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If x is an eigenvector of a matrix A, then show that its corresponding eigenvalue is given by $\lambda=\dfrac{Ax\cdot x}{x\cdot x}$

I tried starting from $(A-\lambda{I})x=0$.

$Ax-\lambda{Ix}=0$

$\lambda{Ix}=Ax$

$\lambda=\dfrac{Ax}{Ix}$. This now is a bit confusing. Any help?

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4 Answers 4

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The identity you got $\lambda=\frac{Ax}{Ix}$ is not only confusing but also a wrong expression in general since one can not divide a vector by another one.

Focus on what you want to prove $$ \lambda=\frac{Ax\cdot x}{x\cdot x} $$ which is equivalent to $$ Ax\cdot x=\lambda (x\cdot x). $$ All you need here is to observe that $(\lambda x)\cdot x=\lambda(x\cdot x)$ and the definition of eigenvectors: $Ax=\lambda x$.

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Your intuition starts off right, it is a good idea to begin with the definition of an eigenvalue, i.e. $A x = \lambda x$. Now the problem is that on both sides of the equation you have vectors, so you cannot divide by $x$. The idea is to "transform" the equation into an equation that only involves scalar quantities. A good way to do this is to take the scalar product with another vector $y$, i.e. (I will write the scalar product $x \cdot y$ as $\langle x, y \rangle$ to avoid confusing it with the normal multiplication of real numbers) $$ \langle Ax, y \rangle = \lambda \langle x,y \rangle. $$ Now the question arises: which $y$ should we choose? To solve the equation for $\lambda$, we obviously would like to divide by $\langle x,y \rangle$ (which is a real number now!). But this number might be zero if we choose the wrong vector $y$ (e.g. if $y = 0$ or $y \perp x$). So we need to make sure that $\langle x,y \rangle \neq 0$. The only thing we know is that $x$ is an eigenvector. But by definition this means that $x \neq 0$, and by the properties of scalar products, you know that $\langle x, x \rangle \neq 0$ if and only if $x \neq 0$.

So $y = x$ looks like a good choice!

Then we get $$ \langle Ax, x \rangle = \lambda \langle x,x \rangle \Rightarrow \lambda = \frac{\langle Ax, x \rangle}{\langle x,x \rangle} $$ or in your notation $$ \lambda = \frac{Ax \cdot x}{x \cdot x}. $$

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  • $\begingroup$ +1 All the answers are correct and nice, but this is the best because of the generality for inner products! $\endgroup$
    – Xoque55
    Jul 25, 2017 at 22:51
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If $x$ is an eigenvector of $A$ then it satisfies, with $\lambda$ its associated eigenvalue :

$Ax = \lambda x$

From there, it is easy to see that :

$${Ax\cdot x\over x \cdot x} = {\lambda x\cdot x\over x\cdot x} = \lambda$$

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In order to avoid confusion between scalar and vector multiplication, I use the notation $\langle x,y\rangle$ to denote the inner product $x\cdot y$ between vectors $x$ and $y$ in the following.

If $x$ is an eigenvector of $A$ to the eigenvalue $\lambda$, then $Ax = \lambda \cdot x$. Taking the inner product with $x$, this implies $\langle Ax,x\rangle= \lambda \cdot \langle x, x\rangle$, which is a scalar. Hence $$\frac{\langle Ax,x\rangle}{\langle x,x\rangle} = \frac{\langle \lambda\cdot x, x\rangle }{\langle x,x\rangle}= \frac{\lambda\cdot\langle x,x\rangle}{\langle x,x\rangle} = \lambda.$$

Note that in your argument you divide by the vector $I x$, which is something you cannot do! Taking the inner product with $x$ circumvents this issue, while preserving your intuition of the proof.

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    $\begingroup$ Perhaps using the dot for the multiplication between a scalar and a vector can be confusing here, where it is also used as the scalar product $\endgroup$
    – chi
    Jul 25, 2017 at 22:54

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