# linear algebra bases and dimensions

I've had a go at this question, could anyone please tell me if I'm on the right lines. thanks in advance.

Define a real linear transformation $L_1 : R^4 \rightarrow R^2$ by: $$L_1(x_1, x_2, x_3, x_4) = (3x_1 + x_2 + 2x_3 − x_4, 2x_1 + 4x_2 + 5x_3 − x_4)$$ and let $U_1$ denote the kernel of $L_1$.

Define a real linear transformation $L_2 : R^4 \rightarrow R^2$ by: $$L_2(x_1, x_2, x_3, x_4) = (5x_1 + 7x_2 + 11x_3 + 3x_4, 2x_1 + 6x_2 + 9x_3 + 4x_4)$$ and let $U_2$ denote the kernel of $L_2$.

Construct bases for $U_1$, $U_2$, $U1 \cap U2$, and $U_1 + U_2$.

This is what I got: \begin{align*} U_1&=[(-3,-11,10,0),(3,1,0,10)]&\qquad \dim(U_1)&=2;\\ U_2&=[(-3,-23,16,0),(5,-7,0,8)]&\dim(U_2)&=2;\\ U_1+U_2 &=[(1,0,0,3),(0,1,0,1),(0,0,1,2)]&\dim(U_1+U_2)&=3;\\ \dim(U_1)+\dim(U_2)&= \dim(U_1+U_2)+\dim(U_1 \cap U_2)\\ &\implies 2+2=3+\dim(U_1\cap U_2)\\\ \dim(U1 \cap U2)&=1. \end{align*}

Is this right? I'm not too sure of how to find the base of $\dim(U_1 \cap U_2)$

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Or, better, increase it. – Gerry Myerson Jul 15 '12 at 6:30
what does that mean? – user34742 Jul 15 '12 at 18:17
@user34742: "Accepting an answer" is explained in the FAQ; see here for the process, and here for why you may want to do so. – Arturo Magidin Jul 15 '12 at 19:00

$U_1$ is not equal to an (ordered?) set with two vectors in it. $U_1$ contains an infinite number of vectors. What you actually mean, I think, is that $U_1$ is the span of the vectors you list, and you then assert implicitly that the set is linearly independent and hence you have a basis. But these assertions cannot be done in silence, they must be justified (and unless you use the square brackets as notation for "span", is incorrect as written). Same comment for $U_2$ and for $U_1+U_2$.

Also, you don't actually want to find "the base" of anything, because bases are not unique. There is no such as thing as "the" basis, except in two very special cases (the zero vector space, and a one-dimensional vector space over a field of two elements); we are in neither case here so at best you are trying to find a basis, not the basis.

Finally, it makes no sense to talk about "base of $\dim(U_1\cap U_2)$." $\dim(U_1\cap U_2)$ is a number, not a vector space, so it does not have a basis. You mean, of course, "a basis for $U_1\cap U_2$", but in mathematics we need to be very precise and clear. Sloppy language is almost never forgivable, so it is best to get in the habit of not being wrong through sloppiness.

I'm not sure how you determined the given bases of $U_1$ and $U_2$. I'm guessing you did some sort of Gaussian elimination. I can tell your answers are correct for $U_1$ and $U_2$ because:

1. The dimension of the kernel of $L_1$ is definitely $2$: by the Rank-Nullity Theorem, $4=\dim(U_1)+\dim(\mathrm{range}(L_1))$. We can see that the image of $L_1$ is all of $\mathbb{R}^2$ because $L_1(1,0,0,0) = (3,2)$ and $L_1(0,1,0,0) = (1,4)$. Since $(3,2)$ and $(1,4)$ are linearly independent, the range has dimension at least $2$; since it is a subspace of $\mathbb{R}^2$ the dimension is at most $2$, so the dimension is exactly $2$. The kernel therefore has dimension $2$. The two vectors you give are in the kernel (you can verify by plugging in), and are linearly independent, so that works.

2. A similar argument holds for $L_2$ and $U_2$.

So the first two are correct.

Now, $U_1\cap U_2$ consists of vectors that are in the kernel of $L_1$ and $L_2$. One way to find it is to consider the function $\mathbb{R}^4\to \mathbb{R}^2\times\mathbb{R}^2 = \mathbb{R}^4$ obtained by mapping $\mathbf{x}\longmapsto(L_1(\mathbf{x}),L_2\mathbf{x})$ and find the kernel. That is, solve the system $$\begin{array}{rcccccccl} 3x_1 & + & x_2 &+& 2x_3 & - & x_4 & = & 0\\ 2x_1 & + & 4x_2 & + & 5x_3 & - & x_4 &= & 0\\ 5x_1 & + & 7x_2 & + & 11x_3 & + & 3x_4 & = & 0\\ 2x_1 & + & 6x_2 & + & 9x_3 & + & 4x_4 & = & 0. \end{array}$$ Find the solutions the same way you did with $U_1$ and $U_2$, and that way you can find a basis for $U_1\cap U_2$.

Assuming your basis for $U_1+U_2$ is correct, then the computations for $\dim(U_1\cap U_2)$ are correct.

I don't know how you found the basis for $U_1+U_2$. I would take the two bases we already have, and check if the first basis vector for $U_2$ is not in $U_1$ (it is not), and then check if the second vector of the basis for $U_2$ is in the span of the other three vectors. If not, then the four vectors are linearly independent and the dimension is $4$; if it is, then the basis of $U_1$ together with the single vector from the basis of $U_2$ are a basis for $U_1+U_2$ and the dimension is $3$. It looks to me like you somehow figured out a basis for $U_1+U_2$ from scratch, and I'm not sure how you did that.

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hi, thank you for your response, basically i followed the example that was answered to another question i posted not long ago. i found $U_1+U_2$ by row reducing the vector parametric form of $U_1$ and $U_2$, similar to the way you said to find $U_1\cap U_2$ by combining the bases. i just followed an answer to a question i previously posted – user34742 Jul 15 '12 at 17:55
i row reduced $L_1$ (the 2x4 matrix) and then put it into vector paramentric form to get the bases of $U_1$. i then row reduced $L_2$ (2x4 matrix) and put that into vector parametric form to get the bases. i then put the bases of $L_1$ and $L_2$ together to produce a 4x4 matrix and row reduced that to find the bases of $U_1+U_2$. is that correct? – user34742 Jul 16 '12 at 12:49
@user34742: Yes, that would give you correct answers. And to get a basis for $U_1\cap U_2$, take your last matrix and find a basis for the nullspace. – Arturo Magidin Jul 16 '12 at 12:51