V from V and W. from W with M^{V}_{W.}(f) = B with 
Let us apply this theorem to an illustration of the form f_{A.} where A is an (m × n) matrix with coefficients in K, we get a new proof of Theorem 1 "". Compare the two proofs: Our first proof of Theorem 1 "" used elementary row and column operations and is constructive (if the matrix A is given, then one knows exactly how to do to convert A into the given form bring). The new proof, on the other hand, is elegant but not directly constructive: it is based on concepts such as subspace and base, and above all on the supplementary theorem. 
 Proof: Let f: V → W be linear.
 Be b_{1}, ..., b_{s} a base of kernel (f),
 Continue this with a_{1}, ..., a_{r} to a base of V.
 Then f (a_{1}),...,fa_{r}) a base of image (f)
(This is precisely what the proof of the dimensional formula for linear mappings says; in particular, r = rank (f).)  Put f (a_{1}),...,fa_{r}) by c_{1}, ..., c_{t} on to a base by W.
and W. = {f (a_{1}),...,fa_{r}), c_{1}, ..., c_{t}}.
Then M^{V}_{W.}(f) the specified form.
Don't be a body. Let A, B (m × n) matrices with coefficients in K. Then the following statements are equivalent:

 The equivalence of (1), (2) and (3) follows from the considerations on how to assign a matrix to a linear mapping between finitedimensional vector spaces.
The equivalence between these statements and statement (4) is deeper: If (1) applies, A and B have the same rank, because rank (A) = rank (f) and rank (B) = rank (f), i.e. applies (4).
Conversely, if (4) holds, we have just seen that both f_{A.} as well as f_{B.} have the same representative matrix:
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