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Find an orthogonal basis for the bilinear form over $\mathbb{R}$ given by $(\mathbf{x}, \mathbf{y})\mapsto\mathbf{x}^{t}A\mathbf{y}$ where $A=\begin{bmatrix} 1 & 4 & 4\\ 4 & 4 & 10\\ 4 & 10 & 16 \end{bmatrix}$.

I'm not sure if this is as easy as using Gram-Schmidt, or if there is another way. I used Gram-Schmidt and obtained very complicated fractions in my vectors, so I have a feeling that this is wrong...

Marc van Leeuwen
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Kirk Fogg
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  • There is something not quite right here. If $x=(-1,2,-1)^T$, then $(x,x) = -15$, so this cannot be an inner product (which I presume, based on the notation, it is meant to be?). – copper.hat Jun 24 '13 at 02:18
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    @copper.hat, not quite, it is just a mapping from two copies of $\mathbb R^3$ and may be indefinite as a quadratic form. Not careful notation. What is wanted here seems to be Hermite reduction or similar. If I get ambitious later I will look that up. – Will Jagy Jun 24 '13 at 02:30
  • @WillJagy: I see, thanks! $A$ is, of course, diagonalizable, maybe this is what the OP wants? However, neither the eigenvalues nor eigenvectors have an obvious (Octave) simple form... – copper.hat Jun 24 '13 at 02:37
  • @copper.hat, i put a complete answer. Instead of eigenvectors, you just clear out the off diagonal terms of the symmetric matrix, you get a sum of squares with coefficients. I think you could say that the resulting inverse matrices give an $A$-orthogonal basis, which was very easy to calculate, as all my numbers show up in an upper-triangular matrix, so even the inverse is easy. – Will Jagy Jun 24 '13 at 03:21
  • @WillJagy: Thanks! I was just playing with Octave trying to get a 'nice' answer in a truly indolent way. With $V=\begin{bmatrix} 1 & 0 & 69696 \ 4 & 9 & 46134 \ 4& -5 & -46926\end{bmatrix}$, we have $V^T A V = \operatorname{diag} (705, -176, 4864432320)$. Beats doing the dishes. – copper.hat Jun 24 '13 at 03:42
  • @copper.hat, at this point, I would like to know how bad the eigenvalues are; if they are horrible, I was right to use Hermite. Wait, I have Pari, – Will Jagy Jun 24 '13 at 03:44
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    @WillJagy: You were right, the spectrum is $(-2.615428823365070...,0.598002265195447...,23.017426558169625...)$ – copper.hat Jun 24 '13 at 03:45

2 Answers2

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PREFACE: this is called Hermite reduction. It is, for example, Theorem 23 in The Arithmetic Theory of Quadratic Forms by Burton W. Jones, pages 56-59 primarily. On page 58 you clearly see an upper triangular matrix as the change of variables. The W stands for Wadsworth.

I get $$ x^2 + 4 y^2 + 16 z^2 + 20 y z + 8 z x + 8 x y = (x + 4 y + 4 z)^2 - 12 \left( y + \frac{z}{2} \right)^2 + 3 z^2, $$ so

$$ x^2 + 4 y^2 + 16 z^2 + 20 y z + 8 z x + 8 x y = (x + 4 y + 4 z)^2 - 3 \left( 2y + z \right)^2 + 3 z^2. $$

Note that this was very, very little work. Nowhere near finding eigenvectors. This is called Hermite reduction. The relevant matrix equation is

$$ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 4 & 2 & 0 \\ 4 & 1 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 3 \end{array} \right) \left( \begin{array}{ccc} 1 & 4 & 4 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & 4 & 4 \\ 4 & 4 & 10 \\ 4 & 10 & 16 \end{array} \right). $$

Either that, or they really want the inverses, as in $$ \left( \begin{array}{ccc} 2 & 0 & 0 \\ -4 & 1 & 0 \\ -4 & -1 & 2 \end{array} \right) \left( \begin{array}{ccc} 1 & 4 & 4 \\ 4 & 4 & 10 \\ 4 & 10 & 16 \end{array} \right) \left( \begin{array}{ccc} 2 & -4 & -4 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \end{array} \right) = \left( \begin{array}{ccc} 4 & 0 & 0 \\ 0 & -12 & 0 \\ 0 & 0 & 12 \end{array} \right). $$

Either way, less work than eigenvectors.

EDDDIITTTT: @copper.hat has confirmed that the eigenvalues are horrible, I get the characteristic polynomial as $$ \lambda^3 - 21 \lambda^2 - 48 \lambda + 36 $$ with three irrational real roots, so this is the Casus Irrededucibilis and so the eigenvectors are horrible. Which tells me they really want Hermite.

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Will Jagy
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  • Thanks! One question: I understand that the coefficients in front of the squared terms are on the diagonal of A, but what parts of the matrix do the other coefficients come from? I am clearly rather rusty with bilinear forms! – Kirk Fogg Jun 24 '13 at 02:58
  • @KirkFogg, you put a row vector $(x,y,z)$ on the left of your $A,$ then the column vector with the same three entries (transpose of the row), multiply out, I think you then divide by 2. You need to check some sources, see if this is what they actually want. – Will Jagy Jun 24 '13 at 03:04
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The trick is to use a nonzero diagonal entry to kill all off-diagonal entries below it or on its right, by simultaneous row and column reduction: \begin{align*} \underbrace{\begin{bmatrix} 1 & 0 & 0\\ -4 & 1 & 0\\ -4 & 0 & 1 \end{bmatrix}}_{P^T} \begin{bmatrix} 1 & 4 & 4\\ 4 & 4 & 10\\ 4 & 10 & 16 \end{bmatrix} \underbrace{\begin{bmatrix} 1 & -4 & -4\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}}_P &=\begin{bmatrix} 1 & 0 & 0\\ 0 & -12 & -6\\ 0 & -6 & 0 \end{bmatrix},\\ \underbrace{\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & -\tfrac12 & 1 \end{bmatrix}}_{Q^T} \begin{bmatrix} 1 & 0 & 0\\ 0 & -12 & -6\\ 0 & -6 & 0 \end{bmatrix} \underbrace{\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & -\tfrac12\\ 0 & 0 & 1 \end{bmatrix}}_Q &=\underbrace{\begin{bmatrix} 1 & 0 & 0\\ 0 & -12 & 0\\ 0 & 0 & 3 \end{bmatrix}}_D. \end{align*} Hence $Q^TP^TAPQ=D$, i.e. $A=S^TDS$, where $$ S=(PQ)^{-1} =\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & \tfrac12\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 4 & 4\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 4 & 4\\ 0 & 1 & \tfrac12\\ 0 & 0 & 1 \end{bmatrix}. $$ In other words, if you define $u=Sx$ and $v=Sy$, then $x^TAy=u^TDv=u_1v_1-12u_2v_2+3u_3v_3$.

The above process is an example of diagonalisation via congruence. (See here for another example.) This example is easy, because in the course of diagonalisation, we can always use a nonzero diagonal entries to eliminate the other entries on the same row or the same column. Sometimes, when the diagonal entry is zero, we need to 'borrow' numbers from other off-diagonal entries. See here for such an example.

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