Show that the set $\{A\textbf{x} \mid \|\textbf{x}\|_2=1\}$ is an ellipse, and describe its two semi-axes (direction and length) in terms of $A$.

Question

Let $A$ be a $2 \times 2$ symmetric matrix with all positive eigenvalues. Show that the set $$\{A\textbf{x} \mid \|\textbf{x}\|_2=1\}$$ is an ellipse, and describe its two semi-axes (direction and length) in terms of $A$.

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How can we find two vectors $\bf{v_1}$, $\bf{v_2}$ and two positive numbers $\lambda_1$, $\lambda_2$ such that
$A\bf{v_1} = \lambda_1 f_1$, $A\bf{v_2} = \lambda_2 f_2$, $(\bf{v_i}, \bf{v_j})=\delta_{ij}$, where $\delta_{ij}$ denotes Kronecker's symbol?

Answer 1

Note that since $A$ is symmetric, it is diagonalizable and we can find an orthonormal basis $\{ v_1, v_2 \}$ of eigenvectors with corresponding eigenvalues $\lambda_1$ and $\lambda_2$. For $x \in \mathbb{R}$ such that $\|x \| = 1$ we can find $\theta$ such that

$$x = \cos(\theta) v_1 + \sin(\theta) v_2.$$

Make sure you understand why this is!

Then we have

$$Ax =A(\cos(\theta) v_1 + \sin(\theta) v_2)=\lambda_1 \cos(\theta)v_1+\lambda_2 \sin(\theta)v_2,$$

which traces out an ellipse with axes along $v_1$ and $v_2$ with lengths $\lambda_1$ and $\lambda_2$ as $\theta$ goes from $0$ to $2\pi$.