• DEF Product Space
• DEF DirectSum/OuterProduct Space

Dimension

• THM the maximal LI elements generate a vector space
• THM for a finte vector space number of maximal LI elements is unique. We call this the dimension of the vector space

Basis

• Given a group of LI elements \( < Dim(V)\), you can extend this group up to \(Dim(V)\) elements

Subspaces

• The basis of a subspace is LI in the superspace
• EX A vector subspace may not share the same basis as the containing set. Consider \(\mathbb{R}^2\) whose basis is \((1,0), and (0,1)\) and subspace \(k(1,1)\)

Linear map

• THM \(dim(V) = dim(Kerf) + dim(Imf)\)

  • IDEA: if \(kerf={0}\) then the basis of domain coincides with basis of image due to the LI being linked across f
    • \(0 = k_1*f(b_1) + ... + k_n*f(b_n) = f(k_1*b_1 + ... + k_n*b_n) \leftrightarrow k_1*b_1 + ... + k_n*b_n=0 \)
  • So the objective is to take f and construct a quotient space where the the \(kerf={0}\)

• DEF Quotient Vector Spaces \(V/W\) is a vector space

• THM \(dim(V) = dim(V/W) + dim(W)\)

• let \(dim(V)=n\)
• let \(w_1 ... w_m\) be the basis of \(W\)
• since the basis of \(W\) is also LI we can extend it to n elements \(w_1 ... w_m, b_{m+1}, ..., b_n\)
• now we will show that \(b_{m+1}, ..., b_n\) is the basis of \(V/W\)
  • first \(b_{m+1}, ..., b_n\) is LI in \(V/W\) because it is LI with the basis of W
    • Proof:
    • suppose for contradiction that LD, then \(k_1*b_{m+1}+ ...+ k_n*b_n \in W\) where some \(k_i\) nonzero
    • \(b_{m+1}, ..., b_n\) LI so \(k_1*b_{m+1}+ ...+ k_n*b_n != 0\)
    • since the basis of W generates \(k_1*b_{m+1}+ ...+ k_n*b_n = c_1*w_1+...+c_m*w_m\) where some \(c_i\) nonzero
    • then we have the linear dependence of the terms \(k_1*b_{m+1}+ ...+ k_n*b_n + c_1*w_1+...+c_m*w_m = 0\) which is contradiction
  • this means the basis of \(V/W\) must be greater or equal to \(b_{m+1}, ..., b_n\)
  • second suppose for contradiction that basis it is greater than \(b_{m+1}, ..., b_n\), then we extend our original basis by 1 \(b_{m+1}, ..., b_n, x\)
    • Proof:
    • \(b_{m+1}, ..., b_n, x\) is LI in \(V/W\)
    • \(w_1,...w_m, b_{m+1}, ..., b_n, x\) is LD in \(V\)
    • \(c_1*w_1 +... + c_m*w_m + k_1*b_{m+1} + ... + k_n*b_n + k*x = 0\) where some coefficients are nonzero
      • k cannot be zero ( otherwise LD of remaining terms)
      • some \(k_i\) must be nonzero (otherwise if only \(c_i\) nonzero then \(x\) can be written in terms of \(w_1\). This would implie x is in W and \(b_{m+1}, ..., b_n, x\) is not LI in \(V/W\))
      • if some \(k_i\) nonzero then we have contradiction since \(k_i*b_{m+i} + k*x\) is in W, which means \(b_{m+i}, x\) are LD in \(V/W\)
      • so this cannot be
  • This means length of basis for \(V/W\) must be equal to \(b_{m+1}, ..., b_n\), giving us \(dim(V) = dim(V/W) + dim(W)\)
• Key ideas from above: Basis of subspace is LI in superspace
• Extension of existing LI elements to new basis allows us to draw powerful conclusions because we can infer additional properties
• ??? Is there some general way to capture the relationship between, LI, LD elements instead of doing this case-by-case check?

Matrices and Linear maps

• Matrices/Linear maps induce a change of basis
• THM There is a 1-1 mapping between every linear map and space of matrices (Are you sure it's one to one?)

Gaussian Elimination

• THM Let G generate V, then \(|G| > = dim(V)\)

• Interesting way to prove by contradiction of \(|G| < dim(V)\)
  • \(G={g_1}\) with \(V={v_1, v_2}\)
    • \(v1=a_1*g1\)
    • \(v2=a_2*g1\)
    • then \(k*v1+k*v2=0\)
  • \(G={g_1, g_2}\) with \(V={v_1, v_2, v_3}\)
    • \(v_1=a_1*g_1 + b_1*g_2\)
    • \(v_2=a_2*g_1 + b_2*g_2\)
    • \(v_3=a_3*g_1 + b_3*g_2\)
    • do Guassian elimination and we get
    • \(a*v_1-a*v_2 = g_2\)
    • \(b*v_2-b*v_3 = g_2\)
    • from here we get LD
  • the Gaussian elim idea is also kinda used in determinants

Determinant

• Determinants? A nice way to compare and see the linear dependence/independence of vectors
  • Build it from smaller cases