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Linear Algebra Note 5

Determinants (2021.11.24)

  • Properties
    9.

    |AB|=|A||B|

    • Recall:
      rank(AB)≤min(rank(A), rank(B))
    • Proof
      Let
      A∈Rk×l, B∈Rl×m
      ∀ v→∈C(AB), v→=(AB)x→       where
      x→       is the
      m×1       vector
      Also,
      v→=A(Bx→)       where
      Bx→       is the
      l×1       vector
      ∴ v→∈C(A), i.e. C(AB)⊆C(A)⟹dim(C(AB))≤dim(C(A)), rank(AB)≤rank(A)
      Similary,
      rank(AB)≤rank(B)
      1. when
        |B|=0         ,
        B         is singular
        rank(AB)≤min(rank(A), rank(B))
        ⟹AB         is also singular
        ⟹|AB|=0⟹|AB|=|A||B|=0
      2. When
        |B|≠0         , Let
        D(A)≡|AB||B|
        To prove
        D(A)=|A| (|AB||B|=|A|⟹|AB|=|A||B|)
        show
        D(A)         satisfies (i)-(iii)
        (i)
        |In|=1
        D(A=I)=|IB||B|=|B||B|=1=|A|
        (ii) Exchange of two rows leads to change of sign
        Let A=[a1→Ta2→T⋮an→T], A′=[a2→Ta1→T⋮an→T], B=[b1→b2→⋯bm→]AB=[a1→Ta2→T⋮an→T][b1→b2→⋯bm→]=[a1→Tb1→a1→Tb2→⋯a1→Tbm→a2→Tb1→a2→Tb2→⋯a2→Tbm→⋮⋮⋱⋮an→Tb1→an→Tb2→⋯an→Tbm→]A′B=[a2→Ta1→T⋮an→T][b1→b2→⋯bm→]=[a2→Tb1→a2→Tb2→⋯a2→Tbm→a1→Tb1→a1→Tb2→⋯a1→Tbm→⋮⋮⋱⋮an→Tb1→an→Tb2→⋯an→Tbm→]=row exchange of AB∴|A′B|=−|AB|, D(A′)=|A′B||B|=−|AB||B|=−D(A)
        (iii) The determinant is a linear function of each row separately
        (scalar multiplication)
        A′=[a1→Tta2→T⋮an→T], A′B=[a1→Tb1→a1→Tb2→⋯a1→Tbm→ta2→Tb1→ta2→Tb2→⋯ta2→Tbm→⋮⋮⋱⋮an→Tb1→an→Tb2→⋯an→Tbm→]|A′B|=t|AB|⟹D(A′)=|A′B||B|=t|AB||B|=tD(A)
        (vector addition)
        A″=[a1→T+a1→′Ta2→T⋮an→T], A″B=[a1→Tb1→+a1→′Tb1→a1→Tb2→+a1→′Tb2→⋯a1→Tbm→+a1→′Tbm→a2→Tb1→a2→Tb2→⋯a2→Tbm→⋮⋮⋱⋮an→Tb1→an→Tb2→⋯an→Tbm→]A‴=[a1→′Ta2→T⋮an→T], A‴B=[a1→′Tb1→a1→′Tb2→⋯a1→′Tbm→a2→Tb1→a2→Tb2→⋯a2→Tbm→⋮⋮⋱⋮an→Tb1→an→Tb2→⋯an→Tbm→]|A″B|=|AB|+|A‴B|⟹D(A″)=|A″B||B|=|AB|+|A‴B||B|=|AB||B|+|A‴B||B|=D(A)+D(A‴)
        D(A)         satisfies rule (i)-(iii)
        ⟹D(A)=|A|
    • det(A−1)=1det(A)
      • Proof
        AA−1=I, |AA−1|=|A||A−1|=|I|=1
    1. |AT|=|A|
      • Proof
        1. when
          |A|=0, A           is singular
          ⟹AT           is singular
          ⟹|AT|=0
        2. when
          |A|≠0           , apply
          PA=LU           (
          P           = permutation matrix)
          (PA)T=(LU)T⟹ATPT=UTLT
          from 9,
          |P||A|=|PA|=|LU|=|L||U|, |AT||PT|=|UT||LT|
          ∵PPT=I⟹|P||PT|=1
          Moreover, a permutation matrix
          P           can be regarded as s row-exchanged identity matrix
          |P|=(−1)k|I|=1 or −1∵|P||PT|=1⟹|PT|=|P|
          Also,
          |L|=1=|LT|           due to all-ones diagonal
          |U|=|UT|=           product of the pivots
          |P||A|=|L||U|=|UT||LT|=|AT||PT|, |PT|=|P|⟹|A|=|AT|

  • Summary: How to find determinant (algorithm)
    Use Gaussian elimination

    PA=LU, |P||A|=|L||U|
    |P|=±1, |L|=1, |U|=     product of the pivots
    ∴|A|=(−1)k|U|=(−1)k×(product of the pivots)     ,
    k=     the number of row exchange

Eigenvectors & Eigenvalues (2021.11.26 ~ 2021.12.03)

  • Example

    A=[0.80.30.20.7], x1→=[0.60.4], x2→=[1−1]Ax1→=[0.80.30.20.7][0.60.4]=[0.60.4]=x1→Ax2→=[0.80.30.20.7][1−1]=[0.5−0.5]=0.5x2→
    Ax1→     and
    Ax2→     lie in the same direction with
    x1→     and
    x2→
    Therefore, Ax1→=λ1x1→, where λ1=1Ax2→=λ2x2→, where λ2=0.5
    x1→, x2→     are called eigenvectors and
    λ1, λ2     are the corresponding eigenvalues

  • Define(eigenvector and eigenvalue): Let

    A be a
    n×n     matrix. A non-zero vector
    x→∈V     is called an eigenvector of
    A     if there exists a scalar
    λ     such that
    Ax→=λx→     . This
    λ     is called the eigenvalue corresponding to the eigenvector
    x→

  • Theorem
    A scalar

    λ is an eigenvalue if and only if
    |A−λI|=0

    • Proof
      If
      x→       is an eigenvector of
      A       and
      λ       is the corresponding eigenvalue
      Ax→=λx→⟺Ax→−λx→=0⟺(A−λI)x→=0⟺x→∈N(A−λI)∵x→≠0→⟹N(A−λI)≠{0→}⟺A−λI is singular ⟹|A−λI|=0

  • Example
    (Finding

    λ)
    A=[0.80.30.20.7]     , Let
    |A−λI|=0
    |[0.8−λ0.30.20.7−λ]|=0.56−1.5λ+λ2=0.06=0λ2−1.5λ+0.5=(λ−1)(λ−0.5)=0⟹λ=1 or 0.5
    (Finding eigenvectors)
    Solve
    N(A−λI)     to obtain eigenvectors
    λ=1, A−λI=[−0.20.30.2−0.3]→[1−3200]x1→=N(A−λI)={c[321]∣c∈R}
    λ=0.5, A−λI=[0.30.30.20.2]→[1100]x2→=N(A−λI)={c[−11]∣c∈R}

  • Define(trace): Let

    A=[a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮an1an2⋯ann] be a
    n×n     square matrix,
    trace(A)     or
    tr(A)≜∑i=1naii

  • Theorem
    Let

    A be a square matrix and
    λ1, λ2, ⋯, λn     be its eigenvalues, then

    1. det(A)=λ1×λ2×⋯×λn
    2. tr(A)=∑i=1nλi
    • Proof
      A=[a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮an1an2⋯ann], A−λIn=[a11−λa12⋯a1na21a22−λ⋯a2n⋮⋮⋱⋮an1an2⋯ann−λ]det(A−λIn)=det(A)+⋯+∑i=1naii(−λ)n−1+(−λ)n
      Let
      λ1, λ2, ⋯, λn       be the roots of the n-order polynomial
      p(λ)=d(λ1−λ)(λ2−λ)⋯(λn−λ)=d[(λ1×λ2×⋯×λn)+⋯+∑i=1nλi(−λ)n−1+(−λ)n]
      match
      det(A−λI)       and
      p(λ)
      ⟹det(A)=λ1×λ2×⋯×λn, tr(A)=∑i=1naii=∑i=1nλi

  • Example (Projection matrix)

    P=[12121212], |P−λI|=|[12−λ121212−λ]|=λ2−λ=0λ=1 or 0⟹x1→=[11], x2→=[−11]
    Note that
    P     is a projection matrix onto a subspace spanned by
    [11]
    ∴ ∀ x1→ already in this subspace Px1→=λ1x1→=x1→ (itself)∀ x2→ orthogonal to this subspace Px2→=λ2x2→=0

  • Example (Rotation matrix)

    Q=[0−110] rotates a vector
    ∈R2     by
    90∘
    check
    Qv→=[0−110][ab]=[−ba]
    (Qv→)Tv→=[−ba]T[ab]=−ba+ab=0     (orthogonal)
    (Find
    λ    )
    |Q−λI|=|[−λ−11−λ]|=λ2+1=0, λ=±i
    There doesn't exist any real number
    λ     and vector
    x→     s.t.
    Qx→=λx→
    since NO vector stay in the same direction after rotation by degrees other than
    nπ, n∈Z

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