2
$\begingroup$

Can someone help me with this Linear Subspace problem?

If $U$ and $W$ are linear subspaces of $\Bbb R^3$ defined by:

$$U = \{(x,y,z)\in \Bbb R^3 \mid x+y+z =0\} \\ W = \operatorname{span}[(1,0,1),(0,-1,1)]$$ Determine a linear operator $T$ of $\Bbb R^3$ such that $\operatorname{Im}(T) = U$ and $\operatorname{Ker}(T) = U\cap W$.

$\endgroup$
  • $\begingroup$ Do you have any thoughts? $\endgroup$ – user137731 Jul 9 '16 at 22:46
  • $\begingroup$ i don't get how i can fit U in a R³ operator, if U is only x+y+z=0... $\endgroup$ – Popeye Jul 9 '16 at 22:48
  • $\begingroup$ Do you know what $\operatorname{Im}$ and $\operatorname{Ker}$ mean? $\endgroup$ – user137731 Jul 9 '16 at 22:49
  • $\begingroup$ Yes, i'll just have some trouble trying to explain it in english $\endgroup$ – Popeye Jul 9 '16 at 22:52
  • $\begingroup$ OK. Then consider the linear operator $S$ given by $S(x,y,z) = (x,0,0)$. Then $\operatorname{Im}(S) = \{(x,y,z)\mid y=z=0\}$ is "fit in $\Bbb R^3$" even though it's a smaller space. So that's not a problem. As for what to do in your problem, the first thing you need to do is figure out what $U\cap W$ is. Can you do that part yourself? $\endgroup$ – user137731 Jul 9 '16 at 22:54
0
$\begingroup$

Here’s another way to approach the problem:

You’ve already worked out that $v_1=(0,-1,1)$ spans $U\cap W$, so you know that you must have $T(v_1)=0$. Extend this to a complete basis $(v_1,v_2,v_3)$ of $\mathbb R^3$ and choose a basis $(u_1, u_2)$ for $U$. Define $T(v_2)=u_1$ and $T(v_3)=u_2$. Since $(v_1,v_2,v_3)$ is a basis of $\mathbb R^3$, this specifies $T$ completely. It’s clear from the construction that $\ker(T)=U\cap W$ and $\operatorname{im}(T)=U$. The solution is obviously not unique, since there are many choices for these bases.

As a concrete example, take $u_1=v_1=(0,-1,1)$ and $u_2=v_2=(1,-1,0)$. You can easily verify that this is indeed a basis for $U$. For $v_3$, we know from the definition of $U$ that $(1,1,1)$ annihilates this space, so it will do for a third basis vector. We can immediately write down the matrix of $T$ relative to this basis: $$\pmatrix{0&1&0\\0&0&1\\0&0&0}.$$ A simple change of basis operation gives us the matrix of $T$ relative to the standard basis: $$\pmatrix{0&1&1\\-1&-1&1\\1&0&1}\pmatrix{0&1&0\\0&0&1\\0&0&0}\pmatrix{0&1&1\\-1&-1&1\\1&0&1}^{-1}=\pmatrix{\frac13&\frac13&\frac13\\-1&0&0\\\frac23&-\frac13&-\frac13}.$$ In fact, this construction can be used to find a matrix for $T$ relative to the standard basis in the general case, above: Extend $(u_1,u_2)$ to a basis $(u_1,u_2,u_3)$ of $\mathbb R^3$. The matrix of $T$ relative to the standard basis will then be $$\pmatrix{\mid&\mid&\mid\\u_1&u_2&u_3\\\mid&\mid&\mid}\pmatrix{0&1&0\\0&0&1\\0&0&0}\pmatrix{\mid&\mid&\mid\\v_1&v_2&v_3\\\mid&\mid&\mid}^{-1}=\pmatrix{\mid&\mid&\mid\\0&u_1&u_2\\\mid&\mid&\mid}\pmatrix{\mid&\mid&\mid\\v_1&v_2&v_3\\\mid&\mid&\mid}^{-1}.$$ Note that this matrix is in fact independent of the choice of $u_3$, which makes sense since it’s not part of the image.

$\endgroup$

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.