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 January 18th, 2011, 01:55 PM #1 Newbie   Joined: Jan 2011 Posts: 1 Thanks: 0 Find the basis of the intersection of two vector subspaces Hello, I find hard to understand the following problem in Linear Algebra... Given two vector subspaces of R5: P = linear span of [ {1,2,-1,1,1},{1,0,0,1,0},{-2,2,2,1,-2}], Q = linear span of [{3,2,-3,1,3},{1,1,0,0,0},{1,-4,-1,-2,1}] find the basis of the intersection P and Q. So according to the Dimension Theorem I know that: dim (P intersects Q) + dim (P + Q) = dim P + dim Q I can clearly see that dim P = dim Q = 3. When I write down those vectors into a 5x6 matrix as columns and bring it to row-echelon form, I get: 1 1 -2 3 1 1 0 1 0 0 1 0 0 0 3 -2 -1 -3 0 0 0 0 3 0 0 0 0 0 0 0 From which I can see that dim (P+Q) = 4, so I can compute dim (P intersects Q) = 2 However, the problem is that I dont know how to find those 2 vectors that form the basis of P int. Q. Any help will be appreciated! Thanks in advance.
 January 19th, 2011, 11:07 PM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Find the basis of the intersection of two vector subspac This is an interesting question. The naive way would be to simply write the equation taking one subspace into the other. We expect two free variables. To be honest, I haven't done it explicitly, but it seems valid to me.
 February 3rd, 2011, 09:48 PM #3 Newbie   Joined: Jan 2011 Posts: 19 Thanks: 0 Re: Find the basis of the intersection of two vector subspac This is a simple intersection problem. I will give you a guideline. We want to find elements both sets have in common. Let any element of a <(1,2,-1,1,1)(1,0,0,1,0)(-2,2,2,1,-2)> be written as: $\alpha$(1,2,-1,1,1)+$\beta$*2nd gen.. etc. Equal this to the other set generators using different letters. Make a eq system equalizing every component of the vector.. Take everything and equalize to zero as to triangulate without an expanded system. Then, as you calculated dim 3 of int (I didn't check this), solve the problem by putting 2 variables in terms of 3 variables. Remember solution would be (alfa,beta,gamma,epsilon,omega) for example.. Bye!
 February 4th, 2011, 03:23 AM #4 Newbie   Joined: Feb 2011 Posts: 3 Thanks: 0 Re: Find the basis of the intersection of two vector subspac $P = \left( \begin{array}{ccc} 1 & 1 & -2 \\ 2 & 0 & 2 \\ -1 & 0 & 2 \\ 1 & 1 & 1 \\ 1 & 0 & -2 \end{array} \right)\$ $Q = \left( \begin{array}{ccc} 3 & 1 & 1 \\ 2 & 1 & -4 \\ -3 & 0 & -1 \\ 1 & 0 & -2 \\ 3 & 0 & 1 \end{array} \right)\$ $P\vec{x}= Q\vec{y} \hspace{80 mm} \line(1,0){200}(1)$ $P\vec{x} - Q\vec{y}= 0$ $[P, -Q] \left( \begin{array}{ccc} \vec{x} \\ \vec{y} \end{array} \right)\ = 0\hspace{30}\line(1,0){200}(2)$ let's call matrix $A=[P, -Q]$ $A = \left( \begin{array}{ccc} 1 & 1 & -2 & -3 & -1 & -1 \\ 2 & 0 & 2 & -2 & -1 & 4\\ -1 & 0 & 2 & 3 & 0 & 1\\ 1 & 1 & 1 & -1 & 0 & 2\\ 1 & 0 & -2 & -3 & 0 & -1 \end{array} \right)\$ By its reduced row echelon form, the nullspace matrix of A has been found which is.... $N = \left( \begin{array}{ccc} -1 & 1.6667 \\ 0 & 0 \\ -1 & -0.6667 \\ 0 & 1 \\ 0 & 0 \\ 1 & 0 \end{array} \right)\$ Thus, the narrowed down particular solutions for $\vec{x}$ and $\vec{y}$ are: (Note: Notice that from equations (1) and (2) $\vec{x}$ and $\vec{y}$ are the intersections of nullspace of A with the rowspace of P and Q respectively) $\vec{x} = c_{1}\left( \begin{array}{ccc} -1\\ 0 \\ -1 \end{array} \right)\ + c_{2}\left( \begin{array}{ccc} 1.6667 \\ 0 \\ -0.6667 \end{array} \right)\$ $\vec{y} = c_{1}\left( \begin{array}{ccc} 0\\ 0 \\ 1 \end{array} \right)\ + c_{2}\left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right)\$ Therefore, the SOLUTION of basis vectors for your question can be found by: $P\left( \begin{array}{ccc} -1 & 1.6667\\ 0 & 0\\ -1 & -0.6667 \end{array} \right)\ = \left( \begin{array}{ccc} 1 & 1 & -2 \\ 2 & 0 & 2 \\ -1 & 0 & 2 \\ 1 & 1 & 1 \\ 1 & 0 & -2 \end{array} \right)\ \left( \begin{array}{ccc} -1 & 1.6667\\ 0 & 0\\ -1 & -0.6667 \end{array} \right)\ = \left( \begin{array}{ccc} 1 & 3\\ -4 & 2\\ -1 & -3 \\ -2 & 1\\ 1 & 3\end{array} \right)\$ Where, the columns of $\left( \begin{array}{ccc} 1 & 3\\ -4 & 2\\ -1 & -3 \\ -2 & 1\\ 1 & 3\end{array} \right)\$ form the basis of the intersection. Notice that its rank(dimension) = 2 and they form a plane. Tip: To be comfortable in solving any linear algebra problem, at the very least, one has to be completely aware of the relationship between nullspace, rowspace, left nullspace and columnspace.

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# intersection of vectors basis and dim

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