# For this bilinear form: $q(v)=q(x_1,x_2,x_3)=x_1^2+x_2^2+9x_3^{2}+2x_1x_2-6x_1x_3-5x_2x_3$ find a base $B$ so that $[q]^B_B=D$ diagonalizable matrix

For this bilinear form: $q(v)=q(x_1,x_2,x_3)=x_1^2+x_2^2+9x_3^{2}+2x_1x_2-6x_1x_3-5x_2x_3$ I need to find a base $B$ so that $[q]^B_B=D$ will be diagonalizable matrix. So, I tried to look for eigenvalues after writing this bilinear form as a matrix $\begin{pmatrix} 1 & 1 &-3 \\ 1&1 &-2.5 \\ -3& -2.5 & 9\end{pmatrix}$ and find eigenvalues but it's impossible mission, it's really messy. Is there any other method which with I can find it? maybe something with Jacobi method?

Thank you.

• Write $q(v)$ as a sum of square of independent linear functionals. If i'm not wrong, we find $q(x_1,x_2,x_3)= (x_1+x_2-x_3)^2-\frac{49}8x_2^2+8\left(x_3-\frac 7{16}x_2\right)^2 = ^t(Pv)DPv$, where $D = \mathrm{Diag}(1,-\frac{49}8,8)$ and $P(x,y,z) = (x_1+x_2-x_3,x_2,x_3-\frac 7{16}x_2)$. – Davide Giraudo Aug 16 '11 at 13:44
• Your matrix has a mistake in it, the $x_2 x_3$ term (2.5) should be negative. – rcollyer Aug 16 '11 at 13:51
• Thanks rcollyer. @Davide: Is this Lagrange theorem? I can't use it. – user6163 Aug 16 '11 at 14:09
• @Nir: What do you mean by Lagrange theorem? – Davide Giraudo Aug 16 '11 at 14:21
• The method that you used, Is it lagrange's? – user6163 Aug 16 '11 at 14:24

It is messy because you have misunderstood the problem. While $q(\underline{v})$ is induced by the bilinear form $f(\underline{u}, \underline{v})=\underline{v}^TA\underline{u}$, where $A$ is your $3\times 3$ coefficient matrix, $q$ is quadratic, not bilinear, also not a linear transformation. So, what you are asked to do is to find a decomposition of the form $A = P^TDP$ (where $P$ is invertible and the diagonal of $D$ does not necessarily contain any eigenvalue of $A$), but you have confused this with an eigenvalue decomposition $A = P^{-1}DP$. Surely, as your matrix $A$ is real symmetric, you can do both by performing an orthogonal decomposition $A=Q^TDQ$ where $QQ^T=I$ and $D$ contains the eigenvalues of $A$, but this is simply not required.
In general, you can find a decomposition $A = P^TDP$ by using elementary row/column operations. This is somewhat akin to finding a row-reduced echelon form of a matrix, but here we need to perform both an elementary row operation and a corresponding elementary column operation at each step. In other words, if, in a certain step, you multiply $A$ by an elementary matrix $E$ on the left, you should also mutiply $A$ by $E^T$ on the right.
For the problem you describe, however, simple inspection plus some completing-square trick is enough. Note that $$\begin{eqnarray} &&x_1^2 + x_2^2 + 9x_3^2 + 2x_1x_2 - 6x_1x_3 - 5x_2x_3\\ &=&(x_1 + x_2 - 3x_3)^2 + x_2x_3\\ &=&(x_1 + x_2 - 3x_3)^2 + \frac14[(x_2 + x_3)^2 - (x_2 - x_3)^2]. \end{eqnarray}$$ So you may take $B=\{(x_1 + x_2 - 3x_3),\ (x_2 + x_3),\ (x_2 - x_3)\}$. You may verify that $A = P^TDP$ where $$P=\begin{pmatrix} 1&1&-3\\0&1&1\\0&1&-1 \end{pmatrix}, \ D=\begin{pmatrix} 1\\&\frac14\\&&-\frac14 \end{pmatrix}.$$
There is a general method to solve this kind of exercises, which works for $\mathbb R^n$, not only $\mathbb R^3$. Let $q\colon\mathbb R^n\rightarrow \mathbb R$ a quadratic form. We can write it as a sum of squares of linearly independent linear functionals, namely $\displaystyle q(x_1,\ldots,x_n)=\sum_{j=1}^r\alpha_jl_j(x_1,\ldots,x_n)$ (we may not have $n$ termes, for example look at $q(x_1,\ldots,x_n)=x_1^2$). Then $q(x_1,\ldots,x_n) = ^t(P(x_1,\dots,x_n))DP(x_1,\dots,x_n)$ where $D=\mathrm{Diag}(\alpha_1,\ldots,\alpha_r,0\ldots,0)$ and $P(x_1,\ldots,x_n) = (l_1(x),\ldots,l_r(x),l_{r+1}(x),\ldots,l_n(x))$ ($l_{r+1}\ldots,l_n$ are linear functionals chosen in order to make $P$ invertible).