Diagonalize a 2 by 2 Matrix if Diagonalizable

Problem 477

Determine whether the matrix
$A=\begin{bmatrix} 1 & 4\\ 2 & 3 \end{bmatrix}$ is diagonalizable.

If so, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

(The Ohio State University, Linear Algebra Final Exam Problem)

Solution.

To determine whether the matrix $A$ is diagonalizable, we first find eigenvalues of $A$.
To do so, we compute the characteristic polynomial $p(t)$ of $A$:
\begin{align*}
p(t)&=\begin{vmatrix}
1-t & 4\\
2& 3-t
\end{vmatrix}
=(1-t)(3-t)-8\6pt] &=t^2-4t-5=(t+1)(t-5). \end{align*} The roots of the characteristic polynomial p(t) are eigenvalues of A. Hence the eigenvalues of A are -1 and 5. Since the 2\times 2 matrix A has two distinct eigenvalues, it is diagonalizable. To find the invertible matrix S, we need eigenvectors. Let us find the eigenvectors corresponding to the eigenvalue -1. By elementary row operations, we have \begin{align*} &A-(-1)I=A+I=\begin{bmatrix} 2 & 4\\ 2& 4 \end{bmatrix}\\[6pt] &\xrightarrow{R_2-R_1} \begin{bmatrix} 2 & 4\\ 0& 0 \end{bmatrix} \xrightarrow{\frac{1}{2}R_1} \begin{bmatrix} 1 & 2\\ 0& 0 \end{bmatrix}. \end{align*} It follows that the eigenvectors corresponding to -1 are of the form \[a\begin{bmatrix} -2 \\ 1 \end{bmatrix} for any nonzero scalar $a$.

Similarly, we have
\begin{align*}
A-5I=\begin{bmatrix}
-4 & 4\\
2& -2
\end{bmatrix}
\xrightarrow{\frac{-1}{4}R_1}
\begin{bmatrix}
1 & -1\\
2& -2
\end{bmatrix}
\xrightarrow{R_2-2R_1}
\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}.
\end{align*}
Hence the eigenvectors corresponding to $5$ are of the form
$b\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ for any nonzero scalar $b$.

Thus, $\mathbf{u}=\begin{bmatrix} -2 \\ 1 \end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ are basis vectors of eigenspace $E_{-1}, E_{5}$, respectively.

Define
$S:=\begin{bmatrix} \mathbf{u} & \mathbf{v} \end{bmatrix} =\begin{bmatrix} -2 & 1\\ 1& 1 \end{bmatrix}.$ Then by the general procedure of the diagonalization, we have
\begin{align*}
S^{-1}AS=D,
\end{align*}
where
$D:=\begin{bmatrix} -1 & 0\\ 0& 5 \end{bmatrix}.$

Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

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