Given an isomorphism linear transformation $T:V \to \mathbb C^3$,where $V$ is a vector space over $\mathbb C$,assume $\alpha_1,\alpha_2,\alpha_3,\alpha_4 \in V$ such that: $$T(\alpha_1)=(1,0,i)$$ $$T(\alpha_2)=(-2,1+i,0)$$ $$T(\alpha_3)=(-1,1,1)$$ $$T(\alpha_4)=(\sqrt 2,i,3)$$

  1. Is $\alpha_1$ in the subspace spanned by $\alpha_2,\alpha_3$?
  2. If $W_1$ is the subspace spanned by $\alpha_1,\alpha_2$ and $W_2$ is the subspace spanned by $\alpha_3,\alpha_4$,then find the intersection of $W_1$ and $W_2$.
  3. Find a basis for the subspace spanned by $\alpha_1,\alpha_2,\alpha_3,\alpha_4$.

$\alpha_1$ is the subspace spanned by $\alpha_2,\alpha_3$ if it can be written as a (finite) linear combination of $\alpha_2,\alpha_3$,if there exist $\lambda_2,\lambda_3 \in \mathbb C$ such that $$\alpha_1=\lambda_2\alpha_2+\lambda_3\alpha_3$$

Or equivalently (the equivalence is duo to the fact that $T$ is an isomorphism which follows that it does have an inverse) $$T(\alpha_1)=\lambda_2T(\alpha_2)+\lambda_3T(\alpha_3)$$ $$(1,0,i)=\lambda_2(-2,1+i,0)+\lambda_3(-1,1,1)$$ Which implies that $\lambda_2=i(i+1)/2,\lambda_3=i$,and so the answer to the question is yes.

So far we know that $\text{dim}(W_1)=2=\text{dim}(W_2)$,on the other hand $$ \text{dim}(W_1 \cap W_2)=\text{dim}(W_1)+\text{dim}(W_2)- \text{dim}(W_1+W_2)$$

$T$ is isomorphism ,hence a bijection,so does have an inverse ,but I don't know how to find the dimension of $W_1+W_2$.

Since the $\alpha_i$'s have not been given explicitly,hence I think we need a trick to find such a basis,but I don't know how.

  • $\begingroup$ the linear combination in part 1 can be used to write $\alpha_3$ in terms of $\alpha_1$ and $\alpha_2$ $\endgroup$ – Lozenges Dec 28 '20 at 17:14
  • $\begingroup$ gives you a vector in the intersection of $W_1$ and $W_2$ $\endgroup$ – Lozenges Dec 28 '20 at 17:17

We have

$$\alpha_3 = \frac1{\lambda_3}(\alpha_1-\lambda \alpha_2)$$ so $\alpha_3 \in W_1$. But also $\alpha_3 \in W_2$ by definition so $\alpha_3 \in W_1\cap W_2$. This implies that $\dim (W_1 \cap W_2) \ge 1$.

It remains to see whether $\dim(W_1\cap W_1) = 1$ or $\dim(W_1\cap W_2) = 2$. In the latter case, since $\dim W_1 = \dim W_2 =2$, we would get $W_1 = W_2$. However, it is easy to see that $\alpha_4$ cannot be represented as a linear combination of $\alpha_1$ and $\alpha_2$ so clearly $W_1 \ne W_2$.

Therefore $\dim (W_1 \cap W_2) = 1$ and hence $\dim (W_1+W_2) = 3$. To find a basis for $W_1+W_2$ we have to pick three linearly independent vectors. We can pick $$\{\alpha_1, \alpha_2, \alpha_4\}.$$ Indeed, the first two are linearly independent since $\dim W_1 = 2$ and we already showed that $\alpha_4 \notin W_1.$

  • $\begingroup$ @masaheb We have $1 \le \dim(W_1 \cap W_2) \le \dim W_1 = 2$ so it can only be 1 or 2. For the second statement, we have $W_1 \cap W_2 \subseteq W_1$ so if they have equal dimension, they are equal. Similarly for $W_2$ so we conclude $W_1= W_2$. $\endgroup$ – mechanodroid Dec 29 '20 at 13:46
  • $\begingroup$ @masaheb Yes. Also in this situation if $\dim A=\dim B$ then $A=B$. $\endgroup$ – mechanodroid Dec 29 '20 at 17:39
  • $\begingroup$ @masaheb We know $\dim (W_1 \cap W_2) = 1$ and $\alpha_3 \in W_1\cap W_2$ so the intersection is the linear span of the vector $\alpha_3$. $\endgroup$ – mechanodroid Dec 29 '20 at 19:07
  • $\begingroup$ @masaheb No, $\alpha_2 \in W_1$ is not a scalar multiple of $\alpha_1$. Maybe you meant $W_1 \cap W_2$? That is true. $\endgroup$ – mechanodroid Dec 30 '20 at 18:49
  • $\begingroup$ @masaheb Every linearly independent set with cardinality equal to the dimension of the space is a basis for the space. Or you can show it directly: let $\{b\}$ be a basis for $W_1 \cap W_2$. Since $\alpha_3 \in W_1 \cap W_2$ there exists a scalar $\lambda$ such that $\alpha_3 = \lambda b$. Since both $b$ and $\alpha_3$ are nonzero, it follows $\lambda \ne 0$. Now for every $w \in W_1 \cap W_2$ there is a scalar $\mu$ such that $w = \mu b$. Therefore $w = \frac{\mu}{\lambda} \alpha_3$ so we conclude that $\{\alpha_3\}$ spans $W_1 \cap W_2$. Since it is also linearly independent, it is a basis. $\endgroup$ – mechanodroid Dec 30 '20 at 20:21

If $W_1 = Sp[(1,0,i),(-2,1+i,0)]$ And $W_2 = Sp[(-1,1,1),(\sqrt2,i,3)] $ then
$W_1+W_2=Sp(W_1 \cup W_2)$, PROOF: (a general proof for any $W_1,W_2$)
It is obvious that $Sp(W_1) \subseteq Sp(W_1 \cup W_2), Sp(W_2) \subseteq Sp(W_1 \cup W_2)$.
Since $Sp(W_1 \cup W_2)$ is a subspace it is closed under addition thus, $W_1+W_2 \subseteq Sp(W_1 \cup W_2)_{(1)}$.
Now, let $ v\in Sp(W_1 \cup W_2)$, thus $v=\lambda_1u_1+...+\lambda_nu_n$, where $u_1 ... u_n \in Sp(W_1 \cup W_2)$.
We can see that $1\le \forall i \le n :\space \space $ $\lambda_iu_i\in Sp(W_1) \subseteq W_1+W_2$ Or, $\lambda_iu_i\in Sp(W_2) \subseteq W_1+W_2$ and in both cases $\lambda_iu_i \subseteq W_1+W_2$, since $W_1+W_2$ is a subspace it is closed under addition and thus $v\in W_1+W_2$ and $Sp(W_1 \cup W_2) \subseteq (W_1+W_2)_{(2)}$ From (1) and (2): $W_1+W_2=Sp(W_1 \cup W_2)$.

We can find it's dimension by setting the vectors as rows in a matrix and then check if any of the vectors are a linear combination of the others. $$\begin{pmatrix} 1 & 0 & i\\ -2 & 1+i & 0\\ -1 & 1 & 1\\ \sqrt2 & i & 3\\ \end{pmatrix} \Longrightarrow \begin{pmatrix} 1 & 0 & i\\ 0 & 1+i & 2i\\ 0 & 0 & -i \sqrt2+(4-i)\\ 0 & 0 & 0\\ \end{pmatrix}\Longrightarrow dim(U+W)=3$$

can you find $W_1\cap W_2$?

Notice that we also found a basis for the subspace spanned by $\alpha_1...\alpha_4$.

  • $\begingroup$ @masaheb I've added some explanations to my answer. Setting the vectors as rows in a matrix lets you to use elementary row functions in order to see if the vectors are linearly independent. $\endgroup$ – NirF Dec 28 '20 at 17:49

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