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Topic: **Homework matrix****Question:**
Let A be and m x n matrix.
A. If B is a nonsingular m x m matrix, show that BA and A have the same nullspace.
B. If C is a nonsingular n x n matrix, show that AC and A have the same rank

June 21, 2019 / By Clemency

HOMEWORK! These are easy enough that you need to see how they are done, though. A. The point is that Bx=0 if and only if x=0 since B is non-singular. Thus, BAx=0 is the same as B(Ax)=0, so Ax=0. In other words, the null space of BA is the same as that for A. B. The rank of A is the number of independent rows of the matrix A. But the k^th row of AC will be the k^th row of A multiplied by C. But since C is non-singular, it takes independent vectors to independent vectors and vice versa.

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HOMEWORK! These are easy enough that you need to see how they are done, though. A. The point is that Bx=0 if and only if x=0 since B is non-singular. Thus, BAx=0 is the same as B(Ax)=0, so Ax=0. In other words, the null space of BA is the same as that for A. B. The rank of A is the number of independent rows of the matrix A. But the k^th row of AC will be the k^th row of A multiplied by C. But since C is non-singular, it takes independent vectors to independent vectors and vice versa.

1) just write out. nullspace v :: Av=0 since B is non sigular Bv=0 has only v=0 as solution Aw ; with w nullspace of A is Aw=0 by definition and BAw = 0 because B0=0. AC is not defined ...

1) just write out. nullspace v :: Av=0 since B is non sigular Bv=0 has only v=0 as solution Aw ; with w nullspace of A is Aw=0 by definition and BAw = 0 because B0=0. AC is not defined ...

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To show it’s a basis, we need to show it spans and is lin. independent. Since the dim of M2x2 is 4 we have the correct number of vectors in each set. Therefore we only need to verify one of them. Let’s show they are independent. All these problems start the same way: consider k1u1+k2u2+k3u3+k4u4=0 where the 0 on the right is the zero 2x2 matrix. The question is what kind of solutions do we get for the ks. If the only solution is k1=k2=k3=k4=0, then the system is linearly independent and they are a basis. If there are nontrivial solutions for the ks, the system is dependent and thus not a basis. Multiply each matrix by the k in front of it. Then we have a matrix equation: by the definition of equality, the sum of the elements in row1,colm1 of the 4 matrices on the left has to equal the row1colm1 entry on the right side (which is zero here), and similarly for the other 3 positions. This yields a 4x4 system in the 4 unknowns (k1, k2, k3, k4). The first eq. for example is k1 + k2 + k4=0, etc. Solving the system, I get that the only solution is the trivial one. (This is confirmed by the fact that the det is not zero.) Therefore, the u’s form a basis. If you do the same for the v’s, you’ll get that there are nontrivial solutions, so the v’s don’t form a basis. Now we need to write the matrix: row1=[2 0], row2=[-2 1] (call it matrix A) as a linear combination of the u-basis. Again take k1u1+k2u2+k3u3+k4u4 and set it equal to matrix A. This gives 4 eqs in the 4 unknowns (k’s). Solving I get that A=2u1 +u2 + 2u3 – u4. The coordinate vector with respect to this basis is simply the coefficients of each (in order) and so = (2, 1, 2, -1). Again, an important point to keep in mind in any problem asking if something is a basis is that the starting point to solve it is by writing k1u1 + k2u2 + k3u3 + ...=0 !! Let me know if you need more details.

The matrix version of the subject is AX = b the place A is a 2x2 matrix [4, ok; ok, a million] X is a vector variable [x; y] and b is a vector consistent [7; 0] there'll be a distinctive answer as long as A is non-singular. there won't in any respect be a distinctive answer if A *is* singular. For A to be singular det(A) = 0 the place det() is the "determinant" of a sq. matrix for a 2x2 matrix [a, b; c, d] det([a, b; c, d]) = advert - bc subsequently on your case 4*a million - ok*ok = 0 ok^2 = 4 ok = +/- 2

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