Linear Combination of Eigenvectors is Not an Eigenvector

Problem 258

Suppose that $\lambda$ and $\mu$ are two distinct eigenvalues of a square matrix $A$ and let $\mathbf{x}$ and $\mathbf{y}$ be eigenvectors corresponding to $\lambda$ and $\mu$, respectively.
If $a$ and $b$ are nonzero numbers, then prove that $a \mathbf{x}+b\mathbf{y}$ is not an eigenvector of $A$ (corresponding to any eigenvalue of $A$).

We use the following fact in the proof. Fact: Two eigenvectors corresponding to distinct eigenvalues are linearly independent.

Proof.

Seeking a contradiction, we assume that $a \mathbf{x}+b\mathbf{y}$ is an eigenvector corresponding to an eigenvalue $\zeta$.
Thus we have
\begin{align*}
A(a \mathbf{x}+b\mathbf{y})=\zeta (a \mathbf{x}+b\mathbf{y}). \tag{*}
\end{align*}

We calculate the left hand side of this equality as follows.
We have
\begin{align*}
A(a \mathbf{x}+b\mathbf{y})&=a A\mathbf{x}+bA\mathbf{y}\\
&=a\lambda \mathbf{x}+b\mu \mathbf{y}
\end{align*}
since $A\mathbf{x}=\lambda \mathbf{x}, A\mathbf{y}=\mu \mathbf{y}$ by defintion.

Therefore, from (*) we obtain
\begin{align*}
a(\lambda -\zeta) \mathbf{x}+b(\mu-\zeta)\mathbf{y}=\mathbf{0}.
\end{align*}

Recall that eigenvectors corresponding to distinct eigenvalues are linearly independent. Thus $\mathbf{x}$ and $\mathbf{y}$ are linearly independent.

Thus, the coefficients of the above linear combinations must be zero:
\[a(\lambda -\zeta)=0 \text{ and } b(\mu-\zeta)=0.\]

Since $a\neq0, b\neq 0$, this implies that we have
\[\lambda=\zeta=\mu,\]
and this is a contradiction because $\lambda$ and $\mu$ are supposed to be distinct.

Hence, $a \mathbf{x}+b\mathbf{y}$ cannot be an eigenvector of any eigenvalue of $A$.

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