235 views

Let $A$ be the $(n+1) \times(n+1)$ matrix given below, where $n \geq 1$. For $i \leq n$, the $i$-th row of $A$ has every entry equal to $2i-1$ and the last row, i.e., the $(n+1)$-th row of $A$ has every entry equal to $-n^2$.

$$\left[\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ 3 & 3 & \cdots & 3 \\ \vdots & \vdots & \vdots & \vdots \\ 2 n-1 & 2 n-1 & \cdots & 2 n-1 \\ -n^2 & -n^2 & \cdots & -n^2 \end{array}\right]$$

Which of the following statements is $\text{TRUE}$ for all $n \geq 1?$

1. $A$ has rank $n$
2. $A^2$ has rank $1$
3. All the eigenvalues of $A$ are distinct
4. All the eigenvalues of $A$ are $0$
5. None of the above

Every column of $A$ is identical $\implies A$ has only one linearly independent column $\implies$ rank of $A = 1$.

$\therefore$ option (A) is not True.

\begin{align}A \times A = A^2 \implies A_{ij} &= row_i(A) . column_j(A) = [(x_i) \text{ } (x_i) … (x_i)] \times [1 \text{ } 3 … (2n-1) \text{ } (-n^2)] ^ T \\ &= (x_i)*1 + (x_i)*3 + … + (x_i)*(2n-1)+(x_i)*(-n^2) \\ &=(x_i)(1 + 3 + … + (2n-1)+(-n^2)) \\ &=(x_i)(\frac{n}{2} * (2*1 + (n-1)*2) - n^2) \\ &=(x_i)(n^2-n^2) \\ &= 0 \end{align}

where $x_i$ is an element occupying $row_i(A)$ as every element in a $row$ are same.

$\implies$ rank of $A^2 = 0$

$\therefore$ option (B) is not True.

Since, A does not have a full rank, $det(A) = 0 =$ product of eigenvalues $\implies$ at least one eigenvalue has to be zero.

Consider, $(A-0 I)x = Ax = 0$

We know rank of $A=1$ and nullity of $A=n \implies$ There are $n$ linearly independent solutions of $Ax=0$.

$\therefore A$ has $n$ linearly independent eigenvectors corresponding to $0$.

$\implies$ Algebraic Multiplicity of $0$ must be equal to $n$ or $n+1$.

$\therefore A$ has atleast $n$ eigenvalues are $0$ ie atleast $n$ eigenvalues are repeating.

$\therefore$ option (C) is not True.

Trace of $A = 1 + 3 + … + (2n-1) + (-n^2) = 0$.

Since, we know $n$ eigenvalues are $0$, $(n+1)^{th}$ eigenvalue must also be $0$.

$\therefore$ option (D) is True.

edited
yes
You've taken wrong matrix, check that.