Answer to Question #153656 in Linear Algebra for pallavi

Question

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.

Answer 1

Here, we have

$$T(x,y)=(-y,x)=\begin{bmatrix} -y\\x \end{bmatrix}$$

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

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

$$A=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$

(ii) New basis is:-

$$B=\begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}$$

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.

$$A\begin{bmatrix} 1\\2 \end{bmatrix}=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\cdot \begin{bmatrix} 1\\2 \end{bmatrix}=\begin{bmatrix} -2\\1 \end{bmatrix}$$

Now, solving for a, b where

$$\begin{bmatrix} -2\\1 \end{bmatrix}=a\begin{bmatrix} 1\\2 \end{bmatrix}+b\begin{bmatrix} 1\\-1 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} -2\\1 \end{bmatrix}=\begin{bmatrix} a+b\\2a-b \end{bmatrix}$$

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

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

$$\begin{bmatrix} 1\\1 \end{bmatrix}=\begin{bmatrix} c+d\\2c-d \end{bmatrix}$$

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

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

$$\begin{bmatrix} \frac{-1}{3} & \frac{2}{3} \\\\ \frac{-5}{3} & \frac{1}{3} \end{bmatrix}$$

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

$$|T-cI|\\~\\ = |\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}-c\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}|\\~\\ =|\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}-\begin{bmatrix} c & 0 \\ 0 & c \end{bmatrix}|\\~\\ =|\begin{bmatrix} -c & -1 \\ 1 & -c \end{bmatrix}|\\~\\ =(-c)^2-(-1)\\~\\ =c^2+1\\$$

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

Therefore the matrix $T-cI$ is always invertible $\forall c\in\R$ .