How to prove that eigenvectors from different eigenvalues are linearly independent

Question

How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?

Answer 1

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.

Answer 2

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,

Answer 3

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 $v_1 \neq 0$.

Answer 4

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 $\vec{c}=0$.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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!!!)