In Exercises 13–16, define \(T:{\mathbb{R}^2} \to {\mathbb{R}^2}\) by \(T\left( {\bf{x}} \right) = A{\bf{x}}\). Find a basis \(B\) for \({\mathbb{R}^2}\) with the property that \({\left( T \right)_B}\) diagonal.

13.\(A = \left( {\begin{aligned}0&{}&1\\{ - 3}&{}&4\end{aligned}} \right)\)

A \(n \times n\) matrix \(A\) is said to be diagonalizable if it has \(n\) distinct eigenvalues.

The eigenvalues of any matrix \(A\) can be found by using the formula\(\left| {A - \lambda I} \right| = 0\).

The given matrix is \(A = \left( {\begin{aligned}0&{}&1\\{ - 3}&4\end{aligned}} \right)\).

Find the eigenvalues by using the formula \(\left| {A - \lambda I} \right| = 0\).

\(\begin{aligned}\left| {\left( {\begin{aligned}0&{}&1\\{ - 3}&{}&4\end{aligned}} \right) - \lambda \left( {\begin{aligned}1&{}&0\\0&{}&1\end{aligned}} \right)} \right| &= 0\\\left| {\left( {\begin{aligned}0&{}&1\\{ - 3}&{}&4\end{aligned}} \right) - \left( {\begin{aligned}\lambda &0\\0&\lambda \end{aligned}} \right)} \right| &= 0\\\left| {\begin{aligned}{ - \lambda }&1\\{ - 3}&{}&{4 - \lambda }\end{aligned}} \right| &= 0\\ - \lambda \left( {4 - \lambda } \right) + 3 &= 0\\ - 4\lambda + {\lambda ^2} + 3 &= 0\\{\lambda ^2} - 4\lambda + 3 &= 0\\\left( {\lambda - 1} \right)\left( {\lambda - 3} \right) &= 0\\\lambda &= 1,3\end{aligned}\)

So, 1 and 3 are two eigenvalues, say \({\lambda _1} = 1\) and \({\lambda _2} = 3\).

As the eigenvalues are distinct, the matrix \(A\) is diagonalizable.

Eigenvector: For a \(n \times n\) matrix \(A\), whose eigenvalue is \(\lambda \), then the set of a subspace of \({\mathbb{R}^n}\) is known as an eigenspace, where the set of the subspace is the set of all the solutions of \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

Find \(\left( {A - \lambda I} \right)\) first for \({\lambda _1} = 1\) to solve the equation \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

\(\begin{aligned}A - I &= \left( {\begin{aligned}0&{}&1\\{ - 3}&{}&4\end{aligned}} \right) - \left( {\begin{aligned}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{ - 1}&{}&1\\{ - 3}&{}&3\end{aligned}} \right)\end{aligned}\)

Write the obtained matrix in the form of an augmented matrix for \(\left( {A - \lambda I} \right){\bf{x}} = 0\), where \(A{\bf{x}} = 0\), and the augmented matrix is given by \(\left( {\begin{aligned}A&{}&0\end{aligned}} \right)\).

\(\left( {\begin{aligned}{ - 1}&{}&1&{}&0\\{ - 3}&{}&3&{}&0\end{aligned}} \right)\)

The obtained matrix is not in a reduced form, so reduce it in row echelon form by applying row operations.

Write a system of equations corresponding to the obtained matrix.

\(\begin{aligned}{x_1} - {x_2} = 0\\{x_2},{\rm{ free variable}}\end{aligned}\)

As \({x_2}\) is a free variable, let \({x_2} = 1\). Then,

\(\begin{aligned}{x_1} = 1\\{x_2} = 1\end{aligned}\)

So, the general solution is given as:

\(\begin{aligned}{{\bf{v}}_1} = \left( {\begin{aligned}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ = {x_2}\left( {\begin{aligned}1\\1\end{aligned}} \right)\end{aligned}\)

So \({{\bf{v}}_1} = \left( {\begin{aligned}1\\1\end{aligned}} \right)\) is the eigenvector for the eigenspace for \({\lambda _1} = 1\).

Find \(\left( {A - \lambda I} \right)\) first for \({\lambda _2} = 3\)to solve the equation \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

\(\begin{aligned}A - 3I &= \left( {\begin{aligned}0&1\\{ - 3}&{}&4\end{aligned}} \right) - 3\left( {\begin{aligned}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{ - 3}&{}&1\\{ - 3}&{}&1\end{aligned}} \right)\end{aligned}\)

Write the obtained matrix in the form of an augmented matrix for \(\left( {A - \lambda I} \right){\bf{x}} = 0\), where \(A{\bf{x}} = 0\), and the augmented matrix is given by \(\left( {\begin{aligned}A&{}&0\end{aligned}} \right)\).

\(\left( {\begin{aligned}{ - 3}&{}&1&{}&0\\{ - 3}&{}&1&{}&0\end{aligned}} \right)\)

The obtained matrix is not in a reduced form, so reduce it in row echelon form by applying row operations.

Write the system of equations corresponding to the obtained matrix as shown below:

\(\begin{aligned}{x_1} - \frac{1}{3}{x_2} = 0\\{x_2},{\rm{ free variable}}\end{aligned}\)

As \({x_2}\) is a free variable, let \({x_2} = 1\). Then,

\(\begin{aligned}{x_1} = \frac{1}{3}\\{x_2} = 1\end{aligned}\)

So, the general solution is given as:

\(\begin{aligned}{{\bf{v}}_2} &= \left( {\begin{aligned}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= {x_2}\left( {\begin{aligned}{\frac{1}{3}}\\1\end{aligned}} \right)\end{aligned}\)

So \({{\bf{v}}_2} = \left( {\begin{aligned}{\frac{1}{3}}\\1\end{aligned}} \right)\) or \({{\bf{v}}_2} = \left( {\begin{aligned}1\\3\end{aligned}} \right)\) is the eigenvector for the eigenspace for \({\lambda _2} = 3\).

Diagonal Matrix Representation: Assume that \(A = PD{P^{ - 1}}\), then \(D\) is the \(B\)-matrix for the transformation \({\bf{x}} \mapsto A{\bf{x}}\), and\(B\) is the basis for \({\mathbb{R}^n}\) formed from the columns of \(P\), where \(D\) is the \(n \times n\) diagonal matrix.

The given matrix is diagonalizable; the basis of the given matrix is given by \(B = \left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\).

According to the theorem in the previous step, \(B = \left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) has the property that \(B\)-matrix of the transformation \({\bf{x}} \mapsto A{\bf{x}}\) is a diagonal matrix.

Hence, \(B = \left\{ {\left( {\begin{aligned}1\\1\end{aligned}} \right),\left( {\begin{aligned}1\\3\end{aligned}} \right)} \right\}\).