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The characteristic polynomial of a 3×33×3 matrix can be factorised as (2−λ)2(5−λ)(2−𝜆)2(5−𝜆). Which of the following statements are true? Select all that apply.The geometric multiplicity of the eigenvalue λ=𝜆= 55 could be 22The number of linearly independent eigenvectors for the eigenvalue λ=𝜆= 55 must be 11The geometric multiplicity of the matrix cannot be greater than 22The geometric multiplicity of the eigenvalue λ=𝜆= 22 could be 22The algebraic multiplicity of the matrix is 22The algebraic multiplicity of the eigenvalue λ=𝜆= 22 cannot be determined from the information we are givenThe eigenvalue λ=𝜆= 22 could have infinitely many eigenvector directionsThe algebraic multiplicity of the eigenvalue λ=𝜆= 22 is 2

Question

The characteristic polynomial of a 3×33×3 matrix can be factorised as (2−λ)2(5−λ)(2−𝜆)2(5−𝜆). Which of the following statements are true? Select all that apply.The geometric multiplicity of the eigenvalue λ=𝜆= 55 could be 22The number of linearly independent eigenvectors for the eigenvalue λ=𝜆= 55 must be 11The geometric multiplicity of the matrix cannot be greater than 22The geometric multiplicity of the eigenvalue λ=𝜆= 22 could be 22The algebraic multiplicity of the matrix is 22The algebraic multiplicity of the eigenvalue λ=𝜆= 22 cannot be determined from the information we are givenThe eigenvalue λ=𝜆= 22 could have infinitely many eigenvector directionsThe algebraic multiplicity of the eigenvalue λ=𝜆= 22 is 2

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Solution

The characteristic polynomial of a 3x3 matrix is given as (2−λ)²(5−λ). This tells us that the eigenvalues of the matrix are 2 and 5. The exponent of (2−λ) and (5−λ) in the factorization of the characteristic polynomial gives us the algebraic multiplicity of the corresponding eigenvalue.

  1. The geometric multiplicity of the eigenvalue λ=5 could be 2: False. The geometric multiplicity of an eigenvalue cannot exceed its algebraic multiplicity. Here, the algebraic multiplicity of λ=5 is 1, so its geometric multiplicity can only be 1.

  2. The number of linearly independent eigenvectors for the eigenvalue λ=5 must be 1: True. The number of linearly independent eigenvectors for an eigenvalue is equal to its geometric multiplicity, which is 1 in this case.

  3. The geometric multiplicity of the matrix cannot be greater than 2: True. The geometric multiplicity of a matrix is the maximum geometric multiplicity of its eigenvalues. Here, the maximum geometric multiplicity is 2 (for λ=2).

  4. The geometric multiplicity of the eigenvalue λ=2 could be 2: True. The geometric multiplicity of an eigenvalue cannot exceed its algebraic multiplicity. Here, the algebraic multiplicity of λ=2 is 2, so its geometric multiplicity can be 1 or 2.

  5. The algebraic multiplicity of the matrix is 2: False. The algebraic multiplicity of a matrix is the sum of the algebraic multiplicities of its eigenvalues. Here, the algebraic multiplicities of λ=2 and λ=5 are 2 and 1 respectively, so the algebraic multiplicity of the matrix is 2+1=3.

  6. The algebraic multiplicity of the eigenvalue λ=2 cannot be determined from the information we are given: False. The algebraic multiplicity of λ=2 is given by the exponent of (2−λ) in the characteristic polynomial, which is 2.

  7. The eigenvalue λ=2 could have infinitely many eigenvector directions: False. The number of eigenvector directions for an eigenvalue is equal to its geometric multiplicity, which can be at most 2 in this case.

  8. The algebraic multiplicity of the eigenvalue λ=2 is 2: True. This is given by the exponent of (2−λ) in the characteristic polynomial.

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Which of the following could be the set of distinct eigenvalues for a real 3×33×3 matrix?  Select all that apply.{−2,2,5}{−2,2,5}{4−3i,5i,3i+4}{4−3i,5i,3i+4}{−3,1}{−3,1}{−2i,2i}{−2i,2i}{−2,3,−2i−4}{−2,3,−2i−4}{−2,4−3i,3i+4}{−2,4−3i,3i+4}{−5,1−6i,6i−1}

To find the eigenvalues, computedet3 − λ 0 0−3 4 − λ 90 0 3 − λ = (3 − λ)(4 − λ)(3 − λ).So the eigenvalues are λ = 3 and λ = 4.We can find two linearly independent eigenvectors301 ,130 corresponding to the eigenvalue 3, and oneeigenvector010 with eigenvalue 4. The diagonalized form of the matrix is3 0 0−3 4 90 0 3 =3 1 00 3 11 0 03 0 00 3 00 0 40 0 11 0 −3−3 1 9 .Note that if you chose different eigenvectors, your matrices will be different. The middle matrix should haveentries 3, 3, 4 in some order, and you should multiply out the product to make sure you have the right answer.

Let 𝐴 𝑏𝑒 𝑎𝑛 𝑛 × 𝑛 matrix. The characteristic polynomial of A is a polynomial of degree

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