Question

Let U ≥ 0 be a real upper triangular n × n matrix with zeros on the diagonal. Show that (In + U) t ≤ t n(In + U + U2 +···+ Un−1) for all positive integers t. See Exercises 46 and 47.

Short answer

T's eigenvalues are1,1 and -1, and the associated eigenfunctions are $x_1^2+x_2^2,x_1{x}_2$, and$x_1^2-x_2^2$ .

For any eigenvalues, T can be diagonalized as $A.M=G.M$.

Step-by-step solution

given information

Given:

$T\left(q\left(x_1,x_2\right)\right)=q\left(x_1,2x_2\right)$

the determine the eigenvalues and eigenfunctions

The determine the eigenvalues and eigenfunctions of the matrix of linear transformation to find the eigenvalues and eigenfunctions of the transformation T. Consider the foundation$B=\left\{x_1^2,x_1{x}_2,x_2^2\right\}of{Q}_2$

, then

$$\begin{gathered} T\left(x_1^2\right)=x_2^2=0\left(x_1^2\right)+0\left(x_1{x}_2\right)+1\left(x_2^2\right) \\ T\left(x_1{x}_2\right)=x_2{x}_1=0\left(x_1^2\right)+1\left(x_1{x}_2\right)+0\left(x_2^2\right) \\ T\left(x_2^2\right)=x_1^2=1\left(x_1^2\right)+0\left(x_1{x}_2\right)+0\left(x_2^2\right) \end{gathered}$$

As a result, the linear transformation matrix with respect to B is given as

$[T{]}_B=\left[T\left(x\begin{array}{c}2\\ 1\end{array}\right)T\left(x_1{x}_2\right)T\left(x\begin{array}{c}2\\ 2\end{array}\right)\right]=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$

The eigen values of $[T{]}_B$are given as det,$\left({\left[T\right]}_B-\lambda I\right)=0$that is,

$$\begin{gathered} \left|\begin{array}{ccc}-\lambda & 0& 1\\ 0& 1-\lambda & 0\\ 1& 0& -\lambda \end{array}\right|=0 \\ \Rightarrow -\lambda \left(-\lambda +{\lambda }^2\right)+1(-1+\lambda )=0 \\ \Rightarrow {\lambda }^3-{\lambda }^2-\lambda +1=0 \\ \Rightarrow (\lambda -1{)}^2(\lambda +1)=0 \\ \Rightarrow \lambda =1,1,-1 \end{gathered}$$

For$\lambda =1$ , eigenvector of$[T{]}_B$ are given as$\left({\left[T\right]}_B-I\right)\overrightarrow{v}=\overrightarrow{0}$ , That is,

$\left[\begin{array}{ccc}-1& 0& 1\\ 0& 0& 0\\ 1& 0& -1\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\Rightarrow v_1-v_3=0,v_2\in R$

As a result, there are two linearly independent eigenvectors that correspond to$\lambda =1$ , are$\lambda =1$$\lambda =1are\left\{{(1,0,1)}^T,{(0,1,0)}^T\right\}$ as well as the corresponding$q_1\left(x_1,x_2\right)=x_1^2+x_2^2$

and$q_2\left(x_1,x_2\right)=x_1{x}_2$

Similarly for$\lambda =1$ , the eigenvector of $[T{]}_B$are given as$\left({\left[T\right]}_B+I\right)\overrightarrow{v}=\overrightarrow{0}$ , that is,

$\left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\Rightarrow v_1+v_3=0,v_2=0$

So, the eigenvector corresponding to $\lambda =-1$are $\left\{{(1,0,-1)}^T\right\}$and thus the corresponding eigenfunction is$q_3\left(x_1,x_2\right)=x_1^2-x_2^2$

 Step 3: The Eigen values and Eigen Function

As a result, the eigenvalues and eigenfunctions are written as

$$\begin{gathered} \lambda =1:x_1^2+x_2^2;T\left(x_1^2+x_2^2\right)=1\left(x_1^2+x_2^2\right) \\ \lambda =1:x_1{x}_2;T\left(x_1{x}_2\right)=1\left(x_1{x}_{2}2\right) \\ \lambda =-1:x_1^2-x_2^2;T\left(x_1^2-x_2^2\right)=-1\left(x_1^2-x_2^2\right) \end{gathered}$$

From the above discussion, it is clear that the algebraic and geometric multiplicity of both eigenvalues, namely 1 and -1 , are equal, i.e. 2 and 1. As a result, the linear transformation can be diagonalized.

T's eigenvalues are1,1 and -1, and the associated eigenfunctions are$x_1^2+x_2^2,x_1{x}_2$, and$x_1^2-x_2^2$.

For any eigenvalues, T can be diagonalized as$A.M=G.M$ .