Let A be an 5 by 5 matrix, let v₁ be an eigenvector of A with eigenvalue X₁ and let V₂ be an eigenvector of A with eigenvalue X₂. Select all items below that are true. A. The vector -5v₁ need not be an eigenvector of A. ✔B. If v₁ is a scalar multiple of v₂, then X₁ = √₂. ✓C. The vector -5v₁ is an eigenvector of A. D. For any real number c, cλ₁ is also an eigenvalue of A. ✓E. If 0 is an eigenvalue of A, then A is singular. F. If X₁ X₂, then V₁ + V₂ is an eigenvector of A (as long as it is nonzero). □G. If X₁ = X₂, then v₁ must be a scalar multiple of V₂ (or vice versa). H. It is entirely possible that the zero vector is an eigenvector of A. I. If A is singular, then 0 is an eigenvalue of A.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Let A be an 5 by 5 matrix, let v₁ be an eigenvector of A with eigenvalue X₁ and let V₂ be an eigenvector of A with eigenvalue X₂. Select all items below that are true.
A. The vector -5v₁ need not be an eigenvector of A.
✔B. If v₁ is a scalar multiple of v₂, then X₁ = √₂.
✓C. The vector -5v₁ is an eigenvector of A.
D. For any real number c, cλ₁ is also an eigenvalue of A.
✓E. If 0 is an eigenvalue of A, then A is singular.
F. If X₁
X₂, then V₁ + V₂ is an eigenvector of A (as long as it is nonzero).
□G. If X₁ = X₂, then v₁ must be a scalar multiple of V₂ (or vice versa).
H. It is entirely possible that the zero vector is an eigenvector of A.
I. If A is singular, then 0 is an eigenvalue of A.
Transcribed Image Text:Let A be an 5 by 5 matrix, let v₁ be an eigenvector of A with eigenvalue X₁ and let V₂ be an eigenvector of A with eigenvalue X₂. Select all items below that are true. A. The vector -5v₁ need not be an eigenvector of A. ✔B. If v₁ is a scalar multiple of v₂, then X₁ = √₂. ✓C. The vector -5v₁ is an eigenvector of A. D. For any real number c, cλ₁ is also an eigenvalue of A. ✓E. If 0 is an eigenvalue of A, then A is singular. F. If X₁ X₂, then V₁ + V₂ is an eigenvector of A (as long as it is nonzero). □G. If X₁ = X₂, then v₁ must be a scalar multiple of V₂ (or vice versa). H. It is entirely possible that the zero vector is an eigenvector of A. I. If A is singular, then 0 is an eigenvalue of A.
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