Question

Find a basis B for the domain of T such that the matrix for T relative to B is diagonal. T: R3 → R3: T(x, y, z) = (-3x + 2y – 32, 2x - 62, -* - 2y – z) -4 0 0 0 -4 B = 0 0 X Need Help? Read It Watch It Talk to a Tutor

Answer 1

$\small \text{ Consider the linear transformation }\\ T : R_3\rightarrow R_3\\ T(x,y,z)=(-3x+2y-3z,\; 2x-6z,\; -x-2y-z)\\ \text{Let us assume tha standard basis of }\mathbb{R}^{3}\; \text{is as follows}\\ S=\left \{ (1,0,0) ,\; (0,1,0) ,\; (0,0,1) \right \}\\ T(1,0,0) = (-3,2,-1) =-3(1,0,0)+2(0,1,0)-1 (0,0,1)\\ T(0,1,0) = (2,0,-2) =2(1,0,0)+0(0,1,0)-2 (0,0,1)\\ T(0,0,1) = (-3,-6,-1) =-3(1,0,0)-6(0,1,0)-1 (0,0,1)\\ \text{The matrix of T relative to basis is}\\ T=\begin{bmatrix} 3 & 2 & -3\\ -2 & 0& -6\\ -1& -2& -1 \end{bmatrix}\\ \text{The characteristic polynomial of matrix is as follows}\\ \left | T-\lambda I \right |=0\\ \begin{vmatrix} 3-\lambda & 2 & -3\\ -2 & 0-\lambda & -6\\ -1& -2& -1-\lambda \end{vmatrix}=0\\ -\lambda^3+2\lambda^2+14\lambda-40=0\\$

$\small \Rightarrow\lambda_1=-4,\: \: \lambda_2=3+i,\: \: \lambda_3=3-i\\ \text{The eigen vector corresponding to eigen value};\: \: \lambda_1=-4\\ \left ( A-\lambda_1I \right )\vec{V^1}=0\\ \begin{bmatrix} 3+4 & 2 & -3\\ -2 & 0+4& -6\\ -1& -2& -1+4 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=0\\ \begin{bmatrix} 7& 2 & -3\\ -2 & 4& -6\\ -1& -2& 3 \end{bmatrix}\sim \begin{bmatrix} 7&2&-3\\ 0&\frac{32}{7}&-\frac{48}{7}\\ 0&0&0 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3\end{bmatrix}=0\\ 7v_1+2v_2-3v_3=0\\ 32v_2-48v_3=0\Rightarrow 2v_2=3v_3 \\$

$\small \text{Therefore, the eigen vector corresponding to eigen value};\: \: \lambda_1=-4\\ {\color{Red} {\vec{V^1}=\begin{bmatrix} 0\\ 3\\ 2 \end{bmatrix}}}\\ \text{Similarly the eigen vector corresponding to the other eigen value are as follows}\\ {\color{Red} {\vec{V^2}=\begin{bmatrix} -4+3i\\ -2i\\ 1 \end{bmatrix}}} ,\; \; {\color{Red} {\vec{V^3}=\begin{bmatrix} -4-3i\\ 2i\\ 1 \end{bmatrix}}}\\ \text{So, the required basis B is as follows}\\ B=\left \{ \begin{bmatrix} 0\\ 3\\ 2 \end{bmatrix},\begin{bmatrix} -4+3i\\ -2i\\ 1 \end{bmatrix} ,\begin{bmatrix} -4-3i\\ 2i\\ 1 \end{bmatrix} \right \}$