Concluding on a subspace on a diagonalaized quadratic form

Question

We have a quadratic form in the reals.

$$ q(x,y,z) = 2zx + 4yz - 2xy $$

($q$ is in the standard form of $V = R^3$)

We need to find: $(1)$ A base for $V$ in which $q$ is diagonalized. $(2)$ Find a subspace $W$ of $V$ with maximal dimension such that $q(w) \geq 0, \forall w \in W$

I am stuck in undertanding $(2)$.

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$(1)$ Using elemtaric congruence we get:

Therefore our change of base matrix is:

Now we can show the base in which $q$ is diagonalaized:

NOW THE PROBLEM STARTS

They choose the subspace: $$ W = Sp \{(-1,0.5,0),(2,1,1)\} $$

Now they say: $$ \forall w \in W: [w]_{B} = (0, \lambda_1, \lambda_2) $$

Therefore: $$ q(w) = -1 \cdot 0^2 + \lambda_1^2 + 4 \lambda_2^2 \geq 0 $$

I dont understand

1. How did they know to take that subspace?

2. How $[w]_B = (0, \lambda_1, \lambda_2)$

How did they know?

Thank you.

Answer 1

Take a look at your diagonalization: in the new basis $B$, the quadratic form looks like $\langle -1,1,4\rangle$. You have one negative coefficient, and two positive ones. So if you want your quadratic form to be positive on a subspace, you should take the subspace spanned by the basis vectors corresponding to the positive coefficients.

In your case, these are the second and third basis vectors, so $(-1,0.5,0)$ and $(2,1,1)$. And of course, if you take a vector in this subspace and write it in the new basis $B$, by construction it has a first component equal to $0$.

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EDIT: I see the new answer states that the diagonalization itself is incorrect, which is indeed the case. I did not try to check that before answering. This being said, I leave my answer since it does provide the method to find the correct subspace (once you get the right diagonanization of course).

Answer 2

Your solution to the first part is incorrect. You cannot diagonalize a quadratic form by row-reducing the matrix. Two procedures are involved: "completing the square" and "forcing a non-zero diagonal." Your initial matrix has all zero terms on the diagonal so the first transformation is $$z=X+Z,y=Y,x=X$$ which gives $$2X^2+2ZX+2XY+4YZ$$ $$=2(X^2+XY+XZ)+4YZ$$ $$2((X+\frac{1}{2}Y+\frac{1}{2}Z)^2-\frac{1}{4}Y^2-\frac{1}{4}Z^2-\frac{1}{2}YZ)+4YZ$$ $$=2(X+\frac{1}{2}Y+\frac{1}{2}Z)^2-\frac{1}{2}Y^2-\frac{1}{2}Z^2+3YZ$$ The next transformation is $$X'=X+\frac{1}{2}Y+\frac{1}{2}Z,Y'=Y,Z'=Zi.e.$$ $$X=X'-\frac{1}{2}Y'-\frac{1}{2}Z',Y=Y',Z=Z'$$ so the quadratic form becomes $$2X'^2-\frac{1}{2}Y'^2-\frac{1}{2}Z'^2+3Y'Z'$$ $$=2X'^2-\frac{1}{2}(Y'^2-6Y'Z')-\frac{1}{2}Z'^2$$ $$=2X'^2-\frac{1}{2}((Y'-3Z')^2-9Z'^2)-\frac{1}{2}Z'^2$$ $$=2X'^2-\frac{1}{2}(Y'-3Z')^2+4Z'^2$$ Your final transformation is $$X''=X',Y''=Y'-3Z',Z''=Z',i.e.$$ $$X'=X'',Y'=Y''+3Z'',Z'=Z''$$ giving the quadratic form $$2X''^2-\frac{1}{2}Y''^2+4Z''^2.$$ By writing each of the transformations in matrix form and multiplying the matrices, you should have no problem finding a matrix $M$ such that $$\begin{bmatrix}x\\y\\z\end{bmatrix}=M\begin{bmatrix}X''\\Y''\\Z''\end{bmatrix}$$