Linear independence of linear combinations

I'm having trouble with my homework:

Under which conditions of $α, β, γ, δ$ the two vectors $αx + βy$ and $γx + δy$ are linearly independent if the vectors $x$ and $y$

a) are linearly independent?

b) are linearly dependent?

For a), the only thing I can come to is that since $αx + βy$ and $γx + δy$ are linearly independent, $a(αx + βy) + b(γx + δy) = 0$ with only $a=b=0$. Therefore, the matrices $αx + βy$ and $γx + δy$ must be non-zero and because $x$ and $y$ are linearly independent $α, β, γ, δ$ must also be non-zero. I'm not sure if this is okay and would appreciate any help with both questions, thank you!

Solution 1:

Let us suppose first that $x,y$ are linearly independent. Then, $\mathcal{B}=\{x,y\}$ is a basis of $V:=span(\{x,y\})$. Let $\phi:V\rightarrow V$ be linear such that $x\mapsto \alpha x+\beta y$ and $y\mapsto \gamma x+\delta y$ (This defines a unique map). Observe that $\alpha x+\beta y$ and $\gamma x+\delta y$ are independent $\iff$ $\phi$ is an isomorphism $\iff$ the matrix corresponding to $\phi$ with respect to $\mathcal{B}$, that is the matrix $ \begin{bmatrix} \alpha & \gamma \\ \beta & \delta \end{bmatrix} $ is invertible $\iff$ $\left\lvert \begin{bmatrix} \alpha & \gamma \\ \beta & \delta \end{bmatrix} \right\rvert=\alpha\delta-\beta\gamma\neq0$

If $x$ and $y$ are linearly dependent, then, WLOG, we may assume that $y\in $ span$(x)$ . This implies that $\alpha x+\beta y$, $\gamma x + \delta y \in$ span$(x)$ which is a subspace with dimension strictly less than $2$ of the underlying vector space. Hence, $\alpha x+\beta y$ and $\gamma x + \delta y$ cannot be linearly independent for any choice of scalars $\alpha,\beta,\gamma,\delta$.

Solution 2:

An elementary approach, and essentially Shagchi's solution in plainer language.

(a) Suppose $x$ and $y$ are linearly independent, meaning that $c_1 x + c_2 y = 0$ if and only if $c_1 = c_2 = 0$. Now consider the linear combination $$ d_1 (\alpha x + \beta y) + d_2 (\gamma x + \delta y) = (d_1 \alpha + d_2 \gamma) x + (d_1 \beta + d_2 \delta) y = 0. \tag{1}\label{linindep} $$ We must have, since $x$ and $y$ are linearly independent, $$ d_1 \alpha + d_2 \gamma = d_1 \beta + d_2 \delta = 0. $$ Recording these two as a system of equations $$ \begin{cases} \alpha d_1 + \gamma d_2 = 0 \\ \beta d_1 + \delta d_2 = 0 \end{cases} $$ we can translate this into a homogeneous matrix equation $$ \begin{pmatrix} \alpha & \gamma \\ \beta & \delta \end{pmatrix} \begin{pmatrix} d_1 \\ d_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$ Going back, for $\alpha x + \beta y$ and $\gamma x + \delta y$ to be linearly independent, meaning $\eqref{linindep}$ is true if and only if $d_1 = d_2 = 0$, means that the above system has only the trivial solution. But such a system having only the trivial solution requires the matrix $\begin{pmatrix} \alpha & \gamma \\ \beta & \delta \end{pmatrix}$ to be invertible (since it's square). But a matrix is invertible if and only if its determinant is nonzero, so $$\det \begin{pmatrix} \alpha & \gamma \\ \beta & \delta \end{pmatrix} = \alpha \delta - \beta \gamma \neq 0 $$ is a condition on $\alpha$, $\beta$, $\gamma$, and $\delta$ for which $\alpha x + \beta y$ and $\gamma x + \delta y$ are linearly independent, assuming $x$ and $y$ are linearly independent.

(b) For the second part, if $x$ and $y$ are linearly dependent, we must have (since there are only two of them) something like $x = c y$ for a scalar $c$. In that case $$ \alpha x + \beta y = \alpha c y + \beta y = (\alpha c + \beta) y $$ and $$ \gamma x + \delta y = \gamma c y + \delta y = (\gamma c + \delta) y $$ are scalar multiples of one another, so they're certainly not linearly independent. (If $y = 0$, the above doesn't necessarily work, but then something like $y = c x$ must instead be true, and a small modification of the argument works identically.)