Eigenvalues and Eigenvectors Examples 6

Eigenvalues and Eigenvectors Examples 6

Recall from the Eigenvalues and Eigenvectors page that the number $\lambda \in \mathbb{F}$ is said to be an eigenvalue of the linear operator $T \in \mathcal L (V)$ if $T(u) = \lambda u$ for some nonzero vector $u \in V$. The nonzero vectors $u$ such that $T(u) = \lambda u$ are called eigenvectors corresponding to the eigenvalue $\lambda$.

We will now look at some examples regarding eigenvalues of linear operators and eigenvectors corresponding to eigenvalues.

Example 1

Let $T$ and $S$ be linear operators over the vector space $V$ and such that $S$ is invertible. Prove that then $T$ and $S^{-1}TS$ have the same eigenvalues.

Let $\lambda \in \mathbb{F}$ be an arbitrary eigenvalue of $T$. Then for some nonzero vector $u \in V$ we have that $T(u) = \lambda u$. We want to show that then $\lambda$ is an eigenvalue of $S$, that is, there exists a nonzero vector $v \in V$ such that $S(v) = \lambda v$.

Note that:

(1)
\begin{align} \quad (S^{-1}TS)(S^{-1}(u)) = S^{-1}T(u) = S^{-1}(\lambda u) = \lambda S^{-1}(u) \end{align}

We note that $S^{-1}(u) \neq 0$. If $S^{-1}(u) = 0$, then $S^{-1}(u) = S^{-1}(0)$, and since $S^{-1}$ is invertible, then $S^{-1}$ is injective which implies that $u = 0$ which is a contradiction since $u$ is a nonzero eigenvector to the eigenvalue $\lambda$ of $T$.

Therefore $\lambda$ is an eigenvalue to $S^{-1}TS$ and the corresponding eigenvector is $S^{-1}(u)$.

Note that then if $u$ is an eigenvector corresponding to the eigenvalue $\lambda$ of $T$, then $S^{-1}(u)$ is an eigenvector corresponding to the eigenvalue $\lambda$ of $S^{-1}TS$.

Example 2

Let $S$ and $T$ be linear operators over the finite-dimensional vector space $V$. Prove that $ST$ and $TS$ have the same eigenvalues.

Let $\lambda \in \mathbb{F}$ be an arbitrary eigenvalue of $ST$. Then for some nonzero vector $u \in V$ we will have that $ST(u) = \lambda u$. We want to show that then $\lambda$ is an eigenvalue of $TS$.

We have that:

(2)
\begin{align} \quad TS(T(u)) = T(ST(u)) = T(\lambda u) = \lambda T(u) \end{align}

There are two cases to consider.

If $T(u) \neq 0$ then $\lambda$ is an eigenvalue of $TS$ and the corresponding eigenvector is $T(u)$.

If $T(u) = 0$, then we must have that $\lambda = 0$. This is because $ST(u) = \lambda u$ implies that $S(0) = \lambda u$, and $u$ is a nonzero vector, so $\lambda = 0$.

Also, if $T(u) = 0$ then since $u$ is a nonzero vector we have that $u \in \mathrm{null} (T)$, and so $\mathrm{null} (T) \neq \{ 0 \}$ so $T$ is not injective and hence $T$ is not invertible. Therefore $TS$ is not invertible. Since $TS$ is not invertible, this implies that the eigenvalue $\lambda = 0$ for $TS$.

Therefore $ST$ and $TS$ have the same eigenvalues.