An important view of the geometry of A'A$A'A$ is this (the viewpoint strongly stressed in Strang's book on "Linear Algebra and Its Applications"): Suppose A is an (m x n)$m \times n$-matrix of rank k, representing a linear map A:R^n-->R^m $A: R^n \rightarrow R^m$. Let Col(A) and Row(A) be the column and row spaces of A$A$. Then
(a) As a real symmetric matrix, (A'A):R^n --> R^n$(A'A): R^n \rightarrow R^n$ has a basis {e_1,..., e_n}$\{e_1,..., e_n\}$ of eigenvectors with non-zero eigenvalues d_1,...,d_k. Thus (A'A)(x_1e_1 + ... + x_ne_n) = d_1x_1e_1 + ..$d_1,\ldots,d_k$. + d_kx_ke_kThus:
$(A'A)(x_1e_1 + \ldots + x_ne_n) = d_1x_1e_1 + ... + d_kx_ke_k$.
(b) Range(A) = Col(A), by definition of Col(A). So A|Row(A) maps Row(A) into Col(A).
(c) Kernel(A) is the orthogonal complement of Row(A). This is because matrix multiplication is defined in terms of the dot products (row i)*(col j). (So Av'= 0 <==> v is in Kernel(A) <==> v is in orthogonal complement of Row(A)$Av'= 0 \iff \text{v is in Kernel(A)} \iff v \text{is in orthogonal complement of Row(A)}$
(d) A(R^n)=A(Row(A)) $A(R^n)=A(\text{Row}(A))$ and A|Row(A):Row(A)-->Col(A)$A|\text{Row(A)}:\text{Row(A)} \rightarrow Col(A)$ is an isomorphism.
Reason: If v = r+k (r \in Row(A), k \in Kernel(A),from (c)) then
A(v) = A(r) + 0 = A(r) where A(r) = 0 <==> r = 00$.
[Incidentally gives a proof that Row rank = Column rank!]
(e) Applying (d), A'|:Col(A)=Row(A')-->Col(A')=Row(A)$A'|:Col(A)=\text{Row(A)} \rightarrow \text{Col(A')}=\text{Row(A)}$ is an isomorphism
(f)By (d) and (e): A'A(R^n) = Row(A)$A'A(R^n) = \text{Row(A)}$ and A'A maps Row(A) isomorphically onto Row(A).
