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01Invertible Linear Maps: Injective and Surjective02Equal Dimensions: Injective or Surjective Means Invertible03Isomorphic Spaces and the Dimension of L(V, W)04The Matrix of a Vector: M(Tv) = M(T)M(v)05Change of Basis for an Operator: A = C⁻¹BCProblem set0/10Problem set 20/10
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01Diagonalizable Operators and Direct Sums of Eigenspaces02Four Equivalent Tests for Diagonalizability03Distinct Linear Factors Decide Diagonalizability04The Gershgorin Disk Theorem: Where Eigenvalues Live05Commuting Operators Share a Diagonalizing Basis06Commuting Operators Over C: One Upper-Triangular BasisProblem set0/10Problem set 20/10
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01The Closest Point Is the Projection: Fitting Sine on [−π, π]02The Pseudoinverse T† and Its Three Inverse-Like Properties03T†w: The Best Approximate Solution of Smallest NormProblem set0/10Problem set 20/10
Operators on Inner Product Spaces
01The Adjoint T*: Moving T into the Second Slot of ⟨Tv, w⟩02Null T* = (range T)⊥ and M(T*) as the Conjugate Transpose03Self-Adjoint Operators: Real Eigenvalues and Real ⟨Tv, v⟩04Normal Operators: ‖Tv‖ = ‖T*v‖ and Orthogonal EigenvectorsProblem set0/10Problem set 20/10

Four Equivalent Tests for Diagonalizability

What four conditions are each equivalent to diagonalizability, and how does an eigenvector basis make T^100 easy?


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Your summary note

    1. 1

      The four equivalent conditions (a)–(d)

      State diagonalizable, eigenvector basis, V=E(λ1,T)⊕⋯⊕E(λm,T)V=E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T)V=E(λ1​,T)⊕⋯⊕E(λm​,T) and the dimension sum, and prove the cycle, writing out the independence argument for combined eigenspace bases in (d) implies (b).

    2. 2

      The non-diagonalizable operator T(a,b,c)=(b,c,0)T(a,b,c)=(b,c,0)T(a,b,c)=(b,c,0)

      Show 000 is its only eigenvalue, compute E(0,T)E(0,T)E(0,T), and conclude from the dimension condition that TTT is not diagonalizable over R\mathbf{R}R or C\mathbf{C}C.

    3. 3

      dim⁡V\dim VdimV distinct eigenvalues imply diagonalizable

      Prove the result with one eigenvector per eigenvalue, and record T(x,y,z)=(6x,6y,7z)T(x,y,z)=(6x,6y,7z)T(x,y,z)=(6x,6y,7z) as a diagonalizable operator with only two eigenvalues, so the condition is sufficient but not necessary.

    4. 4

      Computing T100(0,0,1)T^{100}(0,0,1)T100(0,0,1) with an eigenvector basis

      For T(x,y,z)=(2x+y,5y+3z,8z)T(x,y,z)=(2x+y,5y+3z,8z)T(x,y,z)=(2x+y,5y+3z,8z), find eigenvectors for 222, 555, 888, expand (0,0,1)(0,0,1)(0,0,1) in them, and apply Tkv=λkvT^k v=\lambda^k vTkv=λkv through to the final vector.

    Attempt 1 of 2