The proof of the Cauchy-Schwarz inequality in Axler's Linear Algebra Done Right (pg. 104) hinges on showing that

$\|u\|^2 = \|\frac{\langle u,v\rangle}{||v||^2}v\|^2 + \|w\|^2 \tag{1}$

equals $\frac{|\langle u,v\rangle|^2}{\|v\|^2} + \|w\|^2 \tag{2}$

but I'm not quite sure how (1) = (2).

Focusing in on $\|\frac{\langle u,v\rangle}{\|v\|^2}v\|^2$ we can rewrite this as:

$\|\frac{\langle u,v\rangle}{\langle v,v\rangle}v\|^2$ where the fraction is a scalar so we can take it out: $|\frac{\langle u,v\rangle}{\|v\|^2}|^2\|v\|^2$ and then the squared norms of $v$ cancel out, but clearly this wrong somehow.

Many thanks for any help.

  • $\begingroup$ when you say (1)=(2), you mean that $||u||^2 = \frac{|\langle u, v \rangle | }{||v||^2} ^2 ||v||^2 + ||w||^2$, I guess ? $\endgroup$ Mar 3, 2015 at 8:21
  • $\begingroup$ yup that's exactly what I meant $\endgroup$
    – Theodore
    Mar 3, 2015 at 8:25
  • $\begingroup$ $u$ can be expressed in terms of two orthogonal vectors $v$ and $w$. When looking at the distances involved (1) is just a restatement of Pythagora's theorem. to get the C-S inequality you must get from (1) to (2), but I'm not sure how the author does it :( $\endgroup$
    – Theodore
    Mar 3, 2015 at 8:35
  • 1
    $\begingroup$ When you pull out $\frac{1}{||v||^2}$, it's being squared, so really, you're pulling out $\frac{1}{||v||^4}$. $\endgroup$
    – Brent
    Mar 3, 2015 at 8:47
  • $\begingroup$ I get it :D this is what happens when it's 6am and you're still awake... $\endgroup$
    – Theodore
    Mar 3, 2015 at 9:10

2 Answers 2


For some reason, people teaching Linear Algebra forgot to connect inner product with the ordinary dot product in $\mathbb{R}^{3}$ or $\mathbb{R}^{n}$. If you have a vector $u$ and a line through the origin with direction vector $v$, then how would you find the point on that line closest to $u$? Answer: orthogonal projection. The same holds in infinite dimensional inner-product spaces, and even for complex spaces. You want to find a scalar $\alpha$ such that $(u-\alpha v)\perp v$, which gives $$ (u,v)-\alpha(v,v) = 0,\\ \alpha = \frac{(u,v)}{(v,v)} $$ Now you can decompose into a right triangle where $u$ is the hypotenuse, $\alpha v$ is the leg along the line with direction vector $v$ and $u-\alpha v$ is the other leg. Explicitly, you have the following orthogonal decomposition: $$ u = \alpha v + (u-\alpha v),\\ (\alpha v,u-\alpha v) = 0. $$ By the Pythagorean Theorem (which is a direct computation using inner product axioms): $$ \|u\|^{2} =\|\alpha v\|^{2}+\|(u-\alpha v)\|^{2}. $$ The Cauchy-Schwarz inequality is exactly the following $$ \|\alpha v\|^{2} \le \|u\|^{2} \\ \mbox{ with equality iff } \|u-\alpha v\| = 0. $$ Write this out: $$ |\alpha|^{2}\|v\|^{2} \le \|u\|^{2} \\ \frac{|(u,v)|^{2}}{(v,v)^{2}}\|v\|^{2} \le (u,u) \\ |(u,v)|^{2} \le (u,u)(v,v). $$


You've solved the question, so I can't understand your confusion. See this: $$\|\frac{\langle u,v\rangle}{\|v\|^2}v\|^2=|\frac{\langle u,v\rangle}{\|v\|^2}\times \|v\||^2=\frac{|\langle u,v\rangle|^2}{\|v\|^2}$$ Hope this can help you.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .