With $\mathbf{u,v}$ being vectors in $\mathbb{R}^n$ euclidean space, the Cauchy–Schwarz inequality is
$$
{\left(\sum_{i=1}^{n} u_i v_i\right)}^2 \leq \left(\sum_{i=1}^{n} u_i^2\right)\left(\sum_{i=1}^{n} v_i^2\right)
$$
further given that $\mathbf{u}=\lambda\mathbf{v}$, the csi looks like the following:
$$
{\left(\sum_{i=1}^{n} \lambda v_i v_i\right)}^2 \leq \left(\sum_{i=1}^{n} (\lambda v_i)^2\right)\left(\sum_{i=1}^{n} v_i^2\right)
$$
With equality applying in the Cauchy-Schwarz inequality only if $\mathbf{u,v}$ are linear dependent, how do I show that equality is given in this case? A start would be enough, I'm quite new to linear algebra
Edit:
Thanks so far!
Rewriting the last line – following your advice – I get the following
$$
{\lambda^2\left(\sum_{i=1}^{n} v_i^2\right)}^2 \leq \lambda^2\sum_{i=1}^{n} v_i^2\sum_{i=1}^{n} v_i^2
$$
Okay I'm not sure about the following, so make sure you have foul fruit nearby to throw at me:
Canceling $\lambda^2$ this results in
$$
{\left(\sum_{i=1}^{n} v_i^2\right)}^2 \leq \sum_{i=1}^{n} v_i^2\sum_{i=1}^{n} v_i^2
$$
with the inquality being wrong, equality applies… is that evidence enough?
Best Answer
Hints:
$$\left(\sum_{i=1}^n\lambda v_iv_i\right)^2=\lambda^2\left(\sum_{i=1}^n v_i^2\right)^2$$
$$\sum_{i=1}^n (\lambda v_i)^2\sum_{j=1}^n (v_j)^2=\lambda^2\sum_{i=1}^n v_i^2\sum_{j=1}^n v_j^2$$