# Find locus of points by finding eigenvalues

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?

• The eigenvalues are $5\pm\sqrt3$, not $6,2$ because $6+2\ne5+5$ – Shubham Johri Dec 17 '18 at 10:44
• @ShubhamJohri see the edited question. I wrongly typed one entry of $M$. – VanDerWarden Dec 17 '18 at 10:45
• Yes, now the eigenvalues and eigenvectors are fine – Shubham Johri Dec 17 '18 at 10:47

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

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

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

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.

• Not op but I got $3u^2+2v^2=8$. So do I just plot this elipse with axis being the two eigenvectors? And how long shoud $1$ unit be? Say for axis1 with $v_1$, should $1$ unit be normal 1 unit, normal 6 units (eigenvalue) or 1 unit should be $2$ normal units (length of $v1$). I think the last one is correct – Anvit Dec 17 '18 at 11:11
• @Anvit: Yes, the eigenvectors define the directions of the axes of symmetry of the conic. If $\mathbf v$ is an eigenvector, then so is $k\mathbf v$, for any $k \ne 0$, so the eigenvectors can't tell you anything about scale. – bubba Dec 17 '18 at 11:17
• @bubba -- there's little or no mileage in appealing to downvoters to explain or justify their objections. It's best not to betray that you heed them atall. – AmbretteOrrisey Dec 17 '18 at 11:35