Cauchy-Schwarz Proof Question

Question

I'm just doing some revision on the Cauchy-Schwarz inequality for Linear Algebra using this picture from my lecture notes:

The line:

'as a basic consequence of the dot product listed above, f(t) > 0 for all t as an element of the real numbers' has previously assumed that we assume that b is NOT the zero vector. This I understand.

What I don't understand if why no specification of a not being the zero vector is made. Surely if a was the zero vector, for t = 0, (a+tb) . (a+tb) = 0, and this would mean would mean that f(t) > 0 does NOT hold?

Am I missing something really obvious here?

Thanks for your help.

In addition to Clement's answer, note that the case $\mathbf{a} = \mathbf{0}$ is pretty easy to handle. In that case, equality does hold, and $\mathbf{a}$ is a scalar multiple of $\mathbf{b}$ (by $0$). — Matthew Leingang, Feb 03 '20 at 23:22
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. — Clement C., Feb 05 '20 at 06:20
Sorry about that - just have completed that now. Your answer made lots of sense. Thank you. — cthodrums, Jun 22 '20 at 08:35

score 1 · Accepted Answer · answered Feb 03 '20 at 23:20

No claim that $$\forall t\in\mathbb{R},\qquad f(t)>0$$ is ever made (as you note, this would indeed not be true for $t=0$). What is claimed is that $$\forall t\in\mathbb{R},\qquad f(t)\geq 0$$ which is true.

At a higher-level, why is the assumption that $\mathbf{b}\neq \textbf{0}$ is even made? It is only because for the rest of the proof we want $f(t)$ to be a degree-two polynomial (in $t$). Since $f(t) = \lVert \mathbf{b}\rVert^2 t^2 + 2t(\mathbf{a}\cdot\mathbf{b}) + \lVert \mathbf{a}\rVert^2$, this is only true of the coefficient if $t^2$ is non-zero.

Cauchy-Schwarz Proof Question

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