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Let $boldsymbolx=left(beginmatrixx\ yendmatrixright)$ be a vector in two-dimensional real space. By finding the eigenvalues and eigenvectors of $boldsymbolM$, sketch the locus of points $boldsymbolx$ that satisfy $$ boldsymbolx^TMx=4$$
given that
$$boldsymbolM=left(beginmatrix&5 &sqrt3\ &sqrt3 &3endmatrixright). $$

I found two eigenvalues to be $lambda_1 = 6$ and $lambda_2=2$, and the corresponding eigenvectors are
$$ boldsymbolv_1=left(beginmatrixsqrt3\ 1endmatrixright)quadtext and quad boldsymbolv_2=left(beginmatrix1\ -sqrt3endmatrixright)$$
(if I'm not mistaken :) ), but... what now? Frankly, I can't figure out how to make this helpful to find $boldsymbolx^TMx=4$.

Any hints?

The eigenvectors are orthogonal and span $Bbb R^2$. This means $mathbf v=beginbmatrixx\yendbmatrix=c_1mathbf x_1+c_2mathbf x_2$.

$mathbf v^TMmathbf v=(c_1mathbf x_1^T+c_2mathbf x_2^T)M(c_1mathbf x_1+c_2mathbf x_2)\=(c_1mathbf x_1^T+c_2mathbf x_2^T)(c_1lambda_1mathbf x_1+c_2lambda_2mathbf x_2)\=c_1^2lambda_1||mathbf x_1||^2+c_2^2lambda_2||mathbf x_2||^2\=24c_1^2+8c_2^2=4$

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

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 $mathbfM$.




• 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.