matrix help?

Question

1. true/false Every square matrix A has an inverse A^-1 such that A(A^-1)=(A^-1)A=I .
2. true /false If A is an invertible 3x3 matrix and v=[1 2 3]^T , there exists some vector u such that Au=v .
3. true/ false If A is a 3x3 matrix with eigenvalues 0,1,2 , then there are three distinct eigenvectors corresponding to each of these eigenvalues.
4. true/ false If the eigenvalues of a 3x3 matrix A are 0,1,2 , then the eigenvalues of A+I are 1,2,3 if I is the identity matrix.
5. true/false The matrix with ones in every entry is invertible.
6. true/false If A is invertible and B is invertible and AB is invertible then A+B is invertible.

Answer 1

#1: false. Minimal counterexample: the 1×1 matrix [0]
#2: True, and this vector is A⁻¹v.
#3: True -- although the problem wording is a little unusual (it implies 3 eigenvectors per eigenvalue rather than 3 eigenvectors, one for each eigenvalue), but even if interpreted as written, it is still true, since if v is an eigenvector with eigenvalue λ, so too are 2v and 3v.
#4: True. The eigenvalues are the roots of det (A-λI), and if this is zero only when λ=0, 1, or 2, then det (A+I-λI) = det (A-(λ-1)I) is zero only when λ-1=0, 1, or 2, so λ=1, 2, or 3.
#5: False, except when considering the 1×1 matrix. All other matrices with 1s in every entry have at least two identical columns, which make them linearly dependent and thus noninvertible.
6: false. Let A=I, B=-I. Then A is invertible, B is invertible, and AB=B is invertible, but A+B=0, which is not invertible.

Answer 2

1. false, every square matrix is not invertible
2. true, u = A^(-1)v, and A^(-1) exists by the assumption of the problem.
3. true. different eigenvalues always give different eigenvectors
4. true. If Av=ev then (A+I)v=Av+Iv=ev+v=(e+1)v, so if e is an eigenvalue of A then e+1 is an eigenvalue of A+I.
5. false, unless A is the 1x1 matrix. Reason: it is easy to see that
[-1 -1 -1 .. -1 n-1] is an eigenvector with eigenvalue 0, so we have a v such that Av=0, therefore A cannot be invertible.
6. False. One easy example, if A=I and B=-I, then certainly A,B, and AB=-I are all invertible, but A+B=0 is not.

Answer 3

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