GATE 2026 EE Q.35: solution and answer

Official answer: 0Source: Official answer key, GATE 2026 EE (IIT Guwahati)

1. Skew-symmetric means $A^{\mathrm{T}}=-A$. So every diagonal entry is zero ($a_{ii}=-a_{ii}$ gives $a_{ii}=0$), and $a_{ji}=-a_{ij}$ for $i\ne j$.

2. Let $q={\mathbf{x}}^{\mathrm{T}}A\mathbf{x}$. It is a $1\times 1$ matrix, i.e. a scalar, and a scalar equals its own transpose, so $q=q^{\mathrm{T}}$.

3. The transpose of a product reverses the order, and $({\mathbf{x}}^{\mathrm{T}}{)}^{\mathrm{T}}=\mathbf{x}$: $$q^{\mathrm{T}}={\left({\mathbf{x}}^{\mathrm{T}}A\mathbf{x}\right)}^{\mathrm{T}}={\mathbf{x}}^{\mathrm{T}}{A}^{\mathrm{T}}\mathbf{x}={\mathbf{x}}^{\mathrm{T}}(-A)\mathbf{x}=-q$$

4. So $q=-q$, which gives $2q=0$ and $q=0$.

5. Check by expanding: $q={\sum }_i{\sum }_j{a}_{ij}{x}_i{x}_j$. The diagonal terms vanish because $a_{ii}=0$. Each off-diagonal pair cancels because $a_{ij}{x}_i{x}_j+a_{ji}{x}_j{x}_i=(a_{ij}+a_{ji})x_i{x}_j=(a_{ij}-a_{ij})x_i{x}_j=0$.

6. The condition ${\mathbf{x}}^{\mathrm{T}}\mathbf{x}=1$ plays no part. The quadratic form of a real skew-symmetric matrix is zero for every real $\mathbf{x}$, whatever its length. Hence ${\mathbf{x}}^{\mathrm{T}}A\mathbf{x}=0$.