IISc Bangalore: Entrance for Integrated PhD Previous Year Q. 1

Question

Let $n > 2$ and for $1 \leq j \leq n$, define $\mathbf{a}_j$ to be the vector in $\mathbb{R}^n$ with $j^\text{th}$ entry 0 and the remaining entries 1. Then, $\{\mathbf{a}_1, \dots, \mathbf{a}_n\}$

a.  is a linearly dependent set.
b.  is an orthogonal system.
c.  spans a proper subspace of $\mathbb{R}^n$.
d.  is a basis for $\mathbb{R}^n$.

Answer 1

Each vector differs in exactly one coordinate position, so the set is linearly independent. Since thetre are n  such independent vectors in 𝑅n they form a basis.

Comments

LOVELY!!! Well done 👍

Answer 2

 

The vectors are $\mathbf{v}_j = (1, \dots, 1, \underbrace{0}_{j\text{-th}}, 1, \dots, 1)$.

To form the zero vector $\mathbf{0}$ from a linear combination:
\[
c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_n \mathbf{v}_n = \mathbf{0}
\]

Example:  
for linear independence if $n=3$:  
$\mathbf{v}_1 = (0,1,1)$  
$\mathbf{v}_2 = (1,0,1)$  
$\mathbf{v}_3 = (1,1,0)$  
Then: $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0}$

The matrix has 0s on the diagonal and 1s elsewhere.''
The determinant formula for this matrix is
\[\det(M) = (n - 1) \cdot (-1)^{\,n-1}\]
The problem states $n \geq 2$
If $n \geq 2$, then $(n - 1) \neq 0$
Therefore, $\det(M) \neq 0$
Non-zero determinant $\implies$ the columns are linearly independent $\implies$ they form a basis of $\mathbb{R}^n$.

 

Option D = Correct