True/False along with reasons and explanations - A is an nxn matrix that has no inverse, so the only eigenvalues of A is 0. - If the vector v is an eigenvector of the matrix A, then kv is also an eigenvector provided that k ≠ 0. - A is row equivalent to B, then A and B have the same eigenvalues. - It is known that the two eigenvectors are linearly independent, it can be concluded that they come from different eigenspaces. - Every matrix that has an inverse can be diagonalized. - If A is a non-singular matrix that can be diagonalized, then A^T and A^-1 can also be diagonalized. - A matrix of order 3x3. There is a basis R^3 which consists of the eigenvectors of A. Then A can be diagonalized. - If the matrix can be diagonalized, then the matrix is diagonally single. - A and B are similar to each other. B and C are similar, so det(A) = det(C)
True/False along with reasons and explanations
- A is an nxn matrix that has no inverse, so the only eigenvalues of A is 0.
- If the vector v is an eigenvector of the matrix A, then kv is also an eigenvector provided that k ≠ 0.
- A is row equivalent to B, then A and B have the same eigenvalues.
- It is known that the two eigenvectors are linearly independent, it can be concluded that they come from different eigenspaces.
- Every matrix that has an inverse can be diagonalized.
- If A is a non-singular matrix that can be diagonalized, then A^T and A^-1 can also be diagonalized.
- A matrix of order 3x3. There is a basis R^3 which consists of the eigenvectors of A. Then A can be diagonalized.
- If the matrix can be diagonalized, then the matrix is diagonally single.
- A and B are similar to each other. B and C are similar, so det(A) = det(C)
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