Why are the axis of an ellipsoid eigenvectors?

Question

Consider an ellipsoid $\{x| x^TAx = 1\}$.

Let $A$ be a real symmetric matrix, and consider the eigen-decomposition $A=P\Lambda P^{-1} =P\Lambda P^T$, where the matrix $P$ is orthogonal because $A$ is symmetric, and the diagonal matrix $\Lambda$ contains the eigenvalues of $A$.

Then the ellipsoid can be rewritten as: $$x^TP\Lambda P^Tx = (P^Tx)^T\Lambda(P^Tx) = y^T\Lambda y = 1$$

Since the matrix is diagonal, the quadratic form gives: $$y^T\Lambda y =\lambda_1y_1^2+...+\lambda_ny_n^2 = 1$$

This is clearly the equation of ellipsoid.

My question is, why are the axes of this ellipsoid the eigenvectors of $A$?

Answer 1

The axes of this ellipsoid are (the multiples of) $e_1=(1,0,0,\ldots,0)$, $e_2=(0,1,0,\ldots,0)$, …, $e_n=(0,0,0,\ldots,1)$ (and the corresponding eigenvalues are $\lambda_1,\lambda_2,\ldots,\lambda_n$). But $\Lambda=P^{-1}AP$ and $P$ is a change-of-bases matrix. So, the columns of $P$ are eigenvectors of $A$ (and, again, the corresponding eigenvalues are $\lambda_1,\lambda_2,\ldots,\lambda_n$). This was just a change of basis, so, geometrically, the columns of $P$ are the still the vectors $e_1,\ldots,e_n$.

Answer 2

As you mentioned $$\left(\frac{y_1}{\sqrt{\lambda_1}}\right)^2 + \left(\frac{y_2}{\sqrt{\lambda_2}}\right)^2 + \cdots + \left(\frac{y_n}{\sqrt{\lambda_n}}\right)^2 = 1$$ is the equation of an ellipse with semiaxis lengths $\sqrt{\lambda_1},\sqrt{\lambda_2},\ldots,\sqrt{\lambda_n}$ (each $\lambda_i > 0$ because the original matrix $A$ is positive semidefinite). Specifically, this is an axis aligned ellipse in the y-coordinate space. Set any $y_i$ to be $\sqrt{\lambda_i}$ and the rest to $0$ and plug into the equality to convince yourself this is so.

From your question statement we have $y = P^Tx = P^{-1}x$ or equivalently $x = Py$. We know the axes of the ellipse in y-coordinate space are just the standard basis vectors $\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n$.

The axes in x-coordinate space then must be $P\mathbf{e}_1, P\mathbf{e}_2, \ldots, P\mathbf{e}_n$. That is to say, the ellipse axes are just the columns of $P$. The columns of $P$ are of course the eigenvectors of $A$ because of how $P$ was defined to be the matrix diagonalizing $A$.