Ytukyg

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

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 $0$, $alpha_1mathbf{v}_1+alpha_2mathbf{v}_2 = mathbf{0}$. We need to show that $alpha_1=alpha_2=0$.

Applying $T$ to both sides, we get
$$mathbf{0} = T(mathbf{0}) = T(alpha_1mathbf{v}_1+alpha_2mathbf{v}_2) = alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2.$$
Now, instead, multiply the original equation by $lambda_1$:
$$mathbf{0} = lambda_1alpha_1mathbf{v}_1 + lambda_1alpha_2mathbf{v}_2.$$
Now take the two equations,
$$begin{align*}
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_2mathbf{v}_2\
mathbf{0} &= alpha_1lambda_1mathbf{v}_1 + alpha_2lambda_1mathbf{v}_2
end{align*}$$
and taking the difference, we get:
$$mathbf{0} = 0mathbf{v}_1 + alpha_2(lambda_2-lambda_1)mathbf{v}_2 = alpha_2(lambda_2-lambda_1)mathbf{v}_2.$$

Since $lambda_2-lambda_1neq 0$, and since $mathbf{v}_2neqmathbf{0}$ (because $mathbf{v}_2$ is an eigenvector), then $alpha_2=0$. Using this on the original linear combination $mathbf{0} = alpha_1mathbf{v}_1 + alpha_2mathbf{v}_2$, we conclude that $alpha_1=0$ as well (since $mathbf{v}_1neqmathbf{0}$).

So $mathbf{v}_1$ and $mathbf{v}_2$ are linearly independent.

Now try using induction on $n$ for the general case.

Alternative:

Let $j$ be the maximal $j$ such that $v_1,dots,v_j$ are independent. Then there exists $c_i$, $1leq ileq j$ so that $sum_{i=1}^j c_iv_i=v_{j+1}$. But by applying $T$ we also have that

$$sum_{i=1}^j c_ilambda_iv_i=lambda_{j+1}v_{j+1}=lambda_{j+1}sum_{i=1}^j c_i v_i$$ Hence $$sum_{i=1}^j left(lambda_i-lambda_{j+1}right) c_iv_i=0$$ which is a contradiction since $lambda_ineq lambda_{j+1}$ for $1leq ileq j$.

Hope that helps,

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 $v_1 neq 0$.

For eigenvectors $vec{v^1},vec{v^2},dots,vec{v^n}$ with different eigenvalues $lambda_1neqlambda_2neq dots neqlambda_n$ of a $ ntimes n$ matrix $A$.

Given the $ ntimes 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 $ ntimes 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:
$$c_1vec{v^1}+c_2vec{v^2}+...= vec{0^{}}$$
Applying the matrix to this equation gives:
$$c_1lambda_1vec{v^1}+c_2lambda_2vec{v^2}+...+c_nlambda_nvec{v^n}= vec{0^{}}$$
We can write this equation in the form of vectors and matrices:

$$PLambda vec{c^{}}=vec{0^{}}$$

But with since $A$ can be diagonalised to $Lambda$, we know $PLambda=AP$
$$implies APvec{c^{}}=vec{0^{}}$$
since $APneq 0$, we have $vec{c}=0$.