Answer to Question #153656 in Linear Algebra for pallavi

Question #153656

Let T be the linear operator on R2 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
2021-01-08T14:39:27-0500

Here, we have

T(x,y)=(y,x)=[yx]T(x,y)=(-y,x)=\begin{bmatrix} -y\\x \end{bmatrix}

(i) Standard ordered basis for R2is[1001]\> \R^2 \> is \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}

So, for the first column applying the linear operator T we have [01]\begin{bmatrix} 0\\1 \end{bmatrix} and for the second column after applying the linear operator T we have [10]\begin{bmatrix} -1\\0 \end{bmatrix} . So, the matrix for T is A(let):-


A=[0110]A=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}




(ii) New basis is:-


B=[1121]B=\begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}

Now, we have to find the matrix A wrt the basis B.

Now we take the first column of B and left multiply with A.


A[12]=[0110][12]=[21]A\begin{bmatrix} 1\\2 \end{bmatrix}=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\cdot \begin{bmatrix} 1\\2 \end{bmatrix}=\begin{bmatrix} -2\\1 \end{bmatrix}

Now, solving for a, b where


[21]=a[12]+b[11][21]=[a+b2ab]\begin{bmatrix} -2\\1 \end{bmatrix}=a\begin{bmatrix} 1\\2 \end{bmatrix}+b\begin{bmatrix} 1\\-1 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} -2\\1 \end{bmatrix}=\begin{bmatrix} a+b\\2a-b \end{bmatrix}

Solving we get, a=13\frac{-1}{3} and b=53\frac{-5}{3} .


Similarly we do the above steps with the second column of B and find c, d:-


[11]=[c+d2cd]\begin{bmatrix} 1\\1 \end{bmatrix}=\begin{bmatrix} c+d\\2c-d \end{bmatrix}

Solving, we have c=23\frac{2}{3} and d=13\frac{1}{3} .


Therefore the final matrix wrt, the given basis B is:-


[13235313]\begin{bmatrix} \frac{-1}{3} & \frac{2}{3} \\\\ \frac{-5}{3} & \frac{1}{3} \end{bmatrix}


(iii) Here we calculate the TcI|T-cI|



TcI =[0110]c[1001] =[0110][c00c] =[c11c] =(c)2(1) =c2+1|T-cI|\\~\\ = |\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}-c\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}|\\~\\ =|\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}-\begin{bmatrix} c & 0 \\ 0 & c \end{bmatrix}|\\~\\ =|\begin{bmatrix} -c & -1 \\ 1 & -c \end{bmatrix}|\\~\\ =(-c)^2-(-1)\\~\\ =c^2+1\\

Now the determinant of the matrix TcIT-cI is 1+c21+c^2 which is never equal to 0, as cRc \in\R .

Therefore the matrix TcIT-cI is always invertible cR\forall c\in\R .


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