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Considering the matrix

$$A=\begin{bmatrix}{2}&{1}&{0}\\{1}&{0}&{-1}\\{0}&{-1}&{-2}\end{bmatrix}\in M_3(R).$$ And $\phi:R^3 \times R^3\longrightarrow R$ the bilinear form $\phi(v,w)=vAw^t$

Find an orthogonal basis for $\phi.$

Can we find a basis $B$ such that $M(\phi,B)=\begin{bmatrix}{-1}&{1}&{0}\\{1}&{-2}&{1}\\{0}&{1}&{-1}\end{bmatrix}$?

I have already calculated an orthogonal basis, which would be $\langle (1,0,0),(-1/2,1,0),(1,-2,1)\rangle$

However, I am stuck at the second part of the exercise. How can I know if that $B$ basis exists such that $M(\phi,B)=\begin{bmatrix}{-1}&{1}&{0}\\{1}&{-2}&{1}\\{0}&{1}&{-1}\end{bmatrix}$?

And if it exists, how can I find $B$?

Thank you.

J doeoeo
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    Are you sure they are asking for an orthogonal basis for a bilinear form? It makes no sense, bases are for (sub-)spaces, not for bilinear forms. – Marc van Leeuwen Jun 27 '16 at 04:02

1 Answers1

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the correct change for a quadratic form matrix is $P^T H P$ with nonsingular $P.$

It is not possible to solve $P^T AP = B$ because $A$ has one positive eigenvalue, one negative and one zero. However, $B$ has two negative eigenvalues and one zero.

Here is diagonal $P^T A P = D_A$ $$ \left( \begin{array}{rrr} 1 & 0 & 0 \\ -\frac{1}{2} & 1 & 0 \\ 1 & -2 & 1 \end{array} \right) \left( \begin{array}{rrr} 2 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & -2 \end{array} \right) \left( \begin{array}{rrr} 1 & -\frac{1}{2} & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 \end{array} \right) $$

Here is diagonal $Q^T B Q = D_B$ $$ \left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array} \right) \left( \begin{array}{rrr} -1 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right) $$ In $D_A$ and $D_B,$ the nonzero diagonal entries are probably not eigenvalues. In the first edition of Horn and Johnson, Matrix Analysis, on page 220, for real symmetric matrices, when $S$ is nonsingular and $B = SAS^T,$ then $A$ and $B$ are called congruent. They do use the word "diagonalize," but they are very careful, in the table on page 229, to have a column for "That which is to be diagonal" with different expression depending on conditions.

Sylvester's_law_of_inertia

It is not necessary to actually find the eigenvalues to confirm this. It is also not necessary to find the characteristic polynomial. The best way to proceed is to solve $P^T AP = D$ diagonal, where I got the diagonal entries as $(2, -1/2,0).$ For $B$ I got $(-1,-1,0).$

reference for linear algebra books that teach reverse Hermite method for symmetric matrices

Addendum: there seems to be some confusion on Sylvester. Here is an example I made up. The original symmetric matrix $C,$ all entries $\pm 1,$ has four positive entries on its diagonal. However, this disagrees with the counts in $D_C,$ which has three positive but one negative entry on the diagonal, and this is what is important for Inertia. Here is diagonal $R^T C R = D_C$ $$ \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ -1 & 1 & 0 & 0 \\ -1 & -1 & 2 & 0 \\ -1 & -1 & -1 & 3 \end{array} \right) \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ -1 & 1 & -1 & -1 \\ -1 & -1 & 1 & -1 \\ -1 & -1 & -1 & 1 \end{array} \right) \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \end{array} \right) = \left( \begin{array}{rrrr} -8 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 12 & 0 \\ 0 & 0 & 0 & 24 \end{array} \right) $$ Again, the eigenvalues of $C$ are not $-8,4,12,24.$ The eigenvalues of $C$ are $-2,2,2,2.$ In fact, the columns of $R$ are eigenvectors and are perpendicular to each other, it is just that they vary in length, so $R$ is not actually orthogonal. Here is $CR:$

$$ \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ -1 & 1 & -1 & -1 \\ -1 & -1 & 1 & -1 \\ -1 & -1 & -1 & 1 \end{array} \right) \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \end{array} \right) = \left( \begin{array}{rrrr} -2 & -2 & -2 & -2 \\ -2 & 2 & -2 & -2 \\ -2 & 0 & 4 & -2 \\ -2 & 0 & 0 & 6 \end{array} \right) $$

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Will Jagy
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  • From Sylvester's Law of Inertia, do we just have to look at the diagonal and look at how $A$ has one positive entry and one negative entry while $B$ has three negative entries and then conclude that $P^TAP=B$ has no solution? – Noble Mushtak Jun 27 '16 at 13:22
  • @NobleMushtak no, that is not enough – Will Jagy Jun 27 '16 at 16:41
  • But according to the Wiki article, if $D=P^TAP$, then $D$ has the same number of positive elements in the diagonal as $A$ which is clearly not true for $A$ and $B$, so how come noticing this does not automatically show us that $P^TAP=B$ has no solutions? I am sorry if this seems rude; I am just trying to understand this law since I've never seen it before. – Noble Mushtak Jun 27 '16 at 16:55
  • @NobleMushtak that cannot be true. Here $B$ has three negative diagonal entries, but there are actually two negatives and a single zero. Please paste the exact quote in a comment. – Will Jagy Jun 27 '16 at 16:59
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    "Namely, if $A$ is the symmetric matrix that defines the quadratic form, and $S$ is any invertible matrix such that $D = SAS^T$ is diagonal, then the number of negative elements in the diagonal of $D$ is always the same, for all such $S$; and the same goes for the number of positive elements." EDIT: Oh, wait, I did not notice the words "is diagonal." Sorry about that! – Noble Mushtak Jun 27 '16 at 17:29
  • @NobleMushtak I just put in a 4 by 4 example, easy enough to do the calculations by hand. Recommend you do so, both reading things online and having a computer do calculations are rather passive and do not cement ideas in our heads. – Will Jagy Jun 27 '16 at 17:35
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    Yes, I will definitely do this. Thank you for all of your help! – Noble Mushtak Jun 27 '16 at 17:38