Showing every eigenvector is a multiple of a known eigenvector I have the following question:

Let $A$ be an $n \times n$ matrix. Suppose $A$ has distinct eigenvalues $\lambda_1,...,\lambda_n$ and let $v_1, ... , v_n$ be eigenvectors with these eigenvalues. Show that every eigenvector is a multiple of one of the vectors $v_i$. Determine the matrix from the eigenvalues and eigenvectors.

Firstly, I understand that the question is true, but I'm having trouble formulating a decent proof.
Suppose we have an eigenvector $v_k \neq v_i, \ \ \forall i \in \left\{1,...,n\right\}$
Then, as $v_k$ is an eigenvector, we have that it satisfies 
$Av_k = \mu v_k$ 
Where $\mu$ is the eigenvalue associated with the eigenvector $v_k$. However, $A$ has $n$ eigenvalues at most and so therefore $\mu = \lambda_m$ for some $m \in  \left\{1,...,n\right\}$
Hence
$Av_k = \lambda_m v_k$
We also know that:
$Av_m = \lambda_m v_m$
So:
$A(v_k + v_m) = \lambda_m (v_k + v_m)$
It's obvious to me that this can only be true if $v_k$ is a multiple of $v_m$, say: $v_k = \alpha v_m$, as this is will allow the $\alpha$ to cancel on both sides - but I can't seem to show this rigorously (I get the impression it's a very simple step I'm failing to realise here).
 A: Your argument is, in fact, invalid! If we consider two-dimensional matrices and set $A=I$, for example, then the standard basis vectors $e_1$ and $e_2$ are both eigenvectors with eigenvalue $1$ and we have an equation
$$A(e_1 + e_2) = e_1 + e_2$$
however, $e_1$ is not a multiple of $e_2$!
Of course, $I$ doesn't have distinct eigenvalues, so this isn't a counterexample. Your proof has to make use of that hypothesis somehow.

As an aside, your proof starts out problematically. $v$ was given to you in the problem: it a sequence of $n$ fixed eigenvectors. So when you say that you have $v_k$, that means you have one of the given eigenvectors, which is exactly the opposite of what you meant to suppose!
Instead, start by saying something like

Suppose we have an eigenvector $w$ such that $w \neq v_i$ ...

or even just 

Suppose we have an eigenvector $w$ 

There is no harm in allowing $w$ to be one of the given eigenvectors.
A: Your reasoning up to $Av_k = \lambda_m v_k$ is correct. To continue, you need to show that $v_k$ is a multiple of $v_m$. Here is a way of doing it: Since the eigenvectors form a basis for $\mathbb{C}^n$, you can write $v_k = \sum_i \alpha v_i$. Multiplying across by $A$ gives $A v_k = \lambda_m v_k = \sum_i \alpha_i \lambda_m v_i = \sum_i \alpha_i \lambda_i v_i$. Here is where linear independence comes in, subtracting gives $\sum_i \alpha_i (\lambda_i-\lambda_m) v_i = 0$, from which we have $\alpha_i (\lambda_i-\lambda_m) = 0$ for all $i$. It follows that if $\lambda_i \neq \lambda_m$, then $\alpha_i = 0$, hence $v_k = \alpha_m v_m$.
Here is another way to do it:
Since the eigenvalues are distinct, the eigenvectors are linearly independent, and since there are $n$ of them, they form a basis for $\mathbb{C}^n$. Hence the matrix $V = \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}$ is invertible. It is easy to see that $A V = V \Lambda$, where $\Lambda$ is a diagonal matrix with entries $\lambda_1,...,\lambda_n$. So we have $V^{-1} A V = \Lambda$.
Now suppose $A v = \lambda v$, for some $v \neq 0$. We can write $A V V^{-1} v = \lambda V V^{-1} v$, and premultiplying across by $V^{-1}$ gives $\Lambda u = \lambda u$, where $u = V^{-1} v$. If we rewrite this as $(\Lambda -\lambda I) u = 0$, we can see that we must have $\lambda = \lambda_k$ for some $k$ (otherwise the matrix $\Lambda -\lambda I$ would be invertible and that would imply $u=0$). Furthermore, we must have $u = \alpha e_k$, where $e_k$ is the unit vector with zeros everywhere except a one in the $k$th position, and $\alpha \neq 0$. Since $v = V u$, we have $v = \alpha V e_k = \alpha v_k$, the $k$th eigenvector.
The above also shows that $A = V \Lambda V^{-1}$, which gives $A$ in terms of its eigenvalues and eigenvectors.
A: Maybe this is a bit different solution, but as you have $n$ distinct eigenvalues and $n$ eigenvectors, they are surely a basis.
So there is a unique representation for every $v$ as 
$$v=\sum_{k=1}^n a_k v_k$$
As all those $v_k$ are eigenvectors to eigenvalues $\lambda_k$ you know that 
$$A v= \sum_{k=1}^n a_k \cdot \lambda_k v_k$$
On the other hand you say that $v$ is an eigenvector so 
$$A v= \lambda v$$ 
Hence 
$$\lambda v= \sum_{k=1}^n \lambda_k a_k v_k$$
on the other hand we know that 
$$v=\sum_{k=1}^n a_k v_k$$ 
When $\lambda \neq 0$ we can devide it, as the representation is unique we know that $\lambda=\lambda_k$ for some $k\in\{1,\dots,n\}$, 
when $\lambda=0$ all $a_k$ with a possible exception of one must be $0$.
A: Note that the eigenvectors $v_1,v_2,\dots,v_n$ are linearly independent. We will prove this by induction on the number $k$ of eigenvectors we have listed (rather than the total number of them). Obviously if $k=1$ this is true. Hence assume $k>1$ and for the sake of contradiction that $v_i$ can be written in the form $\sum_{j\ne i}\alpha_jv_j$. Then $\sum_j \lambda_i\alpha_jv_j=Av_i=\sum_jA\alpha_jv_j=\sum_j \lambda_j\alpha_jv_j$ implying the $k-1$ vectors $v_1,\dots,v_{i-1},v_{i+1},\dots,v_k$ are linearly dependent. This is a contradiction of the inductive hypothesis.
In fact, $v_1,\dots,v_n$ form a basis for the $n$-dimensional space on which the linear transformation given by the matrix acts. That means that any eigenvector $v$ takes the form $\sum_{i=0}^n \beta_iv_i$. So if $Av=\lambda_k$, which is true for some $k$, then $\sum_{i=0}^n \lambda_k\beta_iv_i=Av=\sum_{i=0}^n \lambda_i\beta_iv_i$. By the linear independence of the vectors $v_i$ we have $(\lambda_i-\lambda_k)\beta_i=0$ for all $i$. This means that the coefficients $\beta_i$ are zero for all $i\ne k$ as desired.
A: The set of eigenvalues and vectors is the set of solutions to the equation $$Av = \lambda v$$
The set is complete, by definition, except that multiples of eigenvectors are also eigenvectors. That is to say, the only eigenvectors in addition to the ones which are paired with the eigenvalues are just multiples of these.
This means that we can reduce the set of eigenvectors to a set of vectors which are all of unit length. That is to say, we can restrict the solutions of the equation to be just those vectors which are unit vectors. The solution produces of $n$ eigenvalues $\lambda_1$ through $\lambda_n$, some of which may be repeated, and it produces a set of unique unit vectors $v_1$ through $v_n$.  Each of these vectors is representative of a direction such that all vectors which point in that same direction are scaled when multiplied by the matrix. Thus each of these unit eigenvectors vectors stands for the entire set of vectors which point in the same direction.
Your question is, essentially: show that there are no eigenvectors other than multiples of these unit vectors.
We can do this by reductio ad absurdum. 
Suppose there is a hitherto undiscovered $v_{n+1}$ which is an eigenvector, but is not a multiple of any of the known eigenvectors.  This means that $\frac{v_{n+1}}{|v_{n+1}|}$ is a unit eigenvector which is not equal to any of $v_1$ through $v_n$.
But this is a contradiction, because the set $v_1$ through $v_n$, along with the corresponding eigenvalues, is the complete solution set to the equation. 
There cannot be more solutions than there are roots of the characteristic equation, which is of degree $n$. 
A: This is a basic result about eigenvectors to be proved very early on, so I must assume you know very little about eigenvalues and eigenvectors. However I do assume you know that for a given value $\lambda$, the set of eigenvectors for $\lambda$ together with the zero vector, that is the set of vectors $v$ with $A\cdot v=\lambda v$, is a linear subspace, because that equation is linear in $v$.
Now if you already know that any family of eigenvectors for distinct eigenvalues is linearly independent, then the answer to the first question is easy. By the mentioned fact $v_1,\ldots,v_n$ are linearly independent, but if $v\neq0$ is an eigenvector for $\lambda$, the $n+1$ vectors $v_1,\ldots,v_n,v$ are linearly dependent (the space is only of dimension $n$), so we must have $\lambda=\lambda_i$ for some $i$. Moreover we have is a dependency relation $v=c_1v_1+\cdots+c_nv_n$, and if $w=c_iv_i-v$ were nonzero, it would be an eigenvector for $\lambda=\lambda_i$ and writing $0=c_1v_1+\cdots+c_{i-1}v_{i-1}+w+c_{i+1}v_{i+1}+\cdots+c_nv_n$ would contradict the linear independence of eigenvectors $v_1,\ldots,v_{i-1},w,v_{i+1},\ldots,v_n$ for distinct eigenvalues $\lambda_1,\ldots,\lambda_n$; therefore one must have $w=0$ and so $v=c_iv_i$ as desired.
But this just shows that the result stated is essentially contained in the fact that eigenvectors for distinct eigenvalues must be linearly independent, so how to show that? Supposing eigenvectors $v_1,\ldots,v_n$ for distinct eigenvalues $\lambda_1,\ldots,\lambda_n$ are linearly dependent, we may take $d$ minimal such that $v_1,\ldots,v_d$ are linearly dependent, and write $v_d=c_1v_1+\cdots+c_{d-1}v_{d-1}$. Now applying $A$ on one hand multiplies $v_d$ and therefore the whole expression by $\lambda_d$, and on the other hand multiplies each term $c_iv_i$ by $\lambda_i$; taking the difference gives $0=(\lambda_1-\lambda_d)c_1v_1+\cdots+(\lambda_{d-1}-\lambda_d)c_{d-1}v_{d-1}$. Here, by independence of $v_1,\ldots,v_{d-1}$, the coefficients $(\lambda_i-\lambda_d)c_i$ are all zero, and since by assumption $\lambda_i\neq\lambda_d$ for all $i$, this gives $c_i=0$ for all $i$. But then $v_d=0$, which is not allowed for an eigenvector, so the assumed linear dependence cannot exist.
