Let \(T\) be the linear operator on \(\RR^2\) defined by \[ T(x_1, x_2) = (-x_2,\; x_1). \]
What is the matrix of \(T\) in the standard ordered basis for \(\RR^2\)?
What is the matrix of \(T\) in the ordered basis \(\mathfrak{B} = \{\alpha_1, \alpha_2\}\), where \(\alpha_1 = (1, 2)\) and \(\alpha_2 = (1, -1)\)?
Prove that for every real number \(c\) the operator \((T - cI)\) is invertible.
Prove that if \(\mathfrak{B}\) is any ordered basis for \(\RR^2\) and \([T]_{\mathfrak{B}} = A\), then \(A_{12}A_{21} \neq 0\).
Solution
(a).
1. \(T(1,0) = (0,1)\) and \(T(0,1) = (-1,0)\).
apply the definition \(T(x_1,x_2) = (-x_2, x_1)\).
2. Hence \([T]_{\text{std}} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\).
the columns are the images of the basis vectors.
(b).
1. \(T(\alpha_1) = T(1,2) = (-2, 1)\) and \(T(\alpha_2) = T(1,-1) = (1, 1)\).
apply \(T\).
2. Solve \((-2,1) = a(1,2) + b(1,-1)\): \(a + b = -2\), \(2a - b = 1\), giving \(a = -1/3\), \(b = -5/3\).
linear system.
3. Solve \((1,1) = c(1,2) + d(1,-1)\): \(c + d = 1\), \(2c - d = 1\), giving \(c = 2/3\), \(d = 1/3\).
linear system.
4. Hence \([T]_{\mathfrak{B}} = \begin{pmatrix} -1/3 & 2/3 \\ -5/3 & 1/3 \end{pmatrix}\).
2 and 3.
(c).
1. \([T - cI]_{\text{std}} = \begin{pmatrix} -c & -1 \\ 1 & -c \end{pmatrix}\).
2 (a).
2. \(\det(T - cI) = c^2 + 1 \neq 0\) for every real \(c\).
\(\det \begin{pmatrix} -c & -1 \\ 1 & -c \end{pmatrix} = c^2 + 1 > 0\).
3. Hence \(T - cI\) is invertible for every real \(c\).
2 (a nonzero determinant means invertible).
(d).
1. \(A = [T]_{\mathfrak{B}}\) is similar to \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\).
change of basis.
2. Hence \(\det A = 1\) and \(\operatorname{tr} A = 0\).
determinant and trace are invariant under similarity, and \(\det \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = 1\), \(\operatorname{tr} = 0\).
3. Write \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\); then \(ad - bc = 1\) and \(a + d = 0\).
2.
4. Suppose \(bc = 0\).Then \(ad = 1\) and \(a + d = 0\), so \(d = -a\) and \(-a^2 = 1\), impossible over \(\RR\).
3; \(a^2 = -1\) has no real solution.
5. Hence \(bc = A_{12}A_{21} \neq 0\).
4.
6. Q.E.D.
2 (a), 4 (b), 3 (c), 5 (d).
Solution
In the standard basis, \[ [T]=J=\begin{bmatrix}0&-1\\1&0\end{bmatrix}. \] For \(\mathfrak B=\{(1,2),(1,-1)\}\) let \[ P=\begin{bmatrix}1&1\\2&-1\end{bmatrix}. \] Then \[ [T]_{\mathfrak B}=P^{-1}JP =\begin{bmatrix}-\frac13&\frac23\\-\frac53&\frac13\end{bmatrix}. \]
Since \(T^2=-I\), for every \(c\in\mathbb R\), \[ (T-cI)(-T-cI)=-(T^2)+c^2I=(1+c^2)I. \] Because \(1+c^2\ne0\), \(T-cI\) is invertible, with inverse \[ (T-cI)^{-1}=\frac{-T-cI}{1+c^2}. \]
Finally let \(A=[T]_{\mathfrak C}\) in any real ordered basis. Then \(A^2=-I_2\). Write \(A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\). If \(b=0\), the \((1,1)\) entry of \(A^2=-I\) gives \(a^2=-1\), impossible over \(\mathbb R\). If \(c=0\), the same conclusion follows from a diagonal entry. Thus \[ A_{12}A_{21}=bc\ne0. \]