Question

Let $P_{7} (x)$ be the real vector space of polynomials, in the variable $x$ with real
coefficients and having degree at most 7, together with the zero polynomial.
Let $T : P_{7} (x) \to P_{7} (x)$ be the linear transformation defined by
$T (f (x)) = f (x) + \frac{d f (x)}{d x} .$
Then, which one of the following is TRUE?

𝑇 is not a surjective linear transformation
There exists 𝑘 ∈ ℕ such that $T^{k}$ is the zero linear transformation
1 and 2 are the eigenvalues of 𝑇
There exists 𝑟 ∈ ℕ such that ${(T - I)}^{r}$ is the zero linear transformation, where 𝐼 is the identity map on $P_{7} (x)$

Answer: There exists 𝑟 ∈ ℕ such that ${(T - I)}^{r}$ is the zero linear transformation, where 𝐼 is the identity map on $P_{7} (x)$

Solution :-

We have $P_{7} (x)$ be the real vector space of polynomials. Basis for this vector space is the set $β$ and

β = {1, x, x^{2}, x^{3}, x^{4}, x^{5}, x^{6}, x^{7}}

Now, we have the linear transformation “T” defined by

T (f (x)) = f (x) + \frac{d f (x)}{d x} .

Let’s find ${[T]}_{β}$ matrix transformation with respect to basis $β$ .

\begin{array}{rclcl} So, T (1) & = & 1 + \frac{d (1)}{d x} & = & 1 \\ T (x) & = & x + \frac{d (x)}{d x} & = & x + 1 \\ T (x^{2}) & = & x^{2} + \frac{d (x^{2})}{d x} & = & x^{2} + 2 x \\ T (x^{3}) & = & x^{3} + \frac{d (x^{3})}{d x} & = & x^{3} + 3 x^{2} \\ T (x^{4}) & = & x^{4} + \frac{d (x^{4})}{d x} & = & x^{4} + 4 x^{3} \\ T (x^{5}) & = & x^{5} + \frac{d (x^{5})}{d x} & = & x^{5} + 5 x^{4} \\ T (x^{6}) & = & x^{6} + \frac{d (x^{6})}{d x} & = & x^{6} + 6 x^{5} \\ T (x^{7}) & = & x^{7} + \frac{d (x^{7})}{d x} & = & x^{7} + 7 x^{6} . \end{array}

∴ {[T]}_{β} = {[\begin{array}{cccccccccc} 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}]}_{8 \times 8}

Here, ${[T]}_{β}$ is a upper triangular matrix with all diagonal entries 1. Hence, only 1 is the eigenvalue of the linear transformation “T”.

Option (1)

T is surjective, because zero is not an eigenvalue of T, so T is invertiable. Hence, T is surjective linear transformation.

So, option (1) is incorrect.

Option (2)

Only 1 is the eigenvalue of T. So, T is not nilpotent. Hence, There does not exists 𝑘 ∈ ℕ such that $T^{k}$ is the zero linear transformation.

So, option (2) is incorrect.

Option (3)

2 is not an eigenvalue of T because only 1 is the eigenvalue of T.

So, option (3) is incorrect.

Option (4)

Only 1 is the eigenvalue of T. So, eigenvalues for $T - I$ is 1 - 1 = 0 only. So, yes $T - I$ is nilpotent and hence, there exists 𝑟 ∈ ℕ such that ${(T - I)}^{r}$ is the zero linear transformation.

So, the correct answer is option-4. $\to$ There exists 𝑟 ∈ ℕ such that ${(T - I)}^{r}$ is the zero linear transformation, where 𝐼 is the identity map on $P_{7} (x)$

#Linear Algebra #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs

Question

Let P7(x) be the real vector space of polynomials, in the variable x with real coefficients and having degree at most 7, together with the zero polynomial. Let T:P7(x)→P7(x) be the linear transformation defined by T⁢(f⁡(x))=f⁡(x)+d⁢f⁡(x)d⁢x. Then, which one of the following is TRUE?

Solution :-