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.


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

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)

Thanks for all the help you give(gave) me. Im stuck...

  • 2
    $\begingroup$ Use \langle , \rangle to produce $\langle , \rangle$. $\endgroup$ Apr 2, 2020 at 15:06

1 Answer 1


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. $x=\sum_{i=1}^n\alpha_i v_i$.

  • $\begingroup$ 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. $\endgroup$
    – Alon
    Apr 2, 2020 at 17:32
  • $\begingroup$ Remember an eigenvector multiplied by a non-zero scalar is still an eigenvector. As a first step, consider $x$ with length 1. $\endgroup$
    – Toni
    Apr 2, 2020 at 17:43

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