GATE 2026 EE Q.49: solution and answer

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

1. Stack the $p$ vectors as the columns of an $n\times p$ matrix $A$. The vectors are linearly independent exactly when $A\mathbf{x}=0$ has only the trivial solution, that is, when $\text{rank}(A)=p$.

2. The rank is bounded by both dimensions: $\text{rank}(A)\le \min (n,p)$. Every vector lies in ${\mathbb{R}}^n$, which has dimension $n$, so no more than $n$ of them can be independent.

3. (A): if $p>n$, then $\text{rank}(A)\le n<p$. So $A\mathbf{x}=0$ has a nontrivial solution (a homogeneous system with more unknowns, $p$, than equations, $n$), and the vectors are dependent whatever they are. (A) is ALWAYS true.

4. (B): by step 3, $p>n$ vectors in ${\mathbb{R}}^n$ are never independent, so (B) is false for every choice of vectors. For example, in ${\mathbb{R}}^2$ any three vectors are dependent.

5. (C): take $p=n=2$ with $\left[\begin{array}{c}1\\ 0\end{array}\right]$ and $\left[\begin{array}{c}1\\ 1\end{array}\right]$. Their dot product is $1\cdot 1+0\cdot 1=1\ne 0$, so they are not orthogonal, even though they are independent. Having $p=n$ does not force orthogonality, or even independence: $\left[\begin{array}{c}1\\ 0\end{array}\right]$ and $\left[\begin{array}{c}2\\ 0\end{array}\right]$ are dependent.

6. (D): take $p=2<n=3$ with $\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$ and $\left[\begin{array}{c}2\\ 4\\ 6\end{array}\right]$. The second vector is twice the first, so they are dependent. Having $p<n$ only allows independence; it does not guarantee it.

7. The only statement that holds for every choice of vectors is (A).