Question #337152

Let T be the linear operator on R


2 defined by

T(x, y) = (−y, x)

i. What is the matrix of T in the standard ordered basis for R2 ?

ii. What is the matrix of T in the ordered basis B = {α1, α2}, where α1 = (1, 2) and α2 = (1, −1)?

iii. Prove that for every real number c the operator (T − cI) is invertible.


1
Expert's answer
2022-05-06T10:47:08-0400

i. The operator T(x,y)T(x,y) can be written in the form: (yx)=(0110)(xy)\left(\begin{array}{ll}-y\\ x\end{array}\right)=\left(\begin{array}{ll}0&-1\\1&0\end{array}\right)\left(\begin{array}{ll}x\\ y\end{array}\right). (0110)\left(\begin{array}{ll}0&-1\\1&0\end{array}\right) is the matrix of the operator in the standard ordered basis. The matrix can be easily found by considering equality: (yx)=(a11a12a21a22)(xy)\left(\begin{array}{ll}-y\\ x\end{array}\right)=\left(\begin{array}{ll}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right)\left(\begin{array}{ll}x\\ y\end{array}\right). From the latter we get: y=a11x+a12y-y=a_{11}x+a_{12}y and x=a21x+a22yx=a_{21}x+a_{22}y. We immediately receive: a11=0,a12=1a_{11}=0,a_{12}=-1 and a21=1,a22=0.a_{21}=1,a_{22}=0.

ii. In order to find the matrix in the ordered basis B={α1,α2}B=\{ \alpha_1,\alpha_2\}, we consider the vector (xy)\left(\begin{array}{ll}x\\ y\end{array}\right) in the basis BB. We receive: (xy)=(12)t1+(11)t2\left(\begin{array}{ll}x\\ y\end{array}\right)=\left(\begin{array}{ll}1\\ 2\end{array}\right)t_1+\left(\begin{array}{ll}1\\ -1\end{array}\right)t_2. We receive: x=t1+t2x=t_1+t_2, y=2t1t2y=2t_1-t_2. Thus, t1=x+y3t_1=\frac{x+y}{3}, t2=2xy3t_2=\frac{2x-y}3. In the same way we consider the vector (yx)=(12)s1+(11)s2\left(\begin{array}{ll}-y\\ x\end{array}\right)=\left(\begin{array}{ll}1\\ 2\end{array}\right)s_1+\left(\begin{array}{ll}1\\ -1\end{array}\right)s_2. We receive: s1=y+x3s_1=\frac{-y+x}{3}, s2=2yx3s_2=\frac{-2y-x}{3}. The aim is to find the matrix satisfying: (y+x32yx3)=(b11b12b21b22)(x+y32xy3)\left(\begin{array}{ll}\frac{-y+x}{3}\\ \frac{-2y-x}{3}\end{array}\right)=\left(\begin{array}{ll}b_{11}&b_{12}\\b_{21}&b_{22}\end{array}\right)\left(\begin{array}{ll}\frac{x+y}{3}\\ \frac{2x-y}3\end{array}\right). We receive: y+x3=b11x+y3+b122xy3\frac{-y+x}{3}=b_{11}\frac{x+y}{3}+b_{12}\frac{2x-y}3, 2yx3=b21x+y3+b222xy3\frac{-2y-x}{3}=b_{21}\frac{x+y}{3}+b_{22}\frac{2x-y}3. We set y=xy=-x in the first equation and receive: b12=23b_{12}=\frac23. Setting y=2xy=2x yields: b11=13b_{11}=-\frac{1}3. We set y=xy=-x in the second equation and get: b22=13b_{22}=\frac13. Setting y=2xy=2x provides: b21=53b_{21}=-\frac53. Thus, matrix TT in the ordered bases BB has the form: (13235313)\left(\begin{array}{ll}-\frac13&\frac23\\-\frac53&\frac13\end{array}\right).

iii. Matrix TcIT-cI has the form: TcI=(c11c)T-cI=\left(\begin{array}{ll}-c&-1\\1&-c\end{array}\right). det(TcI)=c2+1>0det(T-cI)=c^2+1>0. Thus, the matrix is invertible for all real cc. It means, that the operator is invertible.

Answer: i. (0110)\left(\begin{array}{ll}0&-1\\1&0\end{array}\right); ii. (13235313)\left(\begin{array}{ll}-\frac13&\frac23\\-\frac53&\frac13\end{array}\right); iii. det(TcI)=c2+1>0det(T-cI)=c^2+1>0. Therefore, the operator is invertible.


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