University of Tokyo 1S2 Linear Algebra

• Continuation of Introduction to Linear Algebra at the University of Tokyo: Linear Algebra.

Lecture 3:

• Basis

  • In a Vector Space V, there exist vectors called basis that satisfy the following conditions:
    • Definition: “The vectors can generate any vector in V through linear combinations” and “The vectors are linearly independent”.
      • The basis v_1, …, v_n is called the generating system of V.
  • What is the difference between being linearly independent and being a basis..? (blu3mo)
    • When we say a set of linearly independent vectors, is the number of vectors not specified? (blu3mo)
    • It seems that we can define it as “a set of vectors that is linearly independent and has the same number as the dimension of the space”.
      • Oh, but is the concept of dimension not defined yet?
      • Maybe the dimension is defined by the definition of a basis? (blu3mo)
  • When expressing a vector using a certain basis, the representation is uniquely determined.
    • Well, that’s obvious (blu3mo)
    • For an orthonormal basis, there is no other representation of (5,3) except for 5(1,0) + 3(0,1).
    • For a skew-orthogonal basis, there is no other representation of (5,3) except for 2(1,0) + 3(1,1).
  • It is useful to have various bases when considering Diagonalization.
  • In physics, I don’t think I’ve ever considered anything other than an orthonormal basis (blu3mo).
    • I have considered a skew-orthogonal basis for the same unit, but I wonder if there are any advantages to considering a skew-orthogonal basis for vectors like velocity and time.
      • I think I used skew-orthogonal bases in relativity theory (takker).
      • Sounds interesting (blu3mo)(blu3mo)(blu3mo).
    • It is also necessary to discuss curved spaces (takker).
      • Relativity theory is famous, but it is also used when considering the deformation of surfaces in continuum mechanics.

• Extension of Basis

• Dimension

  • I’ve never thought about the definition of dimension (blu3mo).
  • As a premise, the number of bases of the same vector space is constant.
    • Therefore, the number is defined as the dimension.
    • It’s casually mentioned as a premise, but it actually needs to be proven (takker).
      • However, it’s not a difficult proof for finite-dimensional vector spaces.
        • It was proven in the class (blu3mo).
        • That’s good to hear (takker).
      • It seems to be a problem for infinite-dimensional vectors.
        • How can we prove it?
        • Can we simply show that a bijection can be constructed between two bases for the same vector space?
  • Theorem
    • They all seem obvious intuitively.
    • Theorem: If there are n vectors ($n>\dim V$) in a vector V, they are linearly dependent.
    • Theorem: If we consider a subspace W of vector V, $\dim V>\dim W$.

Lecture 2:

• Subspace = Subvector Space
  • When we have a vector space V, a subset that itself becomes a vector space.
    • For example,
      • When considering a subspace of natural numbers, if there is only 3, it does not satisfy one of the axioms, so it does not become a subspace (blu3mo).
        • It becomes a subspace if both 3 and -3 are included (blu3mo).
      • It is not a subspace if it does not contain 0 (zero element).
    • Definition/Requirements of a subspace
      • 1. It contains 0.
        • Is this necessary? (takker)
          • If we exclude the empty set from being a subspace, can it be replaced with “not an empty set”?
      • 2. When there are two elements, their sum is also included in the set.
        • (Without this, addition does not hold)
      • 3. When there is an element, any scalar multiple of it is also included in the set.
        • (Without this, scalar multiplication does not hold)
      • As long as we consider these three, we can determine if it is a subspace (blu3mo)(blu3mo).
        • Since it is a subset of a vector space, we can skip checking things like the associative law (blu3mo).
        • The condition for the inverse element can also be included by considering -1 times in 3) (blu3mo).
  • The subspaces of $R^2$ are only the following three.- > $\{(0,0)\}$, a line passing through the origin, $R^2$
• Each of them is 0-dimensional, 1-dimensional, and 2-dimensional (blu3mo)(blu3mo)
• Most cases fall into either 2) or 3) (blu3mo)
• It might be interesting to also examine subspaces of ${\mathbb{R}}^3$ (takker)
• Linear Independence
  • Linear Combination: When $v$ is an element of the vector space V, it can be expressed in the form $a_1{v}_1+...+a_n{v}_n$
    • a is a scalar, right? (blu3mo)
    • Yes, it is a scalar (takker)
  • Linear Independence: ”$a_1{v}_1+...+a_n{v}_n=0$ implies $a_1=...=a_n=0$”
    • When there are two 2D vectors pointing in different directions, we can say that “if the linear combination is zero, then the scalar coefficients of each vector are also zero”
    • Pay attention to the logic (blu3mo)
    • Well, in this case, the reverse is also true, but it’s important to pay attention to the direction (takker)
  • Linear Dependence: not linearly independent
  • Hypothesis: For n-dimensional vectors, it seems that n different vectors pointing in separate directions are linearly independent (blu3mo)
    • Seems like it (blu3mo)
  • As a theorem, there is the following:

  • Theorem: ”$v_1,v_2$ are linearly independent” is equivalent to ”$O,P_1,P_2$ are not on the same line”

    • (Here, $v_n=O{P}_n$ vector)
    • This can be proven by manipulating the equations
    • It seems that the same can be said for $R^n$ (placing n vectors)
    • Regarding the number n:
      • In the vector space $R^n$, n+1 or more vectors cannot be linearly independent
      • This is because with n linearly independent vectors, any vector can be expressed as a linear combination of them