Linear algera-Question on Isomorphism.

Question

Can someone help me to solve the question of isomorphism

I cannot gain an intuitive feel on it,because I cannot visualize $\mathbb C^3(\mathbb C)$.I am in problem with mainly the $2nd $ part of the problem because I cannot canculate $W_1 \cap W_2$.

Answer 1

One approach is as follows: let $x = (x_1,x_2,x_3) \in \Bbb C^3$ be a non-zero vector solving the system $$ x_1 + 0x_2 + ix_3 = 0\\ -2 x_1 + (1+i)x_2 + 0x_3 = 0 $$ Then the image of $W_1$ under the isomorphism is the solution space of vectors $z = (z_1,z_2,z_3)$ satisfying $x_1z_1 + x_2z_2 + x_3 z_3 = 0$.

Similarly, find a vector $y$ such that the image of $W_2$ is the solution space of vectors $z$ satisfying $y_1 z_1 + y_2 z_2 + y_3 z_3 = 0$.

The intersection of the images of $W_1$ and $W_2$ will be the set of vectors $z$ satisfying both equations. That is, the image of $W_1 \cap W_2$ is the solution space to the system $$ x_1z_1 + x_2z_2 + x_3 z_3 = 0\\ y_1 z_1 + y_2 z_2 + y_3 z_3 = 0 $$

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Following the method that I outline my post here yields a different approach. Row reducing the matrix whose columns are $T(\alpha_1),\dots,T(\alpha_4)$ yields $$ \pmatrix{ 1&0&-i&0\\ 0&1&\frac{1-i}2&0\\ 0&0&0&1 } $$ Which is to say that the only non-trivial relation between these vectors is $$ -i T(\alpha_1) + \frac{1-i}{2}T(\alpha_2) = T(\alpha_3). $$ Thus, $T(\alpha_3)$ forms a basis for the intersection of the images of $W_1$ and $W_2$. Thus, $\alpha_3$ forms a basis the intersection of $W_1$ and $W_2$.