📂Linear Algebra

Eigenvectors Corresponding to Distinct Eigenvalues are Linearly Independent

Theorem¹

Let $V$ be a vector space, $T : V \to V$ be a linear transformation, and $\lambda_{1}, \dots, \lambda_{k}$ be distinct eigenvalues of $T$. If $\mathbf{v}_{1}, \dots, \mathbf{v}_{k}$ are eigenvectors of $T$ corresponding to the eigenvalue $\lambda_{1}, \dots, \lambda_{k}$, then $\left\{ \mathbf{v}_{1}, \dots, \mathbf{v}_{k} \right\}$ are linearly independent.

Corollary

Diagonalization

T is diagonalizable if there exist $n$ linearly independent eigenvectors of $T$.

If $T : V \to V$ is a linear transformation, and assuming $\dim(V) = n$. If $T$ has $n$ distinct eigenvalues, then $T$ is diagonalizable.

Proof

We prove this by mathematical induction. Assume $k=1$. Then, by the definition of eigenvectors, $\mathbf{v}_{1} \ne \mathbf{0}$ holds, and $\left\{ \mathbf{v}_{1} \right\}$ are linearly independent.

Now assume the theorem holds for $k-1$ distinct eigenvalues. And let $a_{1}, \dots, a_{k}$ be constants that satisfy the following equation.

$$a_{1}\mathbf{v}_{1} + \cdots + a_{k}\mathbf{v}_{k} = \mathbf{0}$$

Taking $T - \lambda_{k}I$ on both sides, since $v_{i}$ are eigenvectors,

$$a_{1}(\lambda_{1} - \lambda_{k})\mathbf{v}_{1} + \cdots a_{k-1}(\lambda_{k-1} - \lambda_{k})\mathbf{v}_{k-1} = \mathbf{0}$$

Then, by assumption, the following holds.

$$a_{1}(\lambda_{1} - \lambda_{k}) = \cdots = a_{k-1}(\lambda_{k-1} - \lambda_{k}) = 0$$

Since we assume $\lambda_{i}$ are distinct,

$$a_{1} = \cdots = a_{k-1} = 0$$

Therefore, $a_{k}\mathbf{v}_{k} = \mathbf{0}$, but since $\mathbf{v}_{k} \ne \mathbf{0}$, it follows that $a_{k} = 0$. Hence,

$$a_{1} = \cdots = a_{k} = 0$$

and $\left\{ \mathbf{v}_{1}, \dots, \mathbf{v}_{k} \right\}$ are linearly independent.

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