# Given an isomorphism linear transformation $T:V \to \mathbb C^3$,where $V$ is a vector space

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.

• the linear combination in part 1 can be used to write $\alpha_3$ in terms of $\alpha_1$ and $\alpha_2$ – Lozenges Dec 28 '20 at 17:14
• gives you a vector in the intersection of $W_1$ and $W_2$ – 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 $$\alpha_4 \notin W_1.$$

• @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 $W_1= W_2$. – mechanodroid Dec 29 '20 at 13:46
• @masaheb Yes. Also in this situation if $\dim A=\dim B$ then $A=B$. – mechanodroid Dec 29 '20 at 17:39
• @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 $\alpha_3$. – mechanodroid Dec 29 '20 at 19:07
• @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. – mechanodroid Dec 30 '20 at 18:49
• @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. – 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 $$\alpha_1...\alpha_4$$.

• @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. – NirF Dec 28 '20 at 17:49