Problem: Find the EigenValues and EigenVectors of the matrix associated with quadratic forms $2x^2+6y^2+2z^2+8xz$.

I know how to convert a set of polynomial equations to a matrix but I have no clue how to convert this single quadratic equation with 3 variables into a matrix. Google tells me to use Cayley-Hamilton theorem... how?


2 Answers 2


Here's a start. You want a symmetric matrix. The diagonal elements correspond to the squares, and the non-diagonals to half the mixed monomials: $$ 2x^2+6y^2+2z^2+8xz= \left( \matrix{ x \cr y \cr z \cr } \right)^T \left( \matrix{ 2 & 0 & 4 \cr 0 & 6 & 0 \cr 4 & 0 & 2 \cr } \right) \left( \matrix{ x \cr y \cr z \cr } \right) $$ With $A$ the square matrix above, the characteristic polynomial is then $$ \eqalign{ p(\lambda) &= \det\left(A-\lambda I\right)= \left( \matrix{ \lambda-2 & 0 & -4 \cr 0 & \lambda-6 & 0 \cr -4 & 0 & \lambda-2 \cr } \right) \cr &= (\lambda-2)^2(\lambda-6)+16(\lambda-6) \cr &= (\lambda+2)(\lambda-6)^2 } $$ so there should be one eigenvector for $\lambda=-2$ and two for $\lambda=6$.

As to your question in the comments about where to read more...the simple answer is that you would want to look at the equation of a conic, and the matrix form allows you to identify the axes of symmetry.

There is a beautiful result called Sylvester's law of inertia concerning the solutions of the homogeneous equation $\mathbb{x}^TA\mathbb{x}=1$ for symmetric (i.e. real hermitian) matrices $A$, which is a quadratic form and can be extended to nonhomogeneous equations $\mathbb{x}^TA\mathbb{x}+\mathbb{b}^T\mathbb{x}=1$ (not discussed here). By homogeneous, we mean that all LHS monomials have the same degree $n$, and where the $1$ on the RHS could be replaced by any positive constant $r^n$ to scale the "unit" solution by the factor $r$ to an arbitrary level set of the same shape (multiplying by $-1$ changes the geometry to the opposite signature, explained below). Sylvester's law states that the homogeneous equation describes either an ellipsoid, or a hyperboloid of $k$ sheets, depending on the number $k$ of negative eigenvalues of $A$, which is an invariant under similarity or change of basis transformations. (If $A\mathbb{x}=\lambda\mathbb{x}$ and $B=TAT^{-1}$ for $T$ invertible then $\mathbb{y}=B\mathbb{x}$ satisfies $B\mathbb{y}=\lambda\mathbb{y}$.)

I have oversimplified because actually the geometry of the solution depends on the signature of $A$, an invariant including $k$ but also including the number of zero eigenvalues, counting multiplicity or viewed as the multiset of eigenvalues, called the spectrum of $A$. The signature $\sigma$ of $A$ is thus characterized by the triple $(n_+,n_0,n_-)$ representing the number of positive, zero, and negative eigenvalues, and the number of possible signatures depends on how many ordered partitions there are of $n$ into three nonnegative summands, which is ${n+2}\choose{2}$ by a simple argument, e.g. Example 5 on page 7 here. The opposite signature mentioned above is then just the reversed tuple.

This can be geometrically visualized using the eigendecomposition of $A$. Let $T$ be a matrix whose $i^\text{th}$ column contains a unit eigenvector corresponding to eigenvalue $\lambda_i$, ordered in nonincreasing sequence $\lambda_1\ge\lambda_2\ge\dots\lambda_n$. (We further require the eigenvectors to span the complete eigensubspaces in case there are repeated eigenvalues, which is guaranteed by the Schur decomposition or Jordan normal form for $A$ normal or diagonalizable.) Then $T^{-1}$ transforms the solution set onto the solution set of the equation $\sum_{i=0}^{n_+}x_i^2-\sum_{i=n-n_-}^{n}x_i^2=1$, corresponding to the "normalized" equation $\mathbb{x}^TI_\sigma\mathbb{x}=1$, where $I_\sigma$ is the diagonal matrix with the first $n_+$ diagonal entries $+1$, the middle $n_0$ entries $0$, and the last $n_-$ diagonal entries $-1$. These describe various hypersurfaces of dimension $n-1$ in $\mathbb{R}^n$, or of lower dimension in degenerate cases when $n_0>0$, i.e. when $A$ is singular. The eigenvectors in the columns of $T$ thus point to the axes of symmetry for these ellipsoids and hyperboloids. Witt's theorem(s), isometries and isotropies are also helpful to understand quadratic forms over an arbitrary field.

Sylvestor's law is related to the more general spectral decomposition of a hermitian matrix $A$, which is a special case of both the Schur and singular value decompositions, and of Sylvester's formula, since the orthogonal projections $P_{\lambda_i}$ onto the eigenspaces of each eigenvalue $\lambda_i$ mentioned in the spectral decomposition formula are actually the Frobenius covariants $A_i$ of $A$. There are many other types of matrix decompositions as well. The quadratic form $\mathbb{x}^TA\mathbb{x}$ above can also be viewed as a symmetric bilinear form $\mathbb{x}^TA\mathbb{y}$ with $\mathbb{y}=\mathbb{x}$. For visualization, it's also nice to look at ruled surfaces.

  • $\begingroup$ Thanks! >The diagonal elements correspond to the squares <-- which theorem/property says that and where can I read more? $\endgroup$
    – Nitin
    Commented Feb 19, 2012 at 10:36
  • 1
    $\begingroup$ First, try multiplying out $\left(\matrix{x\cr y}\right)^T\left(\matrix{a&b\cr b&a}\right)\left(\matrix{x\cr y}\right)$ by hand to convince yourself. If not convinced, repeat with a $3\times3$. There's lots of good material searching for "quadratic form matrix". $\endgroup$
    – bgins
    Commented Feb 19, 2012 at 11:00

In terms of just simply converting the quadratic expression into a matrix. You can just do it as follows... If you want to form a symmetric matrix consider the following 3x3 matrix:

                 X  A  B
                 A  Y  C  
                 B  C  Z
  • 1.The position of the X, Y and Z represent the coefficients of X, Y and Z SQUARED, respectively.

    • In your case these are
    • X=2
    • Y=6
    • Z=2
    1. The As represent the equal partition of the xy coefficient.
  • in your case the coefficient of xy =0
  • equally portioning that: 0/2=0
  • so A=0
    1. The Bs represent the equal partition of the xz coefficient
  • in your case the coefficient of xz = 8
  • equally partitioning that: 8/2 =4
  • so B=4
    1. C represents the equal partition of the yz coefficient
  • in your case the coefficient for yz =0
  • equally partitioning that: 0/2 = 0
  • so C =0

So the final matrix you obtain, plugging in all those coefficients:

2  0  4
0  6  0
4  0  2

Then of course remember to x(transpose) and x vector to actually turn in into an equation.


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