Hello !
i try to solve Linear algebra 2 questions(but need them be written properly as mathmatical proofs)
Having A matrice nXn:
1)proove that if A^2=0 the columns of matrice A are vectors in solution space of the system Ax=0 (x and 0 are vectors of course),and show that p(A)>=n/2
2)proove that if p(A^2)<p(A) (p in all cases here means: the rank of the vectors)
so the system Ax=o has a non trivial solution and the System A^2x=0 has solution y which is Ay≠0,,,,
I have the clue but how write it rught,math way i have big problem..
thank you very much
To prove the given statements in mathematical form, we can follow these steps:
1) Proving that if A^2 = 0, then the columns of matrix A are vectors in the solution space of the system Ax = 0:
Step 1: Assume that A is an n x n matrix and A^2 = 0.
Step 2: We want to show that the columns of A, denoted as [A1, A2, ..., An], are vectors in the solution space of the system Ax = 0.
Step 3: Let's consider the i-th column of A, denoted as Ai. We need to show that Ai is a solution to the system Ax = 0.
Step 4: Multiplying A by Ai, we get:
A * Ai = [A1, A2, ..., An] * Ai = A1 * Ai + A2 * Ai + ... + Ai * Ai + ... + An * Ai
Step 5: Since A^2 = 0, we can substitute in the above equation to get:
A * Ai = A1 * Ai + A2 * Ai + ... + Ai * Ai + ... + An * Ai = 0 + 0 + ... + Ai * Ai + ... + 0 = Ai * Ai
Step 6: From the above equation, we can see that Ai * Ai = 0. This implies that Ai is a solution to the system Ax = 0.
Step 7: Since the columns of A are solutions to the system Ax = 0, it means that they are vectors in the solution space of the system.
To show that p(A) >= n/2:
Step 8: By the rank-nullity theorem, we know that p(A) + nullity(A) = n.
Step 9: The nullity(A) represents the dimension of the null space (solution space) of A.
Step 10: From Step 7, we know that the columns of A are vectors in the solution space of Ax = 0.
Step 11: Therefore, the nullity(A) ≥ number of columns of A = n.
Step 12: Substituting this in Step 8, we have: p(A) + nullity(A) = n ≥ p(A) + n.
Step 13: By rearranging the equation, we get: p(A) ≤ n/2.
Step 14: Therefore, it can be concluded that p(A) ≥ n/2.
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2) Proving that if p(A^2) < p(A), then the system Ax = 0 has a non-trivial solution and the system A^2x = 0 has a solution y where Ay ≠ 0:
Step 1: Assume that A is an n x n matrix and p(A^2) < p(A).
Step 2: We want to show that the system Ax = 0 has a non-trivial solution and the system A^2x = 0 has a solution y where Ay ≠ 0.
Step 3: Since p(A^2) < p(A), it means that the rank of A^2 is strictly less than the rank of A.
Step 4: By the rank-nullity theorem, we have: p(A^2) + nullity(A^2) = n and p(A) + nullity(A) = n.
Step 5: Since p(A^2) < p(A), it implies that nullity(A^2) > nullity(A).
Step 6: From Step 5 and Step 4, we can conclude that the nullity of A^2 is greater than the nullity of A.
Step 7: If the nullity(A^2) > nullity(A), it means that there exists a non-trivial solution x to the system A^2x = 0.
Step 8: Let y = Ax, where x is a non-trivial solution to A^2x = 0.
Step 9: Since y = Ax, it means that y satisfies the system Ax = 0.
Step 10: However, y = Ax ≠ 0, because x is a non-trivial solution to A^2x = 0.
Step 11: Therefore, we can conclude that the system Ax = 0 has a non-trivial solution and the system A^2x = 0 has a solution y where Ay ≠ 0.
I hope this explanation helps! Let me know if you have any further questions.