# 2 subspaces of R^4, bases, intersection

• Feb 15th 2009, 08:50 AM
arbolis
2 subspaces of R^4, bases, intersection
Hi MHF,
I appreciate very much all the help I got so far. If I didn't answered my last 2 posts it's because of lack of time. I feel overloaded with linear Algebra and the final exam is coming up next 26th of February.
Here is the exercise I'd like to check out with you.
Let $S_1=\{ (1,0,1,2), (-1,2,3,8), (-1,1,1,3), (0,1,2,5) \}$ and $S_2=\{ (1,1,0,1), (1,-1,0,3), (3,1,2,4), (1,1,2,0) \}$ and let $W_1$ be spanned by $S_1$ and $W_2$ be spanned by $S_2$.
a)Give a base $B_i$ of $W_i$ contained in $S_i$, for i=1, 2.
My attempt : I wrote each vector of the base into a matrix as being column vectors and reduced it. I got that 3 of the 4 vectors of $S_1$ were colinear and then I solve a 2 equations system, getting finally that a base of $W_1$ included into $S_1$ is $\{ (1,0,1,-1), (0,1,-1,1) \}$.
Doing the same for $S_2$, I got that a base of $W_2$ included into $S_2$ is $\{ (1,0,0,0), (0,-1,1,-1) \}$.
b)Describe $W_1$, $W_2$, $W_1 \cap W_2$ and $W_1+W_2$ implicitly.
My attempt : don't know how to start.
c)Give a base of $W_1 \cap W_2$ and of $W_1+W_2$.
My attempt : I wrote the bases I got in part a) as vector column in a matrix. I checked out that $\det A \neq 0$, hence the matrix is invertible and the 4 vectors are linear independent. So I think that $W_1 \cap W_2 =$∅. So there's no base that generates $W_1 \cap W_2$, or I'm wrong?
And as the 4 vectors are linear independent, a basis of $W_1+W_2$ would be the union of the bases of $W_1$ and $W_2$. Or : $\{ (1,0,1,-1), (0,1,-1,1),(1,0,0,0), (0,-1,1,-1) \}$. Am I right?
• Feb 15th 2009, 03:54 PM
math2009
a. $A_1=[\vec{S}_{11}\ \ \vec{S}_{12}\ \ \vec{S}_{13}\ \ \vec{S}_{14}]=\begin{bmatrix} 1&-1&-1&0\\0&2&1&1\\1&3&1&2\\2&8&3&5 \end{bmatrix}$, $A_2=[\vec{S}_{21}\ \ \vec{S}_{22}\ \ \vec{S}_{23}\ \ \vec{S}_{24}]=\begin{bmatrix} 1&1&3&1\\1&-1&1&1\\0&0&2&2\\1&3&4&0 \end{bmatrix}$

$rref(A_1)=\begin{bmatrix} 1&0&-\frac{1}{2}&\frac{1}{2}\\0&1&\frac{1}{2}&\frac{1}{ 2}\\0&0&0&0\\0&0&0&0 \end{bmatrix},\ rref(A_2)=\begin{bmatrix} 1&0&0&-1\\0&1&0&-1\\0&0&1&1\\0&0&0&0 \end{bmatrix}$

$quantity\ of\ vectors\ in\ basis:\ dim(W_1)=rank(A_1)=2, dim(W_2)=rank(A_2)=3$ ,
you can select $\mathfrak{B}_1=(\vec{S}_{11},\vec{S}_{12})\ as\ basis\ of \ W_1,\ \mathfrak{B}_2=(\vec{S}_{21},\vec{S}_{22},\vec{S}_ {23})\ as\ basis\ of \ W_2$

b. we suppose $\vec{v}_1,\cdots ,\vec{v}_p, \vec{w}_1,\cdots ,\vec{w}_q , \vec{b}_1,\cdots ,\vec{b}_m \in R^n \ linear\ independent$ and $W_1=span(\vec{v}_1,\cdots ,\vec{v}_p, \vec{b}_1,\cdots ,\vec{b}_m),\ W_2=span(\vec{w}_1,\cdots ,\vec{w}_q, \vec{b}_1,\cdots ,\vec{b}_m)$

Then, $dim(W_1)=p+m,\ dim(W_2)=q+m,\ dim(W_1+W_2)=p+q+m,\ dim(W_1\cap W_2)=m$
Obviously, $dim(W_1) + dim(W_2) -dim(W_1\cap W_2)=dim(W_1+W_2)=p+q+m$

c. $A_4=\frac{[ \vec{S}_{11}\ \ \vec{S}_{12}}{\mathfrak{B}_1}\ \ \frac{\vec{S}_{21}\ \ \vec{S}_{22}\ \ \vec{S}_{23}]}{\mathfrak{B}_2}, rref(A_4)=\begin{bmatrix} 1&0&0&0&\frac{9}{4}\\0&1&0&0&-\frac{1}{12}\\0&0&1&0&\frac{11}{12}\\0&0&0&1&-\frac{1}{4} \end{bmatrix}$ , $\ dim(W_1+W_2)=4=dim(R^4)$
There are only 4 vectors which are linear independent, $so\ W_1+W_2=span(\vec{S}_{11},\vec{S}_{12},\vec{S}_{21 },\vec{S}_{22})$ $=span(\vec{e}_1,\vec{e}_2,\vec{e}_3,\vec{e}_4)=R^4$

$\because\frac{9}{4}\vec{S}_{11}-\frac{1}{12}\vec{S}_{12}+\frac{11}{12}\vec{S}_{21}-\frac{1}{4}\vec{S}_{22}=\vec{S}_{23}\ \Rightarrow$ $\frac{\frac{9}{4}\vec{S}_{11}-\frac{1}{12}\vec{S}_{12}}{in\ W_1}=\frac{-\frac{11}{12}\vec{S}_{21}+\frac{1}{4}\vec{S}_{22}+ \vec{S}_{23}}{in\ W_2}$
$\therefore W_1 \cap W_2 = span(\frac{9}{4}\vec{S}_{11}-\frac{1}{12}\vec{S}_{12})$ $=span(-\frac{11}{12}\vec{S}_{21}+\frac{1}{4}\vec{S}_{22}+ \vec{S}_{23})=span(\begin{bmatrix} \frac{7}{3}\\ -\frac{1}{6}\\ 2 \\ \frac{23}{6} \end{bmatrix})$

I will update frequently if I find any mistake. So you may find contents changed when refresh Web page.
arbolis, to be careful! Linear Algebra will be a great challenge in final exam.(Wink)
• Feb 15th 2009, 05:23 PM
HallsofIvy
Quote:

Originally Posted by arbolis
Hi MHF,
I appreciate very much all the help I got so far. If I didn't answered my last 2 posts it's because of lack of time. I feel overloaded with linear Algebra and the final exam is coming up next 26th of February.
Here is the exercise I'd like to check out with you.
Let $S_1=\{ (1,0,1,2), (-1,2,3,8), (-1,1,1,3), (0,1,2,5) \}$ and $S_2=\{ (1,1,0,1), (1,-1,0,3), (3,1,2,4), (1,1,2,0) \}$ and let $W_1$ be spanned by $S_1$ and $W_2$ be spanned by $S_2$.
a)Give a base $B_i$ of $W_i$ contained in $S_i$, for i=1, 2.
My attempt : I wrote each vector of the base into a matrix as being column vectors and reduced it. I got that 3 of the 4 vectors of $S_1$ were colinear and then I solve a 2 equations system, getting finally that a base of $W_1$ included into $S_1$ is $\{ (1,0,1,-1), (0,1,-1,1) \}$.

What? These vectors may well span $W_1$ but neither of them is in $S_1$. Did you misunderstand? If $W_1$ is 2 dimensional, then "a basis for $W_1$ contained in $S_1$" must be two of the vectors in $S_1$.

Quote:

Doing the same for $S_2$, I got that a base of $W_2$ included into $S_2$ is $\{ (1,0,0,0), (0,-1,1,-1) \}$.
Same point as before. Neither of these vectors is in $S_2$.

Quote:

b)Describe $W_1$, $W_2$, $W_1 \cap W_2$ and $W_1+W_2$ implicitly.
My attempt : don't know how to start.
c)Give a base of $W_1 \cap W_2$ and of $W_1+W_2$.
My attempt : I wrote the bases I got in part a) as vector column in a matrix. I checked out that $\det A \neq 0$, hence the matrix is invertible and the 4 vectors are linear independent. So I think that $W_1 \cap W_2 =$∅. So there's no base that generates $W_1 \cap W_2$, or I'm wrong?
And as the 4 vectors are linear independent, a basis of $W_1+W_2$ would be the union of the bases of $W_1$ and $W_2$. Or : $\{ (1,0,1,-1), (0,1,-1,1),(1,0,0,0), (0,-1,1,-1) \}$. Am I right?
• Feb 15th 2009, 06:21 PM
arbolis
Thank you very much for the help to both.
I'll check out the answers more in details in some hours.
Quote:

arbolis, to be careful! Linear Algebra will be a great challenge in final exam.(Wink)
It will be a massacre.

Quote:

What? These vectors may well span http://www.mathhelpforum.com/math-he...ce082a7f-1.gif but neither of them is in http://www.mathhelpforum.com/math-he...5727065b-1.gif. Did you misunderstand?
I'm not even sure I misunderstood. I just made an error I'd say. I understand my error.
But I don't know what I've done then.
Yeah, linear Algebra is a tough challenge for me. I'm thinking of taking the final exam in July instead of this month, but if I do so I'll likely graduate later. If I fail too, and I'll get less chance for being accepted into a graduate program.
• Feb 15th 2009, 06:54 PM
math2009
I update solution. Please refresh web page.