Prove or give a counterexample:
1 answer:
Answer: Although A and AT have the same eigenvalues, they may actually have different eigenvectors for that particular eigenvalue. Therefore A and AT have at least one eigenvector that aren't in common.
Step-by-step explanation:
These two attachment explains the provings and counterexamples briefly and simply
You might be interested in
Answer:
Hello,
answer is 204 in base 10 or 21120 in base 3
Step-by-step explanation:
Answer:
Options (A) and (C)
Step-by-step explanation:
A). The given inequality is,
x + y ≤ 2
Solution area of the inequality is the shaded area below the line.
Since (0, 0) lies in the shaded region, ordered pair will be a solution of this inequality.
B). For the inequality y ≤
Since, (0, 0) doesn't lie in the shaded region of this inequality, ordered pair will not be a solution of the inequality.
C). y >
In this graph (0, 0) lies in the shaded region of the inequality therefore, it will be a solution of the given inequality.
Therefore, Options (A) and (C) are the correct options.