Question:
Let \(P\) be a \(3 \times 3\) real symmetric positive definite matrix. Which of the following statements is/are TRUE?
Let \(P\) be a \(3 \times 3\) real symmetric positive definite matrix. Which of the following statements is/are TRUE?
Show Hint
Write \(v^{\top}(Q^{\top}PQ)v = (Qv)^{\top}P(Qv)\) and check what happens to \(Qv\) when \(Q\) is singular versus nonsingular.
Updated On: Aug 14, 2026
- \(Q^{\top}PQ\) is positive definite for all nonzero \(Q \in \mathbb{R}^{3\times3}\).
- \(Q^{\top}PQ\) is positive semidefinite for all \(Q \in \mathbb{R}^{3\times3}\).
- \(Q^{\top}PQ\) is positive definite if \(Q \in \mathbb{R}^{3\times3}\) is nonsingular.
- \(Q^{\top}PQ\) is not positive semidefinite for some \(Q \in \mathbb{R}^{3\times3}\).
Show Solution
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The Correct Option is B, C
Solution and Explanation
Step 1: Set up the link between the two matrices.
A real symmetric matrix \(P\) is positive definite when \(v^{\top}Pv > 0\) for every nonzero vector \(v \in \mathbb{R}^3\), and positive semidefinite when \(v^{\top}Pv \geq 0\) for every vector \(v\), where equality is allowed for some nonzero \(v\).
For any real \(3\times3\) matrix \(Q\) and any vector \(v\), we get
\[ v^{\top}(Q^{\top}PQ)v = (Qv)^{\top}P(Qv) \]
So the sign of the quadratic form of \(Q^{\top}PQ\) at \(v\) depends on where \(Qv\) lands and what \(P\) does there.
Step 2: Test option (A).
Pick a nonzero but singular \(Q\), for example \(Q = \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}\). This \(Q\) is nonzero, but it sends the nonzero vector \(v = (0,0,1)\) to \(0\). Then
\[ v^{\top}(Q^{\top}PQ)v = (Qv)^{\top}P(Qv) = 0^{\top}P0 = 0 \]
So \(Q^{\top}PQ\) is not positive definite for this choice of \(Q\), even though \(Q\) is nonzero. Option (A) claims this holds for every nonzero \(Q\), so one failing case is enough to make (A) FALSE.
Step 3: Test option (C).
Now suppose \(Q\) is nonsingular. Then \(Qv = 0\) can only happen when \(v = 0\), because a nonsingular matrix has trivial null space. So for every nonzero \(v\), the vector \(Qv\) is nonzero, and since \(P\) is positive definite,
\[ (Qv)^{\top}P(Qv) > 0 \]
This gives \(v^{\top}(Q^{\top}PQ)v > 0\) for every nonzero \(v\), which is exactly the definition of \(Q^{\top}PQ\) being positive definite. Option (C) is TRUE.
Step 4: Test options (B) and (D) together.
For any real \(Q\), singular or not, and for any vector \(v\), the vector \(Qv\) is just some vector in \(\mathbb{R}^3\) (possibly zero). Since \(P\) is positive definite, \(w^{\top}Pw \geq 0\) for every vector \(w\), including \(w = Qv\). So
\[ v^{\top}(Q^{\top}PQ)v = (Qv)^{\top}P(Qv) \geq 0 \quad \text{for every } v \]
This holds without any restriction on \(Q\), so \(Q^{\top}PQ\) is always positive semidefinite. That makes option (D), which says \(Q^{\top}PQ\) fails to be positive semidefinite for some \(Q\), FALSE, and it makes option (B) TRUE.
Final Answer:
\(Q^{\top}PQ\) is always positive semidefinite (option B), and it is positive definite precisely when \(Q\) is nonsingular (option C). Options (A) and (D) fail.
\[ \boxed{\text{(B) and (C)}} \]
A real symmetric matrix \(P\) is positive definite when \(v^{\top}Pv > 0\) for every nonzero vector \(v \in \mathbb{R}^3\), and positive semidefinite when \(v^{\top}Pv \geq 0\) for every vector \(v\), where equality is allowed for some nonzero \(v\).
For any real \(3\times3\) matrix \(Q\) and any vector \(v\), we get
\[ v^{\top}(Q^{\top}PQ)v = (Qv)^{\top}P(Qv) \]
So the sign of the quadratic form of \(Q^{\top}PQ\) at \(v\) depends on where \(Qv\) lands and what \(P\) does there.
Step 2: Test option (A).
Pick a nonzero but singular \(Q\), for example \(Q = \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}\). This \(Q\) is nonzero, but it sends the nonzero vector \(v = (0,0,1)\) to \(0\). Then
\[ v^{\top}(Q^{\top}PQ)v = (Qv)^{\top}P(Qv) = 0^{\top}P0 = 0 \]
So \(Q^{\top}PQ\) is not positive definite for this choice of \(Q\), even though \(Q\) is nonzero. Option (A) claims this holds for every nonzero \(Q\), so one failing case is enough to make (A) FALSE.
Step 3: Test option (C).
Now suppose \(Q\) is nonsingular. Then \(Qv = 0\) can only happen when \(v = 0\), because a nonsingular matrix has trivial null space. So for every nonzero \(v\), the vector \(Qv\) is nonzero, and since \(P\) is positive definite,
\[ (Qv)^{\top}P(Qv) > 0 \]
This gives \(v^{\top}(Q^{\top}PQ)v > 0\) for every nonzero \(v\), which is exactly the definition of \(Q^{\top}PQ\) being positive definite. Option (C) is TRUE.
Step 4: Test options (B) and (D) together.
For any real \(Q\), singular or not, and for any vector \(v\), the vector \(Qv\) is just some vector in \(\mathbb{R}^3\) (possibly zero). Since \(P\) is positive definite, \(w^{\top}Pw \geq 0\) for every vector \(w\), including \(w = Qv\). So
\[ v^{\top}(Q^{\top}PQ)v = (Qv)^{\top}P(Qv) \geq 0 \quad \text{for every } v \]
This holds without any restriction on \(Q\), so \(Q^{\top}PQ\) is always positive semidefinite. That makes option (D), which says \(Q^{\top}PQ\) fails to be positive semidefinite for some \(Q\), FALSE, and it makes option (B) TRUE.
Final Answer:
\(Q^{\top}PQ\) is always positive semidefinite (option B), and it is positive definite precisely when \(Q\) is nonsingular (option C). Options (A) and (D) fail.
\[ \boxed{\text{(B) and (C)}} \]
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