### CSIR JUNE 2011 PART C QUESTION 74 SOLUTION (Let $T: \Bbb R^n \to \Bbb R^n$ be a linear transformation such that $T^2 = \lambda T$)

Let $T: \Bbb R^n \to \Bbb R^n$ be a linear transformation such that $T^2 = \lambda T$ for some $\lambda \in \Bbb R$. Then
1)$||T(x)|| = |\lambda| ||x||$ for all $x \in \Bbb R^n$,

2)If $||T(x)|| = ||x||$ for some non-zero vector $x \in \Bbb R^n$ then $\lambda = \pm 1$,
3) $T = \lambda I$ where $I$ is the $n \times n$ identity matrix,
4) If $||T(x)|| > ||x||$ for some non-zero vector $x \in \Bbb R^n$ then $T$ is necessarily singular.
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Solution:
Consider the $2 \times 2$ nilpotent matrix $$\begin{bmatrix}0&1\\0&0\end{bmatrix}.$$ We have $A^2 = 0$ and hence this matrix satisfies the given condition $A^2 = \lambda T$ with $\lambda = 0$. Since $A$ is upper diagonal its eigen values are its diagonal entries. Hence the eigen values of $A$ are $0,0$.

option 1: (False) We use the above matrix $A$ as a counter example. Let $x = \begin{bmatrix}0\\1\end{bmatrix}$ then $Ax = y$ where $y = \begin{bmatrix}1\\0\end{bmatrix}$. Now, $$||Ax|| = ||y|| = \sqrt{1^2 + 0^2} = 1.$$ But $|\lambda| ||x|| = 0$. This shows that, for our matrix $A$, $$||Ax|| \ne |\lambda| ||x||.$$
option 3: (False) We again use the above matrix $A$. We ave seen that this matrix satisfies $A^2 = \lambda A$ with $\lambda =0$. We observe that $A$ is a non-zero matrix whereas $\lambda I = 0I$ is the zero matrix. Hence $A \ne \lambda I$ in this case.
option 4: (False) Consider the matrix $$A = \begin{bmatrix}2&0\\ 0&2\end{bmatrix}.$$ Let $x = \begin{bmatrix}0 \\ 1\end{bmatrix}$ then $$||Ax|| = ||\begin{bmatrix}0\\ 2\end{bmatrix}|| = \sqrt{0^2+2^2} = 2>0$$ but $A$ is clearly invertible (non-singular).
option 2: (False) Consider the matrix $$\begin{bmatrix}\sqrt 2 & 0 \\ 0 & 0\end{bmatrix},$$ then $A^2 = \sqrt 2 A$. Hence this matrix satisfies $A^2 = \lambda A$ with $\lambda = \sqrt 2$. Let $x = \begin{bmatrix}1\\ 1\end{bmatrix}$ then $$||x|| = \sqrt{1^2+1^2} = \sqrt 2 = ||\begin{bmatrix}\sqrt 2 \\ 0\end{bmatrix}|| = ||Ax||.$$ But $\lambda = \sqrt 2 \ne \pm 1$.
All the options are false. This question was wrong and the full mark was given to everybody.

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