Let T : R² - R³ be the linear transformation whose matrix with respect to standard basis {e₁, e₂, e₃) of R³ is \(\begin{bmatrix} 0 & 0 & 1 \\\ 0 & 1 & 0 \\\ 1 & 0 & 0 \end{bmatrix}\) then T

Concept:

Linear transformation: The Linear transformation T : V → W is that for any vectors v1 and v2 in V and scalars a and b of the underlying field, it satisfies following condition

T(av1 + bv2) = a T(v1) + b T(v2).

Calculations:

T = \(\begin{bmatrix} 0 & 0 & 1 \\\ 0 & 1 & 0 \\\ 1 & 0 & 0 \end{bmatrix}\)

| T - λI | = 0

⇒ \(\begin{bmatrix} -\lambda & 0 & 1 \\\ 0 & 1-\lambda & 0 \\\ 1 & 0 & -\lambda\end{bmatrix}=0\) 

⇒ λ²(1 - λ) + 1(λ - 1) = 0

⇒ (λ - 1)(1 - λ²) = 0

⇒ λ = 1, 1, -1.

⇒ T has only zero null space.

| T - I | = 0

⇒ \(\begin{bmatrix} -1 & 0 & 1 \\\ 0 & 0 & 0 \\\ 1 & 0 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) 

⇒ - x + z = 0

Then the eigen vector [0, 1, 0] & [1, 0, 1]

| T+ I | = 0

⇒ \(\begin{bmatrix} 1 & 0 & 1 \\\ 0 & 2 & 0 \\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)

⇒ x + z = 0 and y = 0

⇒ Then the eigen vector [1, 0, -1].

Clearly all are independent.

Hence it spans R³.

Hence, the correct answer is option 3)