# 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*)

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## 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.

- Find All the Eigenvalues of 4 by 4 Matrix
- Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue
- Diagonalize a 2 by 2 Matrix if Diagonalizable (This page)
- Find an Orthonormal Basis of the Range of a Linear Transformation
- The Product of Two Nonsingular Matrices is Nonsingular
- Determine Whether Given Subsets in ℝ4 R 4 are Subspaces or Not
- Find a Basis of the Vector Space of Polynomials of Degree 2 or Less Among Given Polynomials
- Find Values of $a , b , c$ such that the Given Matrix is Diagonalizable
- Idempotent Matrix and its Eigenvalues
- Diagonalize the 3 by 3 Matrix Whose Entries are All One
- Given the Characteristic Polynomial, Find the Rank of the Matrix
- Compute $A^{10}\mathbf{v}$ Using Eigenvalues and Eigenvectors of the Matrix $A$
- Determine Whether There Exists a Nonsingular Matrix Satisfying $A^4=ABA^2+2A^3$

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