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.

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 $0,0$.

Click 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.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me. Visit my blog every day and share the solutions with friends. Follow the blog by email. Thank you.**

**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.****here**for more problems.

**Share to your groups:**

**FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.**

## No comments:

## Post a Comment