1
$\begingroup$

The sub-vector spaces $U_1=span\{(0,1,2),\;(1,1,1),\;(3,5,7)\}$ and $U_2=span\{(1,1,0),\;(-1,2,2),\;(2,-13,-10),\;(2,-1,-2)\}$ of the $\mathbb{R}^3$ are given. Determine the dimension $\dim$ and a base of $U_1$,$U_2$, $U_1+U_2$ and $U_1\cap U_2$ in each case.

In order to determine the Basis of $U_1$, we just need to insert all vectors of $U_1$ into a matrix and do the Gauss-algorithm: $$ \begin{pmatrix} 0 & 1 & 3\\ 1 & 1 & 5\\ 2 & 1 & 7 \end{pmatrix} \iff \begin{pmatrix} 1 & 1 & 5\\ 0 & -1 & -3\\ 0 & 0 & 0 \end{pmatrix} $$ Therefore, $\dim(U_1)=2$ and $B_{U_1}=\{(0,1,2),\;(1,1,1)\}$ is a basis of $U_1$

Same procedure for $U_2\dots\implies\dim(U_2)=2$ and $B_{U_2}=\{(1,1,0),\;(-1,2,2)\}$

Now a question: For $U_1+U_2$: Could I use $B_{U_1}$ and $B_{U_2}$ in order to determine the basis and dimension of the sum of both subspaces? (Like that):

$$ \begin{pmatrix} 1 & -1 & 2 & 2\\ 1 & 2 & -13 & -1\\ 0 & 2 & -10 & -2 \end{pmatrix} \iff \begin{pmatrix} 1 & -1 & 2 & 2\\ 0 & 3 & -15 & -3\\ 0 & 0 & 0 & 0 \end{pmatrix} $$ Therefore $\dim(U_1+U_2)=3$ and one basis is $B_{sum}=\{(0,1,2),(1,1,1),(1,1,0)\}$

Now $U_1 \cap U_2 $:

We need to create a linear equation system: \begin{align} \lambda_1(0,1,2)+\lambda_2(1,1,1)+\lambda_3(3,5,7)&=\lambda_4(1,1,0)+\dots+\lambda_7(2,-1,-2)\iff\\ \lambda_1(0,1,2)+\lambda_2(1,1,1)+\lambda_3(3,5,7)-&\lambda_4(1,1,0)-\dots-\lambda_7(2,-1,-2)&=0 \end{align} Because we can multiply the matrix by $(-1)$, we don't need to be aware of the negative signs.

Inserting this into a matrix: $$ \begin{pmatrix} 0 & 1 & 3 & &\dots& & 2\\ 0 & 1 & 5 & &\dots& & -1\\ 2 & 1 & 7 & 0 & 2 & -10 & -2 \end{pmatrix} \iff \begin{pmatrix} 1 & 1 & 5 & &\dots& & 2\\ 0 & -1 & -3 & &\dots& & -1\\ 0 & 0 & 0 & 0 & 2 & -10 & -2 \end{pmatrix} $$ Now we solve the linear equation system and get $\lambda_7=z,\;\lambda_6 = y, \;\lambda_5 = x,\;\lambda_4=-3x+18y+2z,\;\lambda_3=w,\;\lambda_2=-3w+4x-20y-4z,\;\lambda_1=-2w-3x+15y+3z$ We insert those Lambdas on one side of the equation and simplify the term a bit. After that, we get that the basis should be:

$B=\{x(-4,1,2)+y(20,5,-10)+z(4,1,-2)\mid x,y,z\in \mathbb{R}\}$ Therefore, $\dim(intersection)=3$


Are those concepts right? Can you tell me if the dimensions and the basis are right?

$\endgroup$
  • 1
    $\begingroup$ How can the dimension of the intersection of two spaces be greater than the dimension of either one? $\endgroup$ – amd May 14 '19 at 23:38
1
$\begingroup$

Alarm bells should have been going off in your head when you came up with $\dim(U_1\cap U_2)=3$. The intersection of a set $S$ with any other set is a subset of $S$. How can a subset of $U_1$ have a dimension greater than $\dim U_1$? At any rate, once you know the dimensions of the other three spaces, the formula $\dim(U_1)+\dim(U_2)=\dim(U_1+U_2)+\dim(U_1\cap U_2)$ immediately gives you the dimension of the intersection: $1$.

As egreg points out in his answer, you really only needed to consider your computed basis vectors for $U_1$ and $U_2$ when computing $U_1\cap U_2$, but that’s a minor issue that only led to your doing some unnecessary work. The crucial errors you made lie in what you then did with the solution to the system of equations. Substituting into the left-hand side of the original equation and simplifying was fine, although you made a sign error: the correct expression is $x(-4,-1,2)+y(20,5,-10)+z(4,1,-2)$. Then you say that the set $B$ is a basis for $U_1\cap U_2$, but that can’t be right because $B$ is an infinite set! In fact $B=U_1\cap U_2$. It is not three-dimensional, however, despite there being three free variables in your expression for the typical element of $B$. The elements of a basis must be linearly independent, but $(20,5,-10)$ and $(4,1,-2)$ are obviously scalar multiples of each other, as is the correct third vector $(-4,-1,2)$. So, you can simplify even further: $$x(-4,-1,2)+y(20,5,-10)+z(4,1,-2)=(-x+5y+z)(4,1,-2)$$ hence $U_1\cap U_2=\{k(4,1,-2)\mid k\in\mathbb R\}$.

Incidentally, you could’ve computed $U_1 + U_2$ and its dimension with much less work. With a basis in hand for $U_1$, it’s easy to verify that $(1,1,0) \notin U_1$: Since only $(1,1,1)$ can contribute anything to the first component, we must have $(1,1,0)=(1,1,1)+c(0,1,2)$. From the second component $c=0$, but $(1,1,0)\ne(1,1,1)$. So, $\dim(U_1+U_2)\ge3$, but you’re working in $\mathbb R^3$, so no subspace can have a dimension greater than $3$, and the only subspace of dimension $3$ is $\mathbb R^3$ itself. Therefore, $\dim(U_1+U_2)=3$ and a basis for it is the standard basis of $\mathbb R^3$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why is $x(-4,-1,2)+y(20,5,-10)+z(4,1,-2)=\color{red}{(-x+4y+z)(4,1,-2)}$? $\endgroup$ – Doesbaddel May 15 '19 at 7:08
  • 1
    $\begingroup$ @Doesbaddel Sorry, that should’ve been $5y$. It’s because $(-4,-1,2)=-(4,1,-2)$ and $(20,5,-10)=5(4,1,-2)$. $\endgroup$ – amd May 15 '19 at 7:11
  • $\begingroup$ Alright, I understand it now. $\endgroup$ – Doesbaddel May 15 '19 at 7:13
  • $\begingroup$ Why do I need to write $U_1\cap U_2=\{k(4,1,-2)\mid k\in\mathbb R\}$ instead of $U_1\cap U_2=\{(4,1,-2)\}$? $\endgroup$ – Doesbaddel May 15 '19 at 7:15
  • 1
    $\begingroup$ @Doesbaddel You don’t seem to quite understand what “basis” means, since you make a similar mistake in your question. $U_1\cap U_2$ consists of all scalar multiples of $(4,1,-2)$, i.e., it’s the set $\{k(4,1,-2)\mid k\in\mathbb R\}$. The single vector $(4,1,-2)$ obviously spans this vector space, therefore a basis for the space is the singleton set $\{(4,1,-2)\}$. $\endgroup$ – amd May 15 '19 at 7:20
1
$\begingroup$

Yes, you can merge any spanning sets of $U_1$ and $U_2$ in order to find the dimension of the sum.

Once you have determined bases for $U_1$ and $U_2$, the linear system for finding vectors in the intersection is much simpler: $$ a(0,1,2)+b(1,1,1)=c(1,1,0)+d(-1,2,2) $$ This produces the matrix (Gaussian elimination follows) \begin{align} \begin{pmatrix} 0 & 1 & -1 & -1 \\ 1 & 1 & -1 & 2 \\ 2 & 1 & 0 & 2 \end{pmatrix}&\to \begin{pmatrix} 1 & 1 & -1 & 2 \\ 0 & 1 & -1 & -1 \\ 2 & 1 & 0 & 2 \end{pmatrix}\\&\to \begin{pmatrix} 1 & 1 & -1 & 2 \\ 0 & 1 & -1 & -1 \\ 0 & -1 & 2 & -2 \end{pmatrix}\\&\to \begin{pmatrix} 1 & 1 & -1 & 2 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & -3 \end{pmatrix}\\&\to \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & -3 \end{pmatrix}\\&\to \begin{pmatrix} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & -3 \end{pmatrix} \end{align} and you see that a nonzero solution is $a=-3$, $b=4$, $c=3$, $d=1$. Thus the vector $$ -3(0,1,2)+4(1,1,1)=(4,1,-2) $$ forms a basis of the intersection.

By Grassmann's formula $$ \dim(U_1\cap U_2)=\dim U_1+\dim U_2-\dim(U_1+U_2)=2+2-3=1 $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This method saves a lot of time, nice. $\endgroup$ – Doesbaddel May 15 '19 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.