GO Classes DPP | GATE CS | Linear Algebra | Diagonalization & Invertibility

Answer 1

Represent the linear transformation $T:\mathbb{R}^3\to\mathbb{R}^3$ by a $3\times 3$ matrix $A$.

Then the given condition becomes

$A^3-A^2-A+I=0$.

Now factor the polynomial expression:

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

$=(A^2-I)(A-I)$

$=(A-I)^2(A+I)=0$.

This factorization is useful, but from
$(A-I)^2(A+I)=0$ we cannot say that $A-I=0$ or $A^2-I=0$ separately.

So, we check the two claims directly.

First, let $Av=0$ for some vector $v$.

Then from
$A^3-A^2-A+I=0$,

we get

$(A^3-A^2-A+I)v=0$.

Since $Av=0$, it follows that

$0-0-0+v=0$,

so $v=0$.

Hence, $\ker(A)=\{0\}$.

Therefore, $A$ is one-one, and since $A$ is a linear map from $\mathbb{R}^3$ to $\mathbb{R}^3$, it must be invertible.

So, $T$ is necessarily invertible.

Now we check diagonalizability.

Since
$(A-I)^2(A+I)=0$,

the minimal polynomial of $A$ divides $(x-1)^2(x+1)$.

Because this polynomial has a repeated factor $(x-1)^2$, the matrix need not be diagonalizable.

A counterexample is

$A=\begin{pmatrix}
1&1&0\\
0&1&0\\
0&0&-1
\end{pmatrix}$.

This matrix is invertible, but it is not diagonalizable because the block $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ is a Jordan block.

Also, this matrix satisfies

$(A-I)^2(A+I)=0$,

hence

$A^3-A^2-A+I=0$.

Therefore, $T$ is invertible, but not necessarily diagonalizable.

So, the correct option is $\boxed{\text{B}}$.

Comments

Note :  A Jordan block is a matrix where the same eigenvalue is repeated on the diagonal, and there are $1$'s just above the diagonal.

Example:
$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $

This matrix has eigenvalue $1$, but it is not diagonal.

So, when a matrix contains such a block of size bigger than $1$, it is not diagonalizable.

Answer 2

Answer: B

Answer 3

Repeated root ⇒ diagonalizability not guaranteed
No zero root ⇒ invertible

x^3−x^2−x+1=(x−1)^2(x+1) 

From the factorization: λ=1  (repeated),λ=−1

👉 Is 0 an eigenvalue?  No => T is always invertible 

👉Polynomial has repeated root (1 twice) => So minimal polynomial can be:

• (x−1)(x+1)→ diagonalizable ✅
• OR (x−1)^2(x+1)→ NOT diagonalizable ❌

Therefore, T is invertible, but not necessarily diagonalizable 

Option B Correct Answer