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Let $$A = \pmatrix{s&s&0\\ s&s&0\\ 0&0&t}$$ where $s, t \in \mathbb R$ are parameters, and let $QA : R_ 3 \to R$ be the corresponding quadratic form. Determine for which values of $s$ and $t$, $A$ and $QA$ are negative semidefinite and indefinite.

So far I calculated the product of X transposed AX and gave me that below:

(https://i.stack.imgur.com/9Xd9d.jpg)

I then said that this:

(https://i.stack.imgur.com/JmZUe.jpg)

Is that ok? If so how would I go about saying what values are indefinite. Many thanks for you help.

3 Answers 3

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Your first step looks okay, you write the quadratic form as $$s(x+y)^2 + tz^2.$$

However, now you need to look for $s$ and $t$ such that for all $x, y, z$ the quadratic form is nonpositive ($\leq 0$). This must be a a condition on $s$ and $t$ that is independent of $x, y, z$. However, since $(x+y)^2>0$ for all $x,y$ and $z^2>0$ for all $z$, it is easy: if $s\leq 0$ and $t\leq 0$ you are sure that $$s(x+y)^2 + tz^2\leq 0 \quad \forall x,y,z.$$

Now what happens if $s$ and $t$ have opposite sign?

  • 0
    But couldn't you have more values of s and t included as if x,y,z=1 then you have 4s+t is less than or equal to zero so so s=0.25>0 and t=-1<0 – 2017-01-10
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    The question about definiteness of quadratic forms is a different one though. It doesn't ask whether the quadratic form is negative or positive for particular values of $x, y, z$ (like $x,y,z=1$ as you suggested). Instead it asks whether a QF is negative or positive for **all values** of $x,y,z$. If a QF can satisfy this, it is called definite. It's like asking if a parabola is always above the axis or always below. – 2017-01-10
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If $\mathrm A$ is negative semidefinite then $-\mathrm A$ is positive semidefinite. Using Sylvester's criterion, we obtain $$s, t \leq 0$$

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The char. polynomial of $A$ is given by

$$\lambda(t-\lambda)(\lambda-2s),$$

hence $A$ has the eigenvalues

$t,2s$ and $0.$

Your turn !