# Questions related to vector spaces and rank of a matrix

• Jul 29th 2010, 08:33 AM
Ginsburg
Questions related to vector spaces and rank of a matrix
Hello, I'm new here (Hi)
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.
• Jul 29th 2010, 10:45 AM
Opalg
Quote:

Originally Posted by Ginsburg
Hello, I'm new here (Hi)
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 $\displaystyle \dim U + \dim W = \dim(U+W) + \dim(U\cap W)$. 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 $\displaystyle \mathbb{R}^2$ with the standard basis consisting of $\displaystyle v_1 = (1,0)$ and $\displaystyle v_2 = (0,1)$, and let U be the 1-dimensional subspace spanned by (1,1).

3. This result is true. If rank(B) = 2 then the subspace $\displaystyle \{Bx:x\in V\}$ is 2-dimensional. But the null space of A is only 1-dimensional. So the space $\displaystyle \{ABx:x\in V\}$ cannot just consist of the zero vector.
• Jul 29th 2010, 11:28 AM
Ginsburg
Quote:

Originally Posted by Opalg
1. Use the theorem which says that $\displaystyle \dim U + \dim W = \dim(U+W) + \dim(U\cap W)$. 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 $\displaystyle \mathbb{R}^2$ with the standard basis consisting of $\displaystyle v_1 = (1,0)$ and $\displaystyle v_2 = (0,1)$, and let U be the 1-dimensional subspace spanned by (1,1).

3. This result is true. If rank(B) = 2 then the subspace $\displaystyle \{Bx:x\in V\}$ is 2-dimensional. But the null space of A is only 1-dimensional. So the space $\displaystyle \{ABx:x\in V\}$ 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?