Proving summation given vectors and orthonormal basis using triangle inequality and cauchy-schwartz inequality

Question

Let $\{y_1,y_2,...,y_n\}$ be a set of vectors in $\Bbb R^n$ and let $\{u_1,u_2,...,u_n\}$ be an orthonormal basis of $\Bbb R^n$ prove that for every $\lambda_1,...,\lambda_n$ $\in$ $\Bbb R$

$||\sum_{i=1}^n \lambda_i y_i||^2 \leq (\sum_{i=1}^n |\lambda_i|^2) (\sum_{i=1}^n ||y_i||^2)$

Part 2 of the question - Prove that if the set of vectors $\{y_1,...,y_n\}$ fulfill $\sum_{i=1}^n||y_i||^2 <1$ then the set $\{u_1+y_1,...,u_n+y_n\}$ is linearly independent

Hint - use the triangle inequality and cauchy schwartz inequality

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I am struggling to approach it , I was told first I need to prove the triangle inequality so I did the following: first I did for the basic case , let $a,b \in \Bbb R$ and prove $|a+b| \leq |a|+|b|$

$\forall a,b \in \Bbb R ,(|a|+|b|)^2 = |a|^2+2|a||b|+|b|^2 \geq a^2+2ab+b^2 $ and since $|x|^2=x^2$ we get $(a+b)^2=|a+b|^2$ from here we get $(|a|+|b|)^2 \geq |a+b|^2$ $\iff$ $|a|+|b| \geq |a+b|^2$

and for the general case I proved using induction for $n \geq 2$ we need to prove $|x_1+...+x_n| \leq |x_1|+...+|x_n|$

1. base case for $n=2$: $|x_1+x_2| \leq |x_1|+|x_n|$ is correct as I just shown
2. assume for $n=k$ :$|x_1+...+x_k| \leq |x_1|+...+|x_k|$
3. induction step for $n=k+1$ : $|x_1+...+x_k +x_{k+1}| \leq |x_1|+...+|x_k|+|x_{k+1}|$ let $\alpha = x_1+...+x_k$ and $\beta = |x_1|+...+|x_k|$ from the property of absolute value that $|x| \geq x$ we get that $\alpha \leq \beta$ therefore $|\alpha +x_{k+1}| \leq \beta +|x_{k+1}|$ and according to our assumption the statement is true and it is the triangle inequality.

But I do not know how this helps with the question itself (again I was told that by the professor this is why I proved it first)

how is $\lambda$ related to things here or even the cauchy schwartz inequality ? appreciate any tips and hints so I can update my progress

Edit- if $\{u_1,...,u_n\}$ is a basis and $\{y_1,y_2,...,y_n\}$ a set of vectors then is $y=\sum_{i=1}^n y_iu_i$ an inner product? .

Edit 2 - I tried the following , since the cauchy schwartz inequality is $\sum_{i=1}^nu_i \cdot v_i=|(u,v)|\leq ||u||^2 \cdot ||v||^2$ then we get that if $||\sum_{i=1}^n \lambda_i y_i||$ then from the triangle inequality and inner product homogeneity $||\sum_{i=1}^n \lambda_i y_i|| \leq \sum_{i=1}^n||\lambda_i \cdot y_i|| = \sum_{i=1}^n|\lambda_i| \cdot ||y_i|| $ and if we square it we get $||\sum_{i=1}^n \lambda_i y_i||^2 \leq \sum_{i=1}^n |\lambda_i|^2 \sum_{i=1}^n ||y_i||^2$ according to the cauchy schwartz inequality and triangle inequality

Sorry if my English is not all correct hope it is still understandable

Answer 1

With $U$ and $Y$ being matrices having column j as $\mathbf u_j$ and $\mathbf y_j$ respectively, this is can be viewed through Frobenius norms.

The first statement:
$||\sum_{i=1}^n \lambda_i \mathbf y_i||^2=\big \Vert Y\vec{\lambda}\big\Vert_F^2\leq \big \Vert Y\big\Vert_F^2\big \Vert \vec{\lambda}\big\Vert_F^2=\big \Vert \vec{\lambda}\big\Vert_F^2\big \Vert Y\big\Vert_F^2 = (\sum_{i=1}^n |\lambda_i|^2) (\sum_{i=1}^n ||\mathbf y_i||^2)$
by submultiplicativity of the Frobenius norm. (If you are not aware of the submultiplicativity of Frobenius norms, consider proving it e.g. by application of Cauchy-Schwarz and triangle inequality.)

For the second part
$\big \Vert U^T Y\big \Vert_F^2 =\big \Vert Y\big \Vert_F^2 = \sum_{i=1}^n ||\mathbf y_i||^2 \lt 1$
implies that, when working over $\mathbb C$, all eigenvalues of $U^TY$ have modulus $\lt 1$-- justifying this either in terms of the operator 2 norm or by the inequality of Schur: $\sum_{k=1}^n \vert \lambda_k\vert^2 \leq \big \Vert U^T Y\big \Vert_F^2$. Finally

$\big \vert\det\big(U+Y\big)\big \vert = \big \vert\det\big(U\big)\det\big(I+U^TY\big)\big \vert= \big \vert\det\big(I+U^TY\big)\big \vert $
$= \prod_{k=1}^n \vert(1+\lambda_k)\vert\geq \prod_{k=1}^n (\vert 1\vert-\vert\lambda_k\vert) = \prod_{k=1}^n ( 1-\vert\lambda_k\vert)\gt 0$
by reverse triangle inequality