How to prove that eigenvectors from different eigenvalues are linearly independent How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?
 A: Alternative:
Let $j$ be the maximal $j$ such that $v_1,\dots,v_j$ are independent. Then there exists $c_i$, $1\leq i\leq j$ so that $\sum_{i=1}^j c_iv_i=v_{j+1}$.  But by applying $T$ we also have that 
$$\sum_{i=1}^j c_i\lambda_iv_i=\lambda_{j+1}v_{j+1}=\lambda_{j+1}\sum_{i=1}^j c_i v_i$$  Hence $$\sum_{i=1}^j \left(\lambda_i-\lambda_{j+1}\right) c_iv_i=0$$ which is a contradiction since $\lambda_i\neq \lambda_{j+1}$ for $1\leq i\leq j$.
Hope that helps,
A: For eigenvectors $\vec{v^1},\vec{v^2},\dots,\vec{v^n}$ with different eigenvalues $\lambda_1\neq\lambda_2\neq \dots \neq\lambda_n$ of a $ n\times n$ matrix $A$.
Given the $ n\times n$ matrix $P$ of the eigenvectors (with eigenvectors as the columns).
$$P=\Big[\vec{v^1},\vec{v^2},\dots,\vec{v^n}\Big]$$
Given the $ n\times n$ matrix $\Lambda$ of the eigenvalues on the diagonal (zeros elsewhere):
$$\Lambda = \begin{bmatrix}
    \lambda_1 & 0  & \dots  & 0 \\
    0 & \lambda_2 & \dots  & 0 \\
    \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & \dots  & \lambda_n
\end{bmatrix}
$$
Let $\vec{c}=(c_1,c_2,\dots,c_n)^T$
We need to show that only $c_1=c_2=...=c_n=0$ can satisfy the following:
$$c_1\vec{v^1}+c_2\vec{v^2}+...= \vec{0^{}}$$
Applying the matrix to this equation gives:
$$c_1\lambda_1\vec{v^1}+c_2\lambda_2\vec{v^2}+...+c_n\lambda_n\vec{v^n}= \vec{0^{}}$$
We can write this equation in the form of vectors and matrices:
$$P\Lambda \vec{c^{}}=\vec{0^{}}$$
But with since $A$ can be diagonalised to $\Lambda$, we know $P\Lambda=AP$
$$\implies AP\vec{c^{}}=\vec{0^{}}$$
since $AP\neq 0$, we have $\vec{c}=0$.
A: Suppose $v_k$ are eigenvectors of $A$, that is, $\sum_k c_k v_k=0$ with $v_k\neq0$, and $Av_k=\lambda_k v_k$. We only include in this sum eigenvectors corresponding to distinct eigenvalues, so that $\lambda_i\neq\lambda_j$ for $i\neq j$. We want to prove that this implies $c_k=0$.
Define the operators
$$A^{(i)}\equiv \prod_{k\neq i}(A-\lambda_k I).$$
In particular, note that $A^{(i)}v_k=\delta_{ik} d_i$, where $d_i\equiv \prod_{k\neq i}(\lambda_i-\lambda_k)\neq0$, and $\delta_{ik}$ is the standard Dirac delta function.
Then
$$0 = A^{(i)}(0)
= A^{(i)}\left(\sum_k c_k v_k\right)
=d_i c_i v_i.$$
It follows that $c_i=0$ for all $i$ (because, again, $d_i\neq0$ and $v_i\neq0$), and thus the vectors $\{v_i\}_i$ are linearly independent.
A: Hey I think there's a slick way to do this without induction. Suppose that $T$ is a linear transformation of a vector space $V$ and that $v_1,\ldots,v_n \in V$ are eigenvectors of $T$ with corresponding eigenvalues $\lambda_1,\ldots,\lambda_n \in F$ ($F$ the field of scalars). We want to show that, if $\sum_{i=1}^n c_i v_i = 0$, where the coefficients $c_i$ are in $F$, then necessarily each $c_i$ is zero. 
For simplicity, I will just explain why $c_1 = 0$. Consider the polynomial $p_1(x) \in F[x]$ given as $p_1(x) = (x-\lambda_2) \cdots (x-\lambda_n)$. Note that the $x-\lambda_1$ term is "missing" here. Now, since each $v_i$ is an eigenvector of $T$, we have 
\begin{align*}
 p_1(T) v_i = p_1(\lambda_i) v_i &&
\text{ where} &&  p_1(\lambda_i) = \begin{cases}
0 & \text{ if } i \neq 1 \\
p_1(\lambda_1) \neq 0 & \text{ if } i = 1
\end{cases}.
\end{align*}
Thus, applying $p_1(T)$ to the sum $\sum_{i=1}^n c_i v_i = 0$, we get
$$ p_1(\lambda_1) c_1 v_1 = 0 $$
which implies $c_1 = 0$, since $p_1(\lambda_1) \neq 0$ and $v_1 \neq 0$.  
A: I'll do it with two vectors. I'll leave it to you do it in general.
Suppose $\mathbf{v}_1$ and $\mathbf{v}_2$ correspond to distinct eigenvalues $\lambda_1$ and $\lambda_2$, respectively. 
Take a linear combination that is equal to $0$, $\alpha_1\mathbf{v}_1+\alpha_2\mathbf{v}_2 = \mathbf{0}$. We need to show that $\alpha_1=\alpha_2=0$.
Applying $T$ to both sides, we get
$$\mathbf{0} = T(\mathbf{0}) = T(\alpha_1\mathbf{v}_1+\alpha_2\mathbf{v}_2) = \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_2\mathbf{v}_2.$$
Now, instead, multiply the original equation by $\lambda_1$:
$$\mathbf{0} = \lambda_1\alpha_1\mathbf{v}_1 + \lambda_1\alpha_2\mathbf{v}_2.$$
Now take the two equations,
$$\begin{align*}
\mathbf{0} &= \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_2\mathbf{v}_2\\
\mathbf{0} &= \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_1\mathbf{v}_2
\end{align*}$$
and taking the difference, we get:
$$\mathbf{0} = 0\mathbf{v}_1 + \alpha_2(\lambda_2-\lambda_1)\mathbf{v}_2 = \alpha_2(\lambda_2-\lambda_1)\mathbf{v}_2.$$
Since $\lambda_2-\lambda_1\neq 0$, and since $\mathbf{v}_2\neq\mathbf{0}$ (because $\mathbf{v}_2$ is an eigenvector), then $\alpha_2=0$. Using this on the original linear combination $\mathbf{0} = \alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2$, we conclude that $\alpha_1=0$ as well (since $\mathbf{v}_1\neq\mathbf{0}$).
So $\mathbf{v}_1$ and $\mathbf{v}_2$ are linearly independent.
Now try using induction on $n$ for the general case.
A: Let $A$ be a $n \times n$ matrix with pairwaise different eigenvalues $\lambda_1, \ldots, \lambda_n$ and their eigenvectors $v_1,\ldots,v_n$. Put
$$P=[v_1\;\cdots\;v_n],\qquad
\Lambda=\left[\begin{array}{ccc} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n\end{array}\right],\qquad
V=\left[\begin{array}{cccc} 1 & \lambda_1 & \cdots & \lambda_1^{n-1} \\ \vdots & \vdots & \ddots &\vdots \\ 1 & \lambda_n & \cdots & \lambda_n^{n-1}\end{array}\right]$$
$$C=\left[\begin{array}{c} c_1 \\ \vdots \\ c_n\end{array}\right],\qquad
C'=\left[\begin{array}{ccc} c_1 & & \\ & \ddots & \\ & & c_n\end{array}\right]$$
Suppose $PC=c_1v_1+\ldots+c_nv_n=O_{n,1}$. Our aim is to show that $PC'=O_{n,n}$ (equivalently $c_iv_i=O_{n,1}$ for $i=1,\ldots,n$, which implies $c_i=0$, because $v_i\neq O_{n,1}$ for all $i$).
It is clear that $AP=[Av_1\;\cdots\;Av_n] = [\lambda_1v_1\;\cdots\;\lambda_nv_n] = P\Lambda$. Then
$$O_{n,1} = AO_{n,1} = A(PC) = (AP)C = (P\Lambda)C = P(\Lambda C),$$
so $P(\Lambda C)=O_{n,1}$ and inductively $P(\Lambda^kC)=O_{n,1}$ for $k=0,1,2,\ldots$.
Matrix $V$ is Vandermonde matrix and it is invertible because $\det V = \prod_{1\le i < j \le n}(\lambda_j-\lambda_i)\neq0$.
We have
$$PC' = PC'(VV^{-1}) = P(C'V)V^{-1} = P\left[\begin{array}{cccc} c_1 & \lambda_1c_1 & \cdots & \lambda_1^{n-1}c_1 \\ \vdots & \vdots & \ddots &\vdots \\ c_n & \lambda_nc_n & \cdots & \lambda_n^{n-1}c_n \end{array}\right]V^{-1} \\
= P[C\;\;\Lambda C\;\; \cdots\;\; \Lambda^{n-1}C]V^{-1}
= [PC\;\;P\Lambda C\;\; \cdots\;\; P\Lambda^{n-1}C]V^{-1}
= O_{n,n}V^{-1} = O_{n,n},$$
completing the proof.
A: I would like to add a bit of intuition.
Suppose v1, v2, w are eigenvectors of M with different eigenvalue. and
w = ⍺1v1 + ɑ2v2
w as composition of v1 and v2
Think of M as a transformation, for w to come out in the same direction, each of its components needs to be scaled proportionally (by the same amount). That is, v1 and v2 need to have identical eigenvalue, which contradicts with the assumption.
A: Consider the matrix $A$ with two distinct  eigen values $\lambda_1$ and $\lambda_2$. First note that the eigen vectors cannot be same , i.e., If $Ax = \lambda_1x$ and $Ax = \lambda_2x$ for some non-zero vector $x$. Well , then it means $(\lambda_1- \lambda_2)x=\bf{0}$. Since $\lambda_1$ and $\lambda_2$ are scalars , this can only happen iff $x = \bf{0}$ (which is trivial) or when $\lambda_1 =\lambda_2$ .
Thus now we can safely assume that given two eigenvalue , eigenvector pair say $(\lambda_1, {\bf x_1})$ and $(\lambda_2 , {\bf x_2})$ there cannot exist another  pair $(\lambda_3 , {\bf x_3})$ such that ${\bf x_3} = k{\bf x_1}$ or ${\bf x_3} = k{\bf x_2}$ for any scalar $k$. Now let ${\bf x_3} = k_1{\bf x_1}+k_2{\bf x_2}$ for some scalars $k_1,k_2$
Now,
$$ A{\bf x_3}=\lambda_3{\bf x_3} \\ $$
$$ {\bf x_3} = k_1{\bf x_1} + k_2{\bf x_2} \:\:\: \dots(1)\\  $$
$$ \Rightarrow A{\bf x_3}=\lambda_3k_1{\bf x_1} + \lambda_3k_2{\bf x_2}\\$$
but, $\:\: {\bf x_1}=\frac{1}{\lambda_1}Ax_1$ and ${\bf x_2}=\frac{1}{\lambda_2}Ax_2$. Substituting in above equation we get,
$$A{\bf x_3}=\frac{\lambda_3k_1}{\lambda_1}A{\bf x_1} + \frac{\lambda_3k_2}{\lambda_2}A{\bf x_2} \\$$
$$\Rightarrow {\bf x_3}=\frac{\lambda_3k_1}{\lambda_1}{\bf x_1} + \frac{\lambda_3k_2}{\lambda_2}{\bf x_2} \:\:\: \dots (2)$$
From equation $(1)$ and $(2)$ if we compare coefficients we get $\lambda_3 = \lambda_1$ and  $\lambda_3 = \lambda_2$ , which implies  $\lambda_1 = \lambda_2 = \lambda_3$ but according to our assumption they were all distinct!!! (Contradiction)
NOTE: This argument generalizes in the exact same fashion for any number of eigenvectors. Also, it is clear that if $ {\bf x_3}$ cannot be a linear combination of two vectors then it cannot be a linear combination of any $n >2$ vectors (Try to prove!!!)
