Find locus of points by finding eigenvalues

Question

Let $\boldsymbol{x}=\left(\begin{matrix}x\\ y\end{matrix}\right)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $\boldsymbol{M}$, sketch the locus of points $\boldsymbol{x}$ that satisfy $$ \boldsymbol{x^TMx}=4$$ given that $$\boldsymbol{M}=\left(\begin{matrix}&5 &\sqrt{3}\\ &\sqrt{3} &3\end{matrix}\right). $$

I found two eigenvalues to be $\lambda_1 = 6$ and $\lambda_2=2$, and the corresponding eigenvectors are $$ \boldsymbol{v}_1=\left(\begin{matrix}\sqrt{3}\\ 1\end{matrix}\right)\quad\text{ and }\quad \boldsymbol{v}_2=\left(\begin{matrix}1\\ -\sqrt{3}\end{matrix}\right)$$ (if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $\boldsymbol{x^TMx}=4$.

Any hints?

Answer 1

The eigenvectors are orthogonal and span $\Bbb R^2$. This means $\mathbf v=\begin{bmatrix}x\\y\end{bmatrix}=c_1\mathbf x_1+c_2\mathbf x_2$.

$\mathbf v^TM\mathbf v=(c_1\mathbf x_1^T+c_2\mathbf x_2^T)M(c_1\mathbf x_1+c_2\mathbf x_2)\\=(c_1\mathbf x_1^T+c_2\mathbf x_2^T)(c_1\lambda_1\mathbf x_1+c_2\lambda_2\mathbf x_2)\\=c_1^2\lambda_1||\mathbf x_1||^2+c_2^2\lambda_2||\mathbf x_2||^2\\=24c_1^2+8c_2^2=4$

$\implies6c_1^2+2c_2^2=1$

Answer 2

Some hints (as you requested):

• What we're studying is a second degree equation in $x$ and $y$, so it's a conic section curve (ellipse, hyperbola, or parabola).

• After we find the eigenvalues and eigenvectors, we can use them to diagonalize the matrix $\mathbf{M}$.

• The diagonalization process is really just a change of coordinate system, say from $(x,y)$ coordinates to $(u,v)$ coordinates

• In $(u,v)$ coordinates, since we're now dealing with a diagonal matrix, the given equation takes the form $au^2 + bv^2 = 1$. This is a conic section curve whose geometry you probably understand.