Question

Let \(P\) be a \(3 \times 3\) real symmetric positive definite matrix. Which of the following statements is/are TRUE?

• A. \(Q^{\top}PQ\) is positive definite for all nonzero \(Q \in \mathbb{R}^{3\times3}\).
• B. \(Q^{\top}PQ\) is positive semidefinite for all \(Q \in \mathbb{R}^{3\times3}\).
• C. \(Q^{\top}PQ\) is positive definite if \(Q \in \mathbb{R}^{3\times3}\) is nonsingular.
• D. \(Q^{\top}PQ\) is not positive semidefinite for some \(Q \in \mathbb{R}^{3\times3}\).

Hint:

Write \(v^{\top}(Q^{\top}PQ)v = (Qv)^{\top}P(Qv)\) and check what happens to \(Qv\) when \(Q\) is singular versus nonsingular.

Answer 1

The Correct Option is B, C

A clean way to answer this is to write the positive definite matrix $P$ as $P = R^{\top}R$ for some invertible $3\times3$ matrix $R$ (this is always possible for a symmetric positive definite matrix, for instance through a Cholesky decomposition). Then for any real matrix $Q$,

$$Q^{\top}PQ = Q^{\top}R^{\top}RQ = (RQ)^{\top}(RQ)$$

Any matrix of the form $M^{\top}M$ satisfies $v^{\top}M^{\top}Mv = \|Mv\|^2 \ge 0$ for every vector $v$, so it is automatically positive semidefinite, no matter what $M$ is. Setting $M = RQ$, this immediately tells us $Q^{\top}PQ$ is positive semidefinite for every choice of $Q$, with no extra condition needed.

1. (A): If $Q$ is nonzero but singular, then $RQ$ is also singular (multiplying by the invertible $R$ does not fix a rank deficiency), so there is a nonzero $v$ with $(RQ)v = 0$, giving $v^{\top}(Q^{\top}PQ)v = \|(RQ)v\|^2 = 0$. This breaks strict positivity, so (A) is false.
2. (B): True, shown directly above using $Q^{\top}PQ = (RQ)^{\top}(RQ)$, which holds for every $Q$ including the zero matrix and every singular matrix.
3. (C): If $Q$ is nonsingular, then $RQ$ is a product of two invertible matrices, so $RQ$ is invertible too, and $(RQ)v = 0$ only when $v = 0$. So $v^{\top}(Q^{\top}PQ)v = \|(RQ)v\|^2 > 0$ for every nonzero $v$, meaning $Q^{\top}PQ$ is positive definite. (C) is true.
4. (D): This says $Q^{\top}PQ$ fails to be positive semidefinite for some $Q$. Since we showed it equals $(RQ)^{\top}(RQ)$ for every $Q$, and that form is always positive semidefinite, (D) is false.

So the true statements are (B) and (C): writing $P$ as $R^{\top}R$ turns the whole problem into a statement about $M^{\top}M$ matrices, which are semidefinite by construction and become strictly definite exactly when $Q$ is invertible.

Let's summarize:

• $Q^{\top}PQ$ can be rewritten as $(RQ)^{\top}(RQ)$, a Gram-matrix form that is always positive semidefinite.
• Invertibility of $Q$ is exactly the extra condition that upgrades semidefinite to definite.

The correct options are (B) and (C).

\[ \boxed{\text{(B) and (C)}} \]