Suppose T "\\isin" L(R2) is dened by T(x,y) = (-3y,x). Find the eigenvalues of T.
Given that T is a linear transformation
T(x,y)=(-3y,x)
we know that the basis of R2 is {(1,0),(0,1)}
then,
T(1,0)=(-3(0),1)
T(1,0)=(0,1)
T(1,0)=0(1,0)+1(0,1)...........(1)
And,
T(0,1)=((-3)1,0)=(-3,0)
(0,1)=-3(1,0)+0(0,1)............(2)
Now the matrix representation of given linear transformation with respect to standard basis is given as,
[T]="\\begin{bmatrix}\n 0 & -3 \\\\\n 1 & 0\n\\end{bmatrix}"
to find the eigen values of T we need to find the eigen values of matrix [T]
let A=[T]
then, A="\\begin{bmatrix}\n 0 & -3 \\\\\n 1 & 0\n\\end{bmatrix}"
for eigen values put |A-"\\lambda" I|=0
where"\\lambda" is eigen value of A
then,
"\\begin{bmatrix}\n 0 & -3 \\\\\n 1 & 0\n\\end{bmatrix}" -"\\begin{bmatrix}\n \\lambda & 0 \\\\\n 0 & \\lambda\n\\end{bmatrix}" =A-"\\lambda" I
A-"\\lambda" I="\\begin{bmatrix}\n -\\lambda & -3 \\\\\n 1 & -\\lambda\n\\end{bmatrix}"
put A-"\\lambda" I=0
then,
"\\begin{bmatrix}\n -\\lambda & -3 \\\\\n 1 & -\\lambda\n\\end{bmatrix}" =0
"\\lambda^2" +3=0
"\\lambda^2" =-3
"\\lambda=\\sqrt3i,-\\sqrt3i"
"\\therefore" The eigen values of T are "\\sqrt3i,-\\sqrt3i"
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