Eliminating cross product terms of a symmetric bilinear form.

Question

Let $A$ be a symmetric bilinear form on $\mathbb R^n$.For the basis $\{\alpha_1,...,\alpha_n\}$ we have $A(x,y)=\sum\limits_i\sum\limits_j A(\alpha_i,\alpha_j)x_iy_j$.My question is how to find a new basis $\{\beta_1,...,\beta_n\}$ such that $A(x,y)=\sum\sum A(\beta_i,\beta_i)x_iy_i$.In particular for the bilinear form $A(x,y)=x_1y_1+2x_2y_2+x_1y_2+x_2y_1$ how to find a basis with respect to which the bilinear form will have no cross product term i.e. $x_iy_j$ for $i\neq j$.I think the process is to find an orthogonal basis obtained by Gram-Schmidt orthogonalization,am I correct? I think the basis $\{(1,0),(-1,1)\}$ is the required one.

Answer 1

Here is a general answer in that it applies to any finite-dimensional vector space $V$ over a field of characteristic zero. It is based on Theorem 3 of Chapter 10 in Linear Algebra, second edition, by Hoffman and Kunze. As you suspected, the proof uses the idea of Gram-Schmidt orthogonalization.

If $A = 0$ or $n = 1$, there is nothing to prove. Thus we may suppose $A\ne 0$ and $n > 1$. If $A(x,x) = 0$ for every $x$ in $V$, the associated quadratic form $q$ is identically $0$, and the polarization identity shows that $A = 0$. Thus there is a vector $x$ in $V$ such that $A(x,x) = q(x)\ne 0$. Let $W$ be the one-dimensional subspace of $V$ that is spanned by $x$, and let $W^\perp$ be ths et of all vectors $y$ in $V$ such that $A(x,y) = 0$. Now we claim that $V = W\oplus W^\perp$. Certainly the subspaces $W$ and $W^\perp$ are independent. A typical vector in $W$ is $cx$ where $c$ is a scalar. If $cx$ is also in $W^\perp$, then $A(cx,cx) = c^2 A(x,x)$. But $A(x,x)\ne 0$; thus, $c = 0$. Moreover, each vector in $V$ is the sum of a vector in $W$ and a vector in $W^\perp$. For, let $z$ be any vector in $V$, and put $$y = z -\frac{A(z,x)}{A(x,x)}x.$$ Then $$A(x,y) = A(x,z) -\frac{A(z,x)}{A(x,x)}A(x,x),$$ and because $A$ is symmetric, $A(x,y) = 0$. Thus, $y$ is in the subspace $W^\perp$. The expression $$z =\frac{A(z,x)}{A(x,x)}x + y$$ shows us that $V = W + W^\perp$.

The restriction of $A$ to $W^\perp$ is a symmetric bilinear form on $W^\perp$. Because $W^\perp$ has dimension $n - 1$, we may assume by induction that $W^\perp$ has a basis $\{\beta_2,\dots,\beta_n\}$ such that $A(\beta_i,\beta_j) = 0,\ i\ne j,\ i\ge 2,\ j\ge 2$. Putting $\beta_1 = x$, we obtain a basis $\{\beta_1,\dots,\beta_n\}$ for $V$ such that $A(\beta_i,\beta_j) = 0$ for $i\ne j$.

Answer 2

As Jean Marie indicated in the comments, this can be solved by diagonalizing the matrix of $A$, which is $$M=\begin{pmatrix}1&1\\1&2\end{pmatrix}$$ To do this, you can first compute the eigenvalues as the roots of the characteristic polynomial $$\det(\lambda I-M)=\begin{vmatrix}\lambda-1&-1\\-1&\lambda-2\end{vmatrix}=(\lambda-1)(\lambda-2)-(-1)(-1)=\lambda^2-3\lambda+1$$ The roots are just $\lambda_1=\tfrac{1}{2}(3+\sqrt{5})$ and $\lambda_2=\tfrac{1}{2}(3-\sqrt{5})$. Then solve the corresponding linear systems $Mv=\lambda_1v$ and $Mv=\lambda_2v$ to obtain the eigenvectors $v_1=\bigl(\tfrac{1}{2}(-1+\sqrt{5}),1\bigr)$ and $v_2=\bigl(\tfrac{1}{2}(-1-\sqrt{5}),1\bigr)$. The basis $v_1,v_2$ diagonalizes $A$. For example, you can verify that $$A(v_1,v_2)=\frac{1}{4}(-4)+2-\frac{1}{2}+\frac{\sqrt{5}}{2}-\frac{1}{2}-\frac{\sqrt{5}}{2}=0$$ and similarly $A(v_2,v_1)=0$.

Answer 3

This requires neither diagonalisation nor the Gram-Schmidt procedure (the latter should be obvious since the statement does not involve any inner product; the symmetric bilinear form need not be positive or definite). This can be done using what in France is called Gauss's orthogonalisation procedure (I have not seen this name in English texts; what is being orthogonalised is the coordinate system in which the bilinear form is expressed, where orthogonality means exactly the vanishing of off-diagonal terms).

It is important to note that such an orthogonal coordinate system is far from unique (much further than for instance a basis of diagonalisation of a linear operator). Therefore the procedure can leave a lot of liberty.

We are trying to find independent linear forms $\phi_i\in V^*$, where $V$ is the space of dimension$~n$ that $A$ is a bilinear form on (of course no more than $n$ of them, although one might need less) such that $A(v,w)=\sum_i\mu_i\phi_i(v)\phi_i(w)$ where the $\mu_i$ are some nonzero real numbers (one could limit the choices to $\mu_i\in\{-1,+1\}$ but one can avoid having to take square roots if this restriction is not imposed). Once one has a sequence of such $\phi_i$, one can complete them to a basis of $V^*$, and then form the dual basis, a basis of $V$ form which the $\phi_i$ are coordinate functions; on that basis, the expression for $A$ has no cross terms. From the independence of the $\phi_i$ it follows that their number is equal to the rank of $A$; the procedure will proceed in steps, each of which produces $1$ or $2$ such linear forms, such that subtraction from $A$ of their contributions reduces the rank of the remainder by that number $1$ or $2$. Clearly this stops when the remainder becomes zero.

In matrix terms, this is my favourite formulation. Considering matrix expressing the remainder of $A$ (on the initial basis of $V$); if it has any nonzero diagonal entries a step of producing one linear form is taken, otherwise if the matrix is nonzero a step producing two linear forms is taken. For the step producing a single linear form, pick a nonzero diagonal entry$~d$, let $R$ be the row containing it, and $R^\top$ its column; then $\frac1dR^\top R$ is the unique symmetric matrix of rank$~1$ that coincides with the remainder of $A$ in that row and column. Subtracting that matrix will therefore clear that row and column of $A$, and decrease the rank by$~1$. Here $R$ is the matrix of a linear form $\phi$, and we have contributed a rank$~1$ bilinear form $(v,w)\mapsto\frac1d\phi(v)\phi(w)$.

The case where there are nonzero entries but only off the main diagonal is similar. Pick such an off-diagonal nonzero coefficient $c$ at position $(k,l)$ with $k<l$ (the same coefficient $c$ is at position $(l,k)$ as well), let $R_k$ and $R_l$ be their respective rows; then $\frac1c(R_l^\top R_k+R_k^\top R_l)$ is a rank $2$ matrix that coincides with the remainder of $A$ in rows and columns $k$ and $l$ (this uses that the diagonal entries at $(k,k)$ and $(l,l)$ were zero). Subtracting that matrix will therefore clear those rows and columns of $A$, and decrease the rank by$~2$. If $R_k$ and $R_l$ are matrices of linear forms $\phi,\psi$, then we have contributed a linear forms $\phi+\psi$ with coefficient $\frac1{2d}$, and $\phi-\psi$ with coefficient $-\frac1{2d}$. (The rank$~2$ bilinear form $(v,w)\mapsto\frac1{2d}(\phi+\psi)(v)(\phi+\psi)(w)-(\phi-\psi)(v)(\phi-\psi)(w)$ is the one whose matrix we subtracted.)