Let \(V\) be a vector space over the field of complex numbers, and suppose there is an isomorphism \(T\) of \(V\) onto \(C^3\). Let \(\alpha_1, \alpha_2, \alpha_3, \alpha_4\) be vectors in \(V\) such that \[ T\alpha_1 = (1, 0, i), \quad T\alpha_2 = (-2, 1 + i, 0), \] \[ T\alpha_3 = (-1, 1, 1), \quad T\alpha_4 = (\sqrt{2}, i, 3). \]
Is \(\alpha_1\) in the subspace spanned by \(\alpha_2\) and \(\alpha_3\)?
Let \(W_1\) be the subspace spanned by \(\alpha_1\) and \(\alpha_2\), and let \(W_2\) be the subspace spanned by \(\alpha_3\) and \(\alpha_4\). What is the intersection of \(W_1\) and \(W_2\)?
Find a basis for the subspace of \(V\) spanned by the four vectors \(\alpha_i\).
Solution
Because \(T\) is an isomorphism, linear relations among the \(\alpha_i\) are exactly the corresponding linear relations among their images.
For part (a), direct calculation gives \[ (1,0,i)=-\frac{1+i}{2}(-2,1+i,0)+i(-1,1,1). \] Hence \[ \alpha_1=-\frac{1+i}{2}\alpha_2+i\alpha_3, \] so \(\alpha_1\in\operatorname{span}\{\alpha_2,\alpha_3\}\).
For part (b), suppose \[ a\alpha_1+b\alpha_2=c\alpha_3+d\alpha_4. \] Applying \(T\) and comparing coordinates gives the homogeneous system whose solutions are \[ a=-ic,\qquad b=\frac{1-i}{2}c,\qquad d=0. \] Thus every vector in \(W_1\cap W_2\) is a scalar multiple of \(\alpha_3\). Conversely, \[ \alpha_3=-i\alpha_1+\frac{1-i}{2}\alpha_2, \] so \(\alpha_3\in W_1\), while of course \(\alpha_3\in W_2\). Therefore \[ W_1\cap W_2=\operatorname{span}\{\alpha_3\}. \]
For part (c), the three vectors \(T\alpha_1,T\alpha_2,T\alpha_4\) are linearly independent, since \[ \det\begin{bmatrix} 1&-2&\sqrt2\\ 0&1+i&i\\ i&0&3 \end{bmatrix} =5+\sqrt2+i(3-\sqrt2)\ne0. \] They therefore form a basis of \(\mathbb C^3\). Since \(\alpha_3\) is already a linear combination of \(\alpha_1\) and \(\alpha_2\), a basis for the span of all four vectors is \[ \{\alpha_1,\alpha_2,\alpha_4\}. \]