Problems tagged with eigenbasis

Problems tagged with "eigenbasis"

Problem #45

Tags: linear algebra, eigenbasis, quiz-03, eigenvalues, eigenvectors, lecture-04

Let $\vec{u}^{(1)}, \vec{u}^{(2)}, \vec{u}^{(3)}$ be three unit length orthonormal eigenvectors of a linear transformation $\vec{f}$, with eigenvalues $8$, $4$, and $-3$ respectively.

Suppose a vector $\vec{x}$ can be written as:

\[\vec{x} = 2\vec{u}^{(1)} - 3\vec{u}^{(2)} + \vec{u}^{(3)}\]

What is $\vec{f}(\vec{x})$, expressed in coordinates with respect to the eigenbasis $\{\vec{u}^{(1)}, \vec{u}^{(2)}, \vec{u}^{(3)}\}$?

Solution

$[\vec{f}(\vec{x})]_{\mathcal{U}} = (16, -12, -3)^T$ Using linearity and the eigenvector property:

$$\begin{align*}\vec{f}(\vec{x}) &= \vec{f}(2\vec{u}^{(1)} - 3\vec{u}^{(2)} + \vec{u}^{(3)}) \\&= 2\vec{f}(\vec{u}^{(1)}) - 3\vec{f}(\vec{u}^{(2)}) + \vec{f}(\vec{u}^{(3)}) \\\end{align*}$$

Since $\vec{u}^{(1)}$ is an eigenvector with eigenvalue $8$, we have $\vec{f}(\vec{u}^{(1)}) = 8\vec{u}^{(1)}$. Similarly, $\vec{f}(\vec{u}^{(2)}) = 4\vec{u}^{(2)}$ and $\vec{f}(\vec{u}^{(3)}) = -3\vec{u}^{(3)}$:

$$\begin{align*}&= 2(8\vec{u}^{(1)}) - 3(4\vec{u}^{(2)}) + (-3)\vec{u}^{(3)}\\&= 16\vec{u}^{(1)} - 12\vec{u}^{(2)} - 3\vec{u}^{(3)}\end{align*}$$

In the eigenbasis, this is simply $(16, -12, -3)^T$.

Problem #46

Tags: linear algebra, eigenbasis, quiz-03, eigenvalues, eigenvectors, lecture-04

Let $\vec{u}^{(1)}, \vec{u}^{(2)}, \vec{u}^{(3)}$ be three unit length orthonormal eigenvectors of a linear transformation $\vec{f}$, with eigenvalues $5$, $-2$, and $3$ respectively.

Suppose a vector $\vec{x}$ can be written as:

\[\vec{x} = 4\vec{u}^{(1)} + \vec{u}^{(2)} - 2\vec{u}^{(3)}\]

What is $\vec{f}(\vec{x})$, expressed in coordinates with respect to the eigenbasis $\{\vec{u}^{(1)}, \vec{u}^{(2)}, \vec{u}^{(3)}\}$?

Solution

$[\vec{f}(\vec{x})]_{\mathcal{U}} = (20, -2, -6)^T$ Using linearity and the eigenvector property:

$$\begin{align*}\vec{f}(\vec{x}) &= \vec{f}(4\vec{u}^{(1)} + \vec{u}^{(2)} - 2\vec{u}^{(3)}) \\&= 4\vec{f}(\vec{u}^{(1)}) + \vec{f}(\vec{u}^{(2)}) - 2\vec{f}(\vec{u}^{(3)}) \\\end{align*}$$

Since $\vec{u}^{(1)}$ is an eigenvector with eigenvalue $5$, we have $\vec{f}(\vec{u}^{(1)}) = 5\vec{u}^{(1)}$. Similarly, $\vec{f}(\vec{u}^{(2)}) = -2\vec{u}^{(2)}$ and $\vec{f}(\vec{u}^{(3)}) = 3\vec{u}^{(3)}$:

$$\begin{align*}&= 4(5\vec{u}^{(1)}) + (-2)\vec{u}^{(2)} - 2(3\vec{u}^{(3)}) \\&= 20\vec{u}^{(1)} - 2\vec{u}^{(2)} - 6\vec{u}^{(3)}\end{align*}$$

In the eigenbasis, this is simply $(20, -2, -6)^T$.