(a) let A be a 3x3 matrix and let u,v and w be its column vectors?

Question

suppose that the dimension of Span{u,v,w} is 2
is it then true or false that
(i) the system AX=0 has only the trivial solution
(ii) the matrix A is invertible
Justify your answer
(b) true or false: the sum of invertible matrices is an invertible matrix. justify your answer

Answer 1

If we assume all of the assumptions given about A, then the first one is false because when the reduced row echelon form of A is found, the last column will all be zeros (because if the dimension of the column space is two, then it means that those three are not linearly independent, one of them can be written as a linear combination of the other two) which implies that Ax=0 will have infinite solutions.

The second one will also be false because of the first one. Matrix A being invertible is the same as saying that the system Ax=0 has only one trivial solution which we already know is false.

Part b) is also false. Consider {{1,0},{0,1}} and {{-1,0},{0,-1}}. They are both invertible but when I add them, I get the zero matrix which is invertible.

Answer 2

(a) If the dimensioin of the span {u,v,w} is 2, one on the vectors can be written as a liner combination of the other two. Then rank(A) will be only two, one variable will be free, and Ax = 0, will have nontrivial solutions. Also determinant will be 0, and the matrix is will be uninvertable. All this follows from the dimension being 2, indicating the vectors are not linearly independant.

(b) No, look at this simple counter-example
|1 0|
|0 1| is invertable

|-1 0|
|0 -1| is also invertable.

|1 0| |-1 0| | 0 0|
|0 1| + |0 -1| = | 0 0|, which is not invertable.

Answer 3

I think that you mean the dim of the span is 2. That means that u,v,w are linearly dependent and so there exists a linear combo that equals zero. So, there's a non zero vector that goes to 0.

i, false
ii, false

b) false:
ex:

10
01
+
01
10
= =
11
11

which is not invertible.