Chapter 7.7, Problem 1AP

To determine

Whether the given matrix $A$ is diagonalizable, when possible find the matrix $S$ and the diagonal matrix $D$.

Answer to Problem 1AP

Solution:

The matrix $A=\left[\begin{array}{cc}3& 0\\ 16& -1\end{array}\right]$ is diagonalizable as the matrix $A$ is non defective, the matrix $S$ is $S=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]$ and the diagonal matrix $D$ is $D=\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]$.

Explanation of Solution

Given:

The given matrix $A$ is,

$A=\left[\begin{array}{cc}3& 0\\ 16& -1\end{array}\right]$

Approach:

An $n\times n$ matrix that is similar to a diagonal matrix is said to be diagonalizable.

An $n\times n$ matrix $A$ is similar to a diagonal matrix if and only if $A$ is non defective. In such a case, if $v_1,v_2,\mathrm{.........},v_n$ denote $n$ linearly independent eigenvectors of $A$ and

$S=\left[v_1,v_2,\mathrm{..........},v_n\right]$,

Then,

$S^{-1}AS=\text{diag}\left({\lambda }_1,{\lambda }_2,\mathrm{.........},{\lambda }_n\right)$,

Where ${\lambda }_1,{\lambda }_2,\mathrm{........},{\lambda }_n$ are the eigenvalues of $A$ (non necessarily distinct) corresponding in the eigenvectors $v_1,v_2,\mathrm{............},v_n$.

Calculation:

Consider the given matrix $A$,

$A=\left[\begin{array}{cc}3& 0\\ 16& -1\end{array}\right]$

The order the matrix $A$ is $n=2$.

The characteristic equation of $A$ is given as $\det \left(A-\lambda I\right)=0$,

$\begin{array}{rll}\det \left(A-\lambda I\right)& =& 0\\ \left|\left[\begin{array}{cc}3& 0\\ 16& -1\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right|& =& 0\\ \left|\begin{array}{cc}3-\lambda & -2\\ -2& -1-\lambda \end{array}\right|& =& 0\end{array}$

Solve the above equation for the eigenvalues $\left(\lambda \right)$ of $A$,

$\begin{array}{rll}\left|\begin{array}{cc}3-\lambda & 0\\ 16& -1-\lambda \end{array}\right|& =& 0\\ \left(3-\lambda \right)\left(-1-\lambda \right)-\left(16\right)\left(0\right)& =& 0\\ \left(\lambda -3\right)\left(\lambda +1\right)& =& 0\\ \lambda & =& 3,-1\end{array}$

Therefore, the eigenvalue of $A$ are $3$ and $-1$.

The eigenvector corresponding to the eigenvalue ${\lambda }_1=3$ is determined based on the linear system given as,

$\begin{array}{rll}\left(A-\lambda I\right)v& =& 0\\ \left(\left[\begin{array}{cc}3& 0\\ 16& -1\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right)\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}3-\lambda & 0\\ 16& -1-\lambda \end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

Substitute $3$ for $\lambda$ in the above equation.

$\begin{array}{rll}\left[\begin{array}{cc}3-3& 0\\ 16& -1-3\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}0& 0\\ 16& -4\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ 16v_1-4v_2& =& 0\\ v_2& =& 4v_1\end{array}$

Thus, $v_1=r$ and $v_2=4r$, the general solution to this system is,

$\begin{array}{rll}v& =& {\left[v_1,v_2\right]}^{\text{T}}\\ & =& {\left[r,5r\right]}^{\text{T}}\\ & =& \left[\begin{array}{c}r\\ 4r\end{array}\right]\\ & =& r\left[\begin{array}{c}1\\ 4\end{array}\right]:r\in ℝ\end{array}$

Here, $r$ is a scalar.

The eigenvector corresponding to the eigenvalue ${\lambda }_2=-1$ is determined based on the linear system given as,

$\begin{array}{rll}\left(A-\lambda I\right)v& =& 0\\ \left(\left[\begin{array}{cc}3& 0\\ 16& -1\end{array}\right]-\left[\begin{array}{rl}\lambda & 0\\ 0& \lambda \end{array}\right]\right)\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}3-\lambda & 0\\ 16& -1-\lambda \end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

Substitute $-1$ for $\lambda$ in the above equation.

$\begin{array}{rll}\left[\begin{array}{cc}3-\left(-1\right)& 0\\ 16& -1-\left(-1\right)\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}4& 0\\ 16& 0\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ 4v_1& =& 0\end{array}$

Thus, $v_1=0$ and $v_2=r$, the general solution to this system is,

$\begin{array}{rll}v& =& {\left[v_1,v_2\right]}^{\text{T}}\\ & =& {\left[0,r\right]}^{\text{T}}\\ & =& \left[\begin{array}{c}0\\ r\end{array}\right]\\ & =& r\left[\begin{array}{c}0\\ 1\end{array}\right]:r\in ℝ\end{array}$

Here, $r$ is a scalar.

Eigenspace corresponding to ${\lambda }_1=3$ and ${\lambda }_2=-1$ is,

$\begin{array}{rll}{E}_1& =& \left\{v\in {ℝ}^2:v=r\left(1,4\right),r\left(0,1\right),r\in ℝ\right\}\\ & =& \text{span}\left\{\left(1,4\right),\left(0,1\right)\right\}\end{array}$

Thus,

$\begin{array}{rll}\dim \left[E_1\right]& =& 2\\ & =& 2\left(n\right)\end{array}$

Here, $\dim \left[E_1\right]$ is the number of linearly independent eigenvectors of $A$ and $n$ is the order of the matrix $A$.

As, $\dim \left[E_1\right]=n$, the matrix $A$ is non defective which signifies that the matrix $A$ is diagonalizable.

The matrix $S$ is given as,

$S=\left[v_1,v_2,\mathrm{..........},v_n\right]$

Write each eigenvector in column wise to form the matrix $S$,

$S=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]$

Then,

$S^{-1}AS=\text{diag}\left({\lambda }_1,{\lambda }_2,\mathrm{.........},{\lambda }_n\right)$

Substitute the obtained eigenvalues those are ${\lambda }_1=3$ and ${\lambda }_2=-1$ in the above expression.

$S^{-1}AS=\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]$

Therefore, the matrix $A=\left[\begin{array}{cc}3& 0\\ 16& -1\end{array}\right]$ is diagonalizable as the matrix $A$ is non defective, the matrix $S$ is $S=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]$ and the diagonal matrix $D$ is $D=\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]$.

Conclusion:

Hence, the matrix $A=\left[\begin{array}{cc}3& 0\\ 16& -1\end{array}\right]$ is diagonalizable as the matrix $A$ is non defective, the matrix $S$ is $S=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]$ and the diagonal matrix $D$ is $D=\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]$.