Gate DA 2025 | Question - 13 | Linear Algebra

We are given $A \in \mathbb{R}^{n \times n}$, with each row summing to 1.

• Define $B = A^3 - 2A^2 + A$

• Asked: What can be said about the solutions to $Bx = 0$?

Step-by-step approach:

Step 1: Let’s define the all-ones vector:

$\mathbf{1}=\left[\begin{array}{c}1\\ 1\\ \\ \mathrm{⋮}\\ 1\end{array}\right]\in {\mathrm{ℝ}}^n$

Given that the sum of each row of A is 1, this implies:

$A1=1$

So, 1 is a right eigenvector of A corresponding to the eigenvalue 1.

Step 2: Analyse $B = A^3 - 2A^2 + A$

We simplify $B$ using algebra:

$B=A^3-2A^2+A=A(A^2-2A+I)$

Note that:

$A^2-2A+I=(A-I{)}^2$

So:

$B=A(A-I{)}^2$

Step 3: Apply $B$ to $\mathbf{1}$:

Since $A\mathbf{1} = \mathbf{1} \Rightarrow (A - I)\mathbf{1} = 0$

Then:

$(A-I{)}^2\mathbf{1}=0\Rightarrow B\mathbf{1}=A\cdot 0=0$

So, $\mathbf{1} \in \ker(B)$

That means:

$Bx=0$, has at least one non-trivial solution

✅ Final Conclusion:

• Since 1 ≠ 0 it satisfies $Bx = 0$, the null space of $B$ is non-trivial.

• So the system $Bx = 0$ has infinitely many solutions (because it is homogeneous and has a non-trivial solution ⇒ solution space is a subspace of dimension ≥ 1).

Correct answer (C) The equation $Bx = 0$ has infinitely many solutions