Question

Question: Diagonalize the matrices in Exercises $\mathbf{7}\mathbf{-}\mathbf{-}\mathbf{20}$, if possible. The eigenvalues for Exercises $\mathbf{11}\mathbf{-}\mathbf{-}\mathbf{16}$ are as follows:$\left(\mathbf{11}\right)\lambda \mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}$; $\left(\mathbf{12}\right)\lambda \mathbf{=}\mathbf{2}\mathbf{,}\mathbf{8}$; $\left(\mathbf{13}\right)\lambda \mathbf{=}\mathbf{5}\mathbf{,}\mathbf{1}$; $\left(\mathbf{14}\right)\lambda \mathbf{=}\mathbf{5}\mathbf{,}\mathbf{4}$; $\left(\mathbf{15}\right)\lambda \mathbf{=}\mathbf{3}\mathbf{,}\mathbf{1}$; $\left(\mathbf{16}\right)\lambda \mathbf{=}\mathbf{2}\mathbf{,}\mathbf{1}$. For exercise $\mathbf{18}$, one eigenvalue is $\lambda \mathbf{=}\mathbf{5}$ and one eigenvector is $\left(\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\right)$.

10. $\left(\begin{array}{cc}\mathbf{2}& \mathbf{3}\\ \mathbf{4}& \mathbf{1}\end{array}\right)$

Short answer

The matrix $A$ is diagonalizable with $P=\left(\begin{array}{cc}-\frac{3}{4}& 1\\ 1& 1\end{array}\right)$, $P^{-1}=\left(\begin{array}{cc}-\frac{4}{7}& \frac{4}{7}\\ \frac{4}{7}& \frac{3}{7}\end{array}\right)$ and $D=\left(\begin{array}{cc}-2& 0\\ 0& 5\end{array}\right)$.

Step-by-step solution

Write the Diagonalization Theorem

The Diagonalization Theorem: An $n\times n$ matrix $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors. As $A=PD{P}^{-1}$ which has $D$ a diagonal matrix if and only if the columns of $P$ are $n$ linearly independent eigenvectors of $A$.

Find eigenvalues of the matrix

Consider the given matrix $A=\left(\begin{array}{cc}2& 3\\ 4& 1\end{array}\right)$. We need to diagonalize the given matrix if possible.

Write the characteristic equation:

$$\begin{array}{c}\left|A-\lambda I\right|=0\\ \left|\begin{array}{cc}2-\lambda & 3\\ 4& 1-\lambda \end{array}\right|=0\\ \left(\lambda +2\right)\left(\lambda -5\right)=0\\ \lambda =-2,5\end{array}$$

Find the eigenvectors

Write the matrix form for finding the eigenvector for$\lambda =-2$.

$$\begin{array}{c}\left(A-\lambda I\right)\mathbf{x}=0\\ \left(A-\left(-2\right)I\right)\mathbf{x}=0\\ \left(\left(\begin{array}{cc}2& 3\\ 4& 1\end{array}\right)+2\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}4& 3\\ 4& 3\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$$

Therefore, the augmented matrix is shown below:

$$\begin{array}{c}M=\left(\begin{array}{ccc}4& 3& 0\\ 4& 3& 0\end{array}\right)\\ =\left(\begin{array}{ccc}4& 3& 0\\ 4& 3& 0\end{array}\right)\left\{R_2=R_2-R_1\right\}\end{array}$$

Since there are$2$variables and$1$equitation, consider$x_2$as free.

$$\begin{array}{c}4x_1+3x_2=0\\ 4x_1=-3x_2\\ x_1=-\frac{3}{4}{x}_2\end{array}$$

Write the parametric form of solution.

$$\begin{array}{c}\mathbf{x}=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)\\ =x_2\left(\begin{array}{c}-\frac{3}{4}\\ 1\end{array}\right)\end{array}$$

Therefore, the eigenvector for $\lambda =-2$is$\mathrm{v}=\left(\begin{array}{c}-\frac{3}{4}\\ 1\end{array}\right)$.

Write the matrix form for finding the eigenvector for$\lambda =5$.

$$\begin{array}{c}\left(A-\lambda I\right)\mathbf{x}=0\\ \left(A-\left(5\right)I\right)\mathbf{x}=0\\ \left(\left(\begin{array}{cc}2& 3\\ 4& 1\end{array}\right)-5\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}-3& 3\\ 4& -4\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$$

Therefore, the augmented matrix is shown below:

$$\begin{array}{c}M=\left(\begin{array}{ccc}-3& 3& 0\\ 4& -4& 0\end{array}\right)\\ =\left(\begin{array}{ccc}1& -1& 0\\ 1& -1& 0\end{array}\right)\left\{R_1=-\frac{R_1}{3},R_2=-\frac{R_2}{3}\right\}\\ =\left(\begin{array}{ccc}1& -1& 0\\ 0& 0& 0\end{array}\right)\left\{R_1=R_2-R_1\right\}\end{array}$$

Since there are$2$variables and$1$equitation, consider$x_2$as free.

$$\begin{array}{c}{x}_1-x_2=0\\ x_1=x_2\end{array}$$

Write the parametric form of solution.

$$\begin{array}{c}\mathbf{x}=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)\\ =x_2\left(\begin{array}{c}1\\ 1\end{array}\right)\end{array}$$

Therefore, the eigenvector for $\lambda =5$ is $v=\left(\begin{array}{c}1\\ 1\end{array}\right)$.

Find the matrix $P$

The matrix$P$is formed by eigenvectors

$$P=\left(\begin{array}{cc}-\frac{3}{4}& 1\\ 1& 1\end{array}\right)$$

The inverse of the matrix$P$is shown below:

$$\begin{array}{c}{P}^{-1}={\left(\begin{array}{cc}-\frac{3}{4}& 1\\ 1& 1\end{array}\right)}^{-1}\\ =\frac{1}{1\left(-\frac{3}{4}\right)-1}\left(\begin{array}{cc}1& -1\\ -1& -\frac{3}{4}\end{array}\right)\\ =-\frac{4}{7}\left(\begin{array}{cc}1& -1\\ -1& -\frac{3}{4}\end{array}\right)\\ =\left(\begin{array}{cc}-\frac{4}{7}& \frac{4}{7}\\ \frac{4}{7}& \frac{3}{7}\end{array}\right)\end{array}$$

Find the matrix $D$

As the matrix that diagonalizes$A$is shown below:

$$\begin{array}{c}D=P^{-1}AP\\ =\left(\begin{array}{cc}-\frac{4}{7}& \frac{4}{7}\\ \frac{4}{7}& \frac{3}{7}\end{array}\right)\left(\begin{array}{cc}2& 3\\ 4& 1\end{array}\right)\left(\begin{array}{cc}-\frac{3}{4}& 1\\ 1& 1\end{array}\right)\\ =\left(\begin{array}{cc}\frac{8}{7}& -\frac{8}{7}\\ \frac{20}{7}& \frac{15}{7}\end{array}\right)\left(\begin{array}{cc}-\frac{3}{4}& 1\\ 1& 1\end{array}\right)\\ =\left(\begin{array}{cc}-2& 0\\ 0& 5\end{array}\right)\end{array}$$

Thus, the matrix $A$ is diagonalizable with $P=\left(\begin{array}{cc}-\frac{3}{4}& 1\\ 1& 1\end{array}\right)$, $P^{-1}=\left(\begin{array}{cc}-\frac{4}{7}& \frac{4}{7}\\ \frac{4}{7}& \frac{3}{7}\end{array}\right)$ and $D=\left(\begin{array}{cc}-2& 0\\ 0& 5\end{array}\right)$.