Define the linear operator T: R³→ R³ by T([x1 x2 x3 ] = [ 3x1 − x2 + 2x3 2x1 + 2x3 x1 + 3x2] Show that T is diagonalizable

Question

Define the linear operator T: R³→ R³ by

[latex]T\left(\begin{bmatrix} x_{1} \ x_{2} \ x_{3} \end{bmatrix} ight) = \begin{bmatrix} 3x_{1}-x_{2} + 2x_{3} \ 2x_{1} + 2x_{3} \ x_{1} + 3x_{2} \end{bmatrix} [/latex]

Show that T is diagonalizable

Accepted Answer

Let B = [latex]\begin{Bmatrix}e_{1}, &e_{2}, &e_{3}\end{Bmatrix}[/latex] be the standard basis for R³. Then the matrix for T relative to B is

[latex] \left[T ight] _{B} = \begin{bmatrix} 3&-1&2 \2&0&2 \ 1&3&0 \end{bmatrix}[/latex]

Observe that the eigenvalues of [latex]\left[T ight] _{B} are \lambda _{1} = −2 , \lambda _{2} = 4 , and \lambda _{3} = 1[/latex] with corresponding eigenvectors, respectively

[latex]v_{1}= \begin{bmatrix} 1 \1 \ -2 \end{bmatrix} v_{2}= \begin{bmatrix} 1 \1 \ 1 \end{bmatrix} and v_{3}= \begin{bmatrix} -5 \4 \ 7 \end{bmatrix}[/latex]

Now let [latex]B^{\prime} = \begin{Bmatrix}v_{1}, &v_{2}, &v_{3}\end{Bmatrix}[/latex] and

P =[latex]\begin{bmatrix} 1&1&-5 \1&1&4 \ -2&1&7 \end{bmatrix}[/latex]

Then

[latex]\left[T ight] _{B^{\prime } }= \frac{1}{9} \begin{bmatrix} -1&4&-3 \5&1&3\ -1&1&0 \end{bmatrix} \begin{bmatrix} 3&-1&2 \2&0&2 \ 1&3&0 \end{bmatrix} \begin{bmatrix} 1&1&-5 \1&1&4 \ -2&1&7 \end{bmatrix} = \begin{bmatrix} -2&0&0 \0&4&0\ 0&0&1 \end{bmatrix}[/latex]

Question 5.2.6: Define the linear operator T: R³→ R³ by T([x1 x2 x3 ] = [ 3x...

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