Suppose T € L(R^2) is defined by T(x;y) = ((3y; x). Find the eigenvalues of T.
Let us solve the equation "T(x,y)=(\\lambda x, \\lambda y)" for "\\lambda" :
"\\begin{cases} 3y=\\lambda x \\\\ x=\\lambda y \\end{cases}" , now using the substitution we find that "\\begin{cases} 3y=\\lambda^2y \\\\ x= \\lambda^2x\/3 \\end{cases}".
Therefore, the possible solutions for "\\lambda" are "\\sqrt{3}, -\\sqrt{3}" (as at least one of numbers "x,y" is non-zero).
Another way to calculate it could be expressing "T" in a matrix form :
"T= \\begin{pmatrix} 0 & 3 \\\\ 1 & 0 \\end{pmatrix}"
and find the characteristic polynomial :
"\\lambda^2-3=0"
which yields the same result.
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