# linear algebra linear transformation

I would really appreciate if someone could help me with this question, I know that you have to row reduce each matrix and put it in vector parametric form to find $U_1$ and $U_2$. any help would be much appreciated. thank you

Define a real linear transformation $L_1 : \mathbb R^4 \to \mathbb R^2$ by $L_1(x_1; x_2; x_3; x_4) = (3x_1 + x_2 + 2x_3 - x_4; 2x_1 + 4x_2 + x_3 - x_4)$ and let $U_1$ denote the kernel of $L_1$

Define a real linear transformation $L_2 : \mathbb R^2 \to \mathbb R^4$ by $L_2(y_1; y_2) = (y_1 - y_2; y_1 - 3y_2; 2y_1 - 8y_2; 3y_1 - 27y_2)$ and let $U_2$ denote the image of $L_2$

Construct bases for $U_1$, $U_2$, ${U_1 \cap U_2}$, and, $U_1 + U_2$

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BTW there are a few similar questions, you might have a look on them to get better in this type of exercise: How to find basis for intersection of two vector spaces; Finding a basis for the intersection of two subspaces and Constructing a bases for $U_{1}+U_{2}$. – Martin Sleziak Jul 1 '12 at 17:15

You can find a basis for $U_1$ simply by solving the linear system \begin{align} 3x_1 + x_2 + 2x_3 - x_4&=0\\ 2x_1 + 4x_2 + x_3 - x_4&=0 \end{align} Of course, there are many ways to solve this system, e.g. we can try to use elementary row operations: $$\begin{pmatrix} 3 & 1 & 2 & -1 \\ 2 & 4 & 1 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & -3 & 1 & 0 \\ 2 & 4 & 1 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & -3 & 1 & 0 \\ 1 & 7 & 0 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & -3 & 1 & 0 \\ -1 & 7 & 0 & 1 \\ \end{pmatrix}$$

Now we see that the original system is equivalent to the the system of equations $x_1-3x_2+x_3=0$, $-x_1+7x_2+x_4=0$; i.e. $x_3=-x_1+3x_2$, and $x_4=x_1-7x_2$. We see that for any choice of $x_1$, $x_2$ we obtain a solution of the form $$(x_1,x_2,-x_1+3x_2,x_1-7x_2)=x_1(1,0,-1,1)+x_2(0,1,3,7),$$ which means that $U_1=[(1,0,-1,1),(0,1,3,7)]$. I.e., these two vectors are the basis for $U_1$. (Of course, there are many different basis for the same subspace. So might get different vectors. It is also useful to check whether the vectors in basis fulfill the original system - if not, there must be a mistake somewhere.)

The space $U_2$ is generated by vectors $L_2(1,0)$ and $L_2(0,1)$, hence $U_2=[(1,1,2,3),(-1,-3,-8,-27)]$. These vectors are linearly independent, so they form a basis. (To check whether two vectors are independent you only need to check whether one of them is a multiple of the other one.)

Now let us try $U_1+U_2$. We know that $U_1+U_2=[(1,0,-1,1),(0,1,3,7),(1,1,2,3),(-1,-3,-8,-27)]$, but we don't know whether these vectors are independent. So we try to make row echelon form. $$\begin{pmatrix} 1 & 0 & -1 & 1\\ 0 & 1 & 3 & 7\\ 1 & 1 & 2 & 3\\ -1 &-3 &-8 &-27 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -1 & 1\\ 0 & 1 & 3 & 7\\ 0 & 0 & 0 & 1\\ -1 &-3 &-8 &-27 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -1 & 0\\ 0 & 1 & 3 & 0\\ 0 & 0 & 0 & 1\\ -1 &-3 &-8 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -1 & 0\\ 0 & 1 & 3 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ Now we know that vectors $(1,0,-1,0)$, $(0,1,3,0)$, $(0,0,0,1)$ form a basis of $U_1+U_2$ and that $d(U_1+U_2)=3$.

Since $d(U_1)+d(U_2)=d(U_1+U_2)+d(U_1\cap U_2)$, we know that $d(U_1\cap U_2)=1$.

If we notice that $(2,4,10,30)=(1,1,2,3)+(1,3,8,27)$ and $2(1,0,-1,1)+4(0,1,3,7)=(2,4,10,30)$, we see that the vector $(1,2,5,10)$ belongs to $U_1\cap U_2$. Since $U_1\cap U_2$ is one-dimensional, this implies that $U_1\cap U_2=[(1,2,5,10)]$.

Now let's try to think how we could find the basis for $U_1\cap U_2$ without guesswork.

One possibility is to have a closer look at what we have done when doing row operations. If we denote the rows of the above matrix by $\vec a_1$, $\vec a_2$, $\vec b_1$, $\vec b_2$, then we have can find out (using row operation similar to the above) that: $\vec a_1+\vec a_2-\vec b_1=(0,0,0,5)$ $\vec a_1+3\vec a_2+\vec b_2=(0,0,0,-5)$ Obviously by combining the above two vectors we get $2\vec a_1+4\vec a_2+\vec b_2-\vec b_1=\vec 0$, which is the same as $$2\vec a_1+4\vec a_2 = \vec b_1-\vec b_2,$$ hence this vector belongs to the intersection.

The above was still based on the lucky fact that we only had to find one vector. Can we approach this in a more systematic way?

Of course we can. To find vectors in $U_1\cap U_2$ we can solve the linear system of equations given by $x_1\vec a_1+x_2 \vec a_2=y_1\vec b_1+y_2 \vec b_2$. Once we know all possible pairs $(x_1,x_2)$, we can obtain all vectors belonging to $U_1\cap U_2$.

Based on your comment I've guess you got the following when solving the system: $$\begin{pmatrix} 3 & 1 & 2 & -1 \\ 2 & 4 & 1 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 7/10 & -3/10 \\ 0 & 1 & -1/10 & -1/10 \end{pmatrix}$$

From the last matrix you can get that the solutions of this system form the subspace $U_1=[(-7/10,1/10,1,0),(3/10,1/10,0,1)]=[(-7,1,10,0),(3,1,0,10)]$.

This is another basis for the same subspace $U_1$. (And this is also correct.)

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thank you very much for answering my question, you've been amazing help, just one question though, how did you find those vectors for $U_1$? i row reduced it and got [1,0,7/10,-3/10] and [0,1,-1/10,-1/10] and i then applied it to the rest of the outline you gave me and it didnt really work out. the rest i completely understand thank you – user34742 Jul 2 '12 at 14:00
@user34742 From your comment I understand that you have tried to solve the system in a different way and there were some problems. I've tried to address this in the last part of my (edited) post. – Martin Sleziak Jul 2 '12 at 16:08
thank you very much for your help it is much appreciated – user34742 Jul 5 '12 at 13:55