Be $V=R^3$ with the bilinear form:

$\phi((x,y,z),(x',y',z')=(x,y,z)\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\\\end{pmatrix} (x',y',z')^t$

Find an orthogonal basis for V.

To do this, I take 3 linear independent vectors $v_1=(1,1,0)$, $v_2=(0,1,0)$, $v_3=(0,0,1)$,

And I do the Gram-Schmidt process:




The problem comes when I try to calculate $\phi(v_2,v_2)$, which equals 0 (I can't divide by 0) Why is this happening and how can I avoid this situation? Also, I have noticed that $v_3'=v_3-\frac{\phi(v_1,v_3)v}{\phi(v_1,v_1}$ is not orthogonal to $v_1$, what am I doing wrong?

Thank you

  • $\begingroup$ Your bilinear form is not definite. $\endgroup$
    – Bernard
    Jun 15 '16 at 0:06

The given bilinear form is indefinite, so the Gram-Schmidt process won’t work. However, the matrix $A$ of the bilinear form is symmetric, so there is an orthogonal basis relative to the standard Euclidean scalar product $\langle\cdot,\cdot\rangle$ consisting of eigenvectors of $A$.

Suppose that we have such a basis $(v_1, v_2, v_3)$ with corresponding eigenvalues $\lambda_1$, $\lambda_2$, $\lambda_3$. Then for $i\ne j$, $$\phi(v_i,v_j)=\langle v_i,Av_j\rangle=\langle v_i,\lambda_j v_j\rangle=\lambda_j\langle v_i,v_j\rangle=0$$ so this basis is $\phi$-orthogonal as well. Thus, diagonalizing $A$ will give you the basis you’re looking for. Since one of the eigenvalues of the given matrix is zero, $\phi(v,v)=0$ for one of these vectors, but that’s why you’re not being asked to find an orthonormal basis.


As a reminder that a symmetric matrix defined an scalar product iff all its eigenvalues are strictly positive.

the matrix A is not reversible so can not definite scalar product, and we can not proceed to the method of Gram Schmidt.

  • $\begingroup$ What should I do, then, to find an orthogonal basis for V? Thank you for your answer. $\endgroup$
    – J doeoeo
    Jun 15 '16 at 0:15
  • $\begingroup$ apply the Gauss method for the quadratic form $\Phi((x,y,z),(x,y,z))$ associated to the symmetric matrix A and you find the desired orthogonal basis $\endgroup$
    – m.idaya
    Jun 15 '16 at 0:57

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