# Orthonormal basis quadratic form in canonical form

I am very confused and don't know what to do, can you give me any suggestions? This is my task:

In orthonormal basis quadratic form F(x) is calculated like this: $$F(x) = 17x^2 _1 + 17x^2_2 + 11x ^2_3 − 16x_1x_2 + 8x_1x_3 − 8x_2x_3.$$ Find transformation matrix to another orthonormal basis, in which $$F$$ is in canonical form, and find this canonical form.

• Do you mean "canonical form of quadratic form" = only $\;\pm1,\,0\;$ on the main diagonal, or "canonical form" = diagonal form with eigenvalues on the main diagonal? Many times the first want is what is meant with quadratic forms, e.g. in Sylverster's Law of Inertia, yet sometimes it is enough with the "usual* diagonalization. Both cases are nasty this times, though...In fact, it seems it is unusually nasty! Are you sure that is the quadratic form you were given? – DonAntonio Jun 21 at 17:30
• i mean "canonical form of quadratic form" – Edvards Zakovskis Jun 21 at 17:33
• I started calculating the canonical form of the quadratic form but there were very nasty numbers so i tought i was doing something wrong- may be there is something i don't see – Edvards Zakovskis Jun 21 at 17:35
• @Ed Nop, it really looks nasty in a rather extreme way. Even the usually nice Lagrange's Method of completing the square seems to be nasty here , since it would begin as $$17\left(x-\frac8{18}x_2+\frac4{17}x\right)^2+\ldots$$ When something begins as above I usually retire. Good luck, most probably you are in the right track... – DonAntonio Jun 21 at 17:50
• @DonAntonio The form with $\pm1,0$ on the diagonal can’t be achieved with the required transformation. – amd Jun 21 at 18:57

Your quadratic form is represented by a unique symmetric matrix $$S$$ in the sense that $$F(x)=x\cdot Sx$$. In this case, the matrix $$S$$ is $$S=\left[\begin{array}{rrr} 17 & -8 & 4 \\ -8 & 17 & -4 \\ 4 & -4 & 11 \end{array}\right]$$ To construct this $$S$$, note that the $$(i, j)$$th entry is half the coefficient of $$x_ix_j$$ in $$F(x)$$ if $$i\neq j$$ and equal to the coefficient of $$x_ix_j$$ if $$i=j$$.
The process of putting $$F$$ in "canonical form" is sometimes referred to as "completing the square." The strategy here is to diagonalize $$S$$ as $$S=QDQ^\top$$ where $$Q$$ is orthogonal ($$Q^\top=Q^{-1}$$). We then "change variables" by defining $$y=Q^\top x$$. Our quadratic form can then be written as $$F(x) = x\cdot Sx = x^\top QDQ^\top x = (Q^\top x)^\top D (Q^\top x) = y^\top D y = \lambda_1\cdot y_1^2 + \lambda_2\cdot y_2^2 + \lambda_3\cdot y_3^2$$ where the $$\lambda_i$$'s are the diagonal entries of $$D$$ (which are also the eigenvalues of $$S$$).
To go about this process, we need to find orthonormal bases for the eigenspaces of $$S$$. It's not terribly difficult to show that this $$S$$ has two eigenvalues $$\lambda_1=27$$ and $$\lambda_2=9$$. Bases for the eigenspaces are given by \begin{align*} E_{27} &= \operatorname{Null}(S-27\cdot I_3)=\operatorname{Span}\{\left\langle2,\,-2,\,1\right\rangle\} \\ E_{9} &= \operatorname{Null}(S-9\cdot I_3) = \operatorname{Span}\{\left\langle1,\,0,\,-2\right\rangle, \left\langle0,\,1,\,2\right\rangle\} \end{align*} Note, however, that these bases are not orthonormal. To make them orthonormal, we can apply the Gram-Schmidt algorithm. Can you carry out this algorithm and finish "completing the square"?
• We only need GS method for the basis of $\;E_9\;$ , as eigenvectors corresponding to different eigenvalues are already orthogonal. Still, there's the work to find out the eigenvalues. Not so nice with this matrix, though... – DonAntonio Jun 21 at 19:02
• @DonAntonio The eigenvector doesn’t instantly jump out at me like it does in other exercises, but having $4$ and $-4$ at the bottom of the first two columns suggested trying $(1,1,0)^T$. – amd Jun 21 at 19:23
• @DonAntonio I like to remind my students that Gram-Schmidt in dimension one is the same as normalization. So applying Gram-Schmidt to $E_{27}$ is the same as normalizing the given non-unit basis vector. – Brian Fitzpatrick Jun 24 at 0:07