91Ó°ÊÓ

$\mathrm{T}(\mathrm{q}\left({\mathrm{x}}_1{\mathrm{x}}_2\right))=\mathrm{q}({\mathrm{x}}_1,2{\mathrm{x}}_2)$

We shall determine the eigenvalues and eigenfunctions of the matrix of linear transformation to find the eigenvalues and eigenfunctions of the transformation . Consider the ${\mathrm{Q}}_2$basis $\mathrm{B}=\left\{x_1^2,x_1{x}_2,x_2^2\right\}$ Then,

$$\begin{gathered} \mathrm{T}\left({\mathrm{x}}_1^2\right)={\mathrm{x}}_1^2=1\left({\mathrm{x}}_1^2\right)+0\left({\mathrm{x}}_1{\mathrm{x}}_2\right)+0\left({\mathrm{x}}_2^2\right) \\ \mathrm{T}\left({\mathrm{x}}_1{\mathrm{x}}_2\right)=2{\mathrm{x}}_1{\mathrm{x}}_2=0\left({\mathrm{x}}_1^2\right)+0\left({\mathrm{x}}_1{\mathrm{x}}_2\right)+0\left({\mathrm{x}}_2^2\right) \\ \mathrm{T}\left({\mathrm{x}}_2^2\right)=4{\mathrm{x}}_2^2=1\left({\mathrm{x}}_1^2\right)+0\left({\mathrm{x}}_1{\mathrm{x}}_2\right)+4\left({\mathrm{x}}_2^2\right) \end{gathered}$$

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

${\left[T\right]}_B=\left[T\left(x_1^2\right)T\left(x_1{x}_2\right)T\left(x_1^2\right)\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 4\end{array}\right]$

The eigenvalues oflocalid="1660213837126" ${\left[\mathrm{T}\right]}_B$are given as det $\left({\left[\mathrm{T}\right]}_{\mathrm{B}}-\mathrm{λ\pm ô}\right)=0$, that is,

$\left|\begin{array}{ccc}1-\mathrm{λ}& 0& 0\\ 0& 2-\mathrm{λ}& 0\\ 0& 0& 4-\mathrm{λ}\end{array}\right|=0$

$$\begin{gathered} ⇒\left(1-\mathrm{λ}\right)\left(2-\mathrm{λ}\right)\left(4-\mathrm{λ}\right)=0 \\ ⇒\mathrm{λ}=1,2,4 \end{gathered}$$

For $\mathrm{λ}=1$, eigenvector of${\left[\mathrm{T}\right]}_{\mathrm{B}}$are given as $\left({\left[\mathrm{T}\right]}_{\mathrm{B}}-\mathrm{l}\right)\stackrel{→}{\mathrm{v}}=\stackrel{→}{0}$, that as,

$\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 3\end{array}\right]\left[\begin{array}{c}{\mathrm{v}}_1\\ {\mathrm{v}}_2\\ {\mathrm{v}}_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]⇒{\mathrm{v}}_2={\mathrm{v}}_3=0{\mathrm{v}}_1∈\mathrm{R}$

So, for $\mathrm{λ}=1$, the linearly independent eigenvectors are $\left\{{\left(1,0,0\right)}^{\mathrm{T}}\right\}$, and the corresponding eigenfunction is ${\mathrm{q}}_1\left({\mathrm{x}}_1{\mathrm{x}}_2\right)={\mathrm{x}}_1^2$.

Now $\mathrm{λ}=2$, the eigenvector of ${\left[\mathrm{T}\right]}_{\mathrm{B}}$are given as $({\left[\mathrm{T}\right]}_{\mathrm{B}}-2\mathrm{l})\stackrel{→}{\mathrm{V}}=\stackrel{→}{0}$, that is,

$\left[\begin{array}{ccc}-1& 0& 0\\ 0& 0& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{c}{\mathrm{v}}_1\\ {\mathrm{v}}_2\\ {\mathrm{v}}_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]⇒{\mathrm{v}}_1={\mathrm{v}}_3=0{\mathrm{v}}_2∈\mathrm{R}$

So, for $\mathrm{λ}=2$, the linearly independent eigenvectors are $\left\{{\left(0,1,0\right)}^{\mathrm{T}}\right\}$, and the corresponding eigenfunction is ${\mathrm{q}}_2\left({\mathrm{x}}_1{\mathrm{x}}_2\right)={\mathrm{x}}_1{\mathrm{x}}_2$

Now $\mathrm{λ}=4$, the eigenvector of ${\left[\mathrm{T}\right]}_{\mathrm{B}}$are given as $\left({\left[\mathrm{T}\right]}_{\mathrm{B}}-4\mathrm{l}\right)\stackrel{→}{\mathrm{v}}=\stackrel{→}{0}$, that is,

$\left[\begin{array}{ccc}-3& 0& 0\\ 0& -2& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{\mathrm{v}}_1\\ {\mathrm{v}}_2\\ {\mathrm{v}}_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]⇒{\mathrm{v}}_1={\mathrm{v}}_2=0{\mathrm{v}}_3∈\mathrm{R}$

So, for$\mathrm{λ}=4$ , the linearly independent eigenvectors are$\left\{{\left(0,0,1\right)}^{\mathrm{T}}\right\}$ , and the corresponding eigenfunction is${\mathrm{q}}_3\left({\mathrm{x}}_1{\mathrm{x}}_2\right)={\mathrm{x}}_2^2$

As a result, the eigenvalues and eigenfunctions are written as

$$\begin{gathered} \mathrm{λ}=1:x_1^2;T\left(x_1^2\right)=1\left(x_1^2\right) \\ \mathrm{λ}=2:x_1{x}_2;T\left(x_1{x}_2\right)=2\left(x_1{x}_2\right) \\ \mathrm{λ}=4:x_2^2;T\left(x_2^2\right)=4\left(x_2^2\right) \end{gathered}$$

From the above discussion, it is clear that the algebraic and geometric multiplicity of all eigenvalues, namely 1, 2, and 4, are equal (A.M.=G.M.= 1 for all $\mathrm{λ}$). As a result of the linear transformation

The eigenvalues of T are 1, 2, and 4, and the associated eigenfunctions are $x_1^2,x_1{x}_2\mathrm{and}{\mathrm{x}}_2^2$.

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