Question

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.

15.\(A = \left( {\begin{aligned}4&{}&{ - 2}\\{ - 1}&{}&3\end{aligned}} \right)\)

Short answer

The basis \(B\) for\({\mathbb{R}^2}\) is \(B = \left\{ {\left( {\begin{aligned}1\\1\end{aligned}} \right),\left( {\begin{aligned}{ - 2}\\1\end{aligned}} \right)} \right\}\).

Step-by-step solution

Write the theorem 

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

Formula to find eigenvalues \(P\) -matrix

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

Find the eigenvalues

The given matrix is \(A = \left( {\begin{aligned}4&{}&{ - 2}\\{ - 1}&{}&3\end{aligned}} \right)\). Find the eigenvalues by using the formula \(\left| {A - \lambda I} \right| = 0\).

\(\begin{aligned}\left| {\left( {\begin{aligned}4&{}&{ - 2}\\{ - 1}&{}&3\end{aligned}} \right) - \lambda \left( {\begin{aligned}1&{}&0\\0&{}&1\end{aligned}} \right)} \right| &= 0\\\left| {\left( {\begin{aligned}4&{}&{ - 2}\\{ - 1}&{}&3\end{aligned}} \right) - \left( {\begin{aligned}\lambda &0\\0&\lambda \end{aligned}} \right)} \right| &= 0\\\left| {\begin{aligned}{4 - \lambda }&{}&{ - 2}\\{ - 1}&{}&{3 - \lambda }\end{aligned}} \right| &= 0\\\left( {4 - \lambda } \right)\left( {3 - \lambda } \right) - 2 &= 0\\{\lambda ^2} - 7\lambda + 10 &= 0\\\left( {\lambda - 5} \right)\left( {\lambda - 2} \right) &= 0\\\lambda &= 5,2\end{aligned}\)

So, 2 and 5 are two eigenvalues, say \({\lambda _1} = 2\) and \({\lambda _2} = 5\).

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

Write the definition

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 a set of the subspace is the set of all the solutions of \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

Determine basis vector for the eigenspace for \({\lambda _1} = 2\) 

Find \(\left( {A - \lambda I} \right)\) for \({\lambda _1} = 2\) to solve the equation \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

\(\begin{aligned}A - 2I &= \left( {\begin{aligned}4&{}&{ - 2}\\{ - 1}&{}&3\end{aligned}} \right) - 2\left( {\begin{aligned}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}2&{}&{ - 2}\\{ - 1}&{}&1\end{aligned}} \right)\end{aligned}\)

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

\(\left( {\begin{aligned}2&{}&{ - 2}&{}&0\\{ - 1}&{}&1&{}&0\end{aligned}} \right)\)

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

divyanshi@acadedigital.com

The equation is not getting converted.

Thu Jan 19 2023 16:56:20 GMT+0530 (India Standard Time)

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 \({\lambda _1} = 2\).

Determine basis vector for the eigenspace for \({\lambda _2} = 5\) 

Find \(\left( {A - \lambda I} \right)\) first for \({\lambda _2} = 5\)to solve the equation \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

\(\begin{aligned}A - 5I &= \left( {\begin{aligned}4&{}&{ - 2}\\{ - 1}&{}&3\end{aligned}} \right) - 5\left( {\begin{aligned}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{ - 1}&{}&{ - 2}\\{ - 1}&{}&{ - 2}\end{aligned}} \right)\end{aligned}\)

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

\(\left( {\begin{aligned}{ - 1}&{}&{ - 2}&{}&0\\{ - 1}&{}&{ - 2}&{}&0\end{aligned}} \right)\)

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

divyanshi@acadedigital.com

Equation is not getting converted.

Thu Jan 19 2023 16:55:12 GMT+0530 (India Standard Time)

Write a system of equations corresponding to the obtained matrix.

\(\begin{aligned}{x_1} + 2{x_2} = 0\\{x_2},{\rm{ free variable}}\end{aligned}\)

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

\(\begin{aligned}{x_1} = - 2\\{x_2} = 1\end{aligned}\)

So, the general solution is given as:

\(\begin{aligned}{{\bf{v}}_2} &= \left( {\begin{aligned}{{x_1}}\\{{x_2}}\end{aligned}} \right)\\ &= {x_2}\left( {\begin{aligned}{ - 2}\\1\end{aligned}} \right)\end{aligned}\)

So \({{\bf{v}}_2} = \left( {\begin{aligned}{ - 2}\\1\end{aligned}} \right)\) is the eigenvector for the eigenspace for \({\lambda _2} = 5\).

Write the theorem

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.

Find the basis

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, \(B = \left\{ {\left( {\begin{aligned}1\\1\end{aligned}} \right),\left( {\begin{aligned}{ - 2}\\1\end{aligned}} \right)} \right\}\).