# Prove $\frac{\left\langle Ax,x\right\rangle}{\left\langle x,x\right\rangle} \leq \lambda$ for $A$ symmetric and $\lambda$ the biggest eigenvalue

Let $$A$$ be a real $$n\times n$$ symmetric matrix.

Let $$\lambda$$ be the biggest eigenvalue of $$A$$.

Prove that

$$0 \neq \forall x \in \mathbb{R}^n: \frac{\left\langle Ax,x \right\rangle}{\left\langle x,x \right\rangle} \leq \lambda$$

in $$\mathbb{R}^n$$ with the standard inner product.

And Prove that every $$0 \neq x \in \mathbb{R}^n$$ which satisfies:

$$\frac{\left\langle Ax,x \right\rangle}{\left\langle x,x \right\rangle} = \lambda$$

is an eigenvector.

## My try:

But my try this time is bad... i have many assumptions that i couldnt prove. I will write it as we go in the prove.

$$\frac{\left\langle Ax,x\right\rangle }{\left\langle x,x\right\rangle} \leq \frac{\left\langle \lambda x, x\right\rangle}{\left\langle x,x\right\rangle}$$

We will prove it.

Assume there is $$d \in \mathbb{R}$$ such that: $$\frac{\left\langle \lambda x, x\right\rangle}{\left\langle x,x\right\rangle} < \frac{\left\langle Ax,x\right\rangle}{\left\langle x,x\right\rangle} = \frac{\left\langle d x, x\right\rangle}{\left\langle x,x\right\rangle}$$

Therefore, we can conclude that : (This conclution is also if it satisfies $$\forall x$$ but we need to prove that its satisfies it for all $$x$$ we cant assume this. )

$$\lambda x < Ax = ex, e \in \mathbb{R}$$

Therefore $$e$$ is a bigger eigenvalue than $$\lambda$$ in contradiction.

Therefore:

$$\frac{\left\langle Ax,x\right\rangle}{\left\langle x,x\right\rangle} \leq \frac{\left\langle \lambda x, x\right\rangle}{\left\langle x,x\right\rangle}, 0 \neq \forall x \in \mathbb{R}^n$$

If $$\frac{\left\langle Ax,x\right\rangle}{\left\langle x,x\right\rangle} = \frac{\left\langle \lambda x, x\right\rangle}{\left\langle x,x\right\rangle}, \forall x \in \mathbb{R}^n$$

Than we can conclude:

$$\left\langle Ax - \lambda x, x\right\rangle= 0$$

But we know that $$x$$ is not the zero vector. Therefore;

$$(A-\lambda)x = 0, x \neq 0, \lambda - \ eigenvalue$$

Therefore, $$x$$ is an eigenvector.

Another try was to use diagonaized matrix, as: $$A = Q^tDQ$$ and say that $$\left\langle Ax,x\right\rangle = \left\langle Q^tDQ,x\right\rangle \leq \left\langle \lambda Ix, x\right\rangle = \lambda \left\langle x,x\right\rangle$$ Thats because $$\lambda$$ is the biggest eigenvalues, therefore we get: $$\frac{\left\langle Ax,x\right\rangle}{\left\langle x,x\right\rangle} \leq \frac{ \lambda \left\langle x,x\right\rangle}{\left\langle x,x\right\rangle} = \lambda$$

As needed, but still i dont think i can prove its, or if its true anyway...

I would like comments and hints (not a solution - those are my homeworks)

• Use \langle , \rangle to produce $\langle , \rangle$. Apr 2, 2020 at 15:06
Hint: Since $$A$$ is a symmetric matrix, there exists an orthonormal basis $$v_1,\ldots,v_n$$ of eigenvectors of A (spectral theorem). Write $$x$$ in that basis, i.e. $$x=\sum_{i=1}^n\alpha_i v_i$$.
• Great idea and worked well for me, what i have left now is the second part, which ask to prove the if the equality holds, so $x$ is eigenvector, i get to: $$\frac{\sum_{i = 1}^{n} |a_i|^2\lambda_i}{\sum_{i = 1}^{n} |a_i|^2} = \lambda_{max}$$ And here i am stuck again, almost done but still... another hint for this? Thank you alot.
• Remember an eigenvector multiplied by a non-zero scalar is still an eigenvector. As a first step, consider $x$ with length 1.