# Math Help - Questions related to vector spaces and rank of a matrix

1. ## Questions related to vector spaces and rank of a matrix

Hello, I'm new here
I collected some tough questions regarded vector spaces and rank of a matrix that I couldn't answer, and ask your help, please.
1. let U, W be different sub-spaces of a vector space V; dim U= dimW = n-1, dim V = n; find dim(U intersect W).
2. Prove: if B = {v1,...,vn} is a basis of V and if U is a subspace of V with dimension k, k<=n, then there are k vectors in B that spans U.
3. Prove or disprove: if A,B are 3x3 matrices such that rank(A)=rank(B)=2 then AB != 0. I think it's not true but I can't come up with an example that disproves this.

2. Originally Posted by Ginsburg
Hello, I'm new here
I collected some tough questions regarded vector spaces and rank of a matrix that I couldn't answer, and ask your help, please.
1. let U, W be different sub-spaces of a vector space V; dim U= dimW = n-1, dim V = n; find dim(U intersect W).
2. Prove: if B = {v1,...,vn} is a basis of V and if U is a subspace of V with dimension k, k<=n, then there are k vectors in B that spans U.
3. Prove or disprove: if A,B are 3x3 matrices such that rank(A)=rank(B)=2 then AB != 0. I think it's not true but I can't come up with an example that disproves this.
1. Use the theorem which says that $\dim U + \dim W = \dim(U+W) + \dim(U\cap W)$. Notice that if U and W are different then U+W is bigger than either U or W. Since U and W have dimension n–1 it follows that U+W must be the whole of V.

2. This result is false. For example, take V to be 2-dimensional space $\mathbb{R}^2$ with the standard basis consisting of $v_1 = (1,0)$ and $v_2 = (0,1)$, and let U be the 1-dimensional subspace spanned by (1,1).

3. This result is true. If rank(B) = 2 then the subspace $\{Bx:x\in V\}$ is 2-dimensional. But the null space of A is only 1-dimensional. So the space $\{ABx:x\in V\}$ cannot just consist of the zero vector.

3. Originally Posted by Opalg
1. Use the theorem which says that $\dim U + \dim W = \dim(U+W) + \dim(U\cap W)$. Notice that if U and W are different then U+W is bigger than either U or W. Since U and W have dimension n–1 it follows that U+W must be the whole of V.

2. This result is false. For example, take V to be 2-dimensional space $\mathbb{R}^2$ with the standard basis consisting of $v_1 = (1,0)$ and $v_2 = (0,1)$, and let U be the 1-dimensional subspace spanned by (1,1).

3. This result is true. If rank(B) = 2 then the subspace $\{Bx:x\in V\}$ is 2-dimensional. But the null space of A is only 1-dimensional. So the space $\{ABx:x\in V\}$ cannot just consist of the zero vector.
1. why the dimension of U+W is bigger then either U or W when they are diferent?
3. I didn't understand the conenction you made between the space {ABx : x is in V} to the question, can you explain in more details, please?