# Determine the Values of $a$ such that the 2 by 2 Matrix is Diagonalizable

## Problem 459

Let

\[A=\begin{bmatrix}

1-a & a\\

-a& 1+a

\end{bmatrix}\]
be a $2\times 2$ matrix, where $a$ is a complex number.

Determine the values of $a$ such that the matrix $A$ is diagonalizable.

(*Nagoya University, Linear Algebra Exam Problem*)

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## Proof.

To find eigenvalues of the matrix $A$, we determine the characteristic polynomial $p(t)$ of $A$ as follows.

We have

\begin{align*}

p(t)&=\det(A-tI)=\begin{vmatrix}

1-a-t & a\\

-a& 1+a-t

\end{vmatrix}\\[6pt]
&=(1-a-t)(1+a-t)+a^2\\

&=(1-t)^2-a^2+a^2=(1-t)^2.

\end{align*}

Thus, the eigenvalue of $A$ is $1$ with algebraic multiplicity $2$.

Let us determine the geometric multiplicity (the dimension of the eigenspace $E_1$).

We have

\begin{align*}

A-I=\begin{bmatrix}

-a & a\\

-a& a

\end{bmatrix}

\xrightarrow{R_2-R_1}

\begin{bmatrix}

-a & a\\

0& 0

\end{bmatrix} .

\end{align*}

If $a\neq 0$, then we further reduce it and get

\begin{align*}

\begin{bmatrix}

-a & a\\

0& 0

\end{bmatrix}\xrightarrow{\frac{-1}{a}R_1}

\begin{bmatrix}

1 & -1\\

0& 0

\end{bmatrix}.

\end{align*}

Hence the eigenspace corresponding to the eigenvalue $1$ is

\begin{align*}

E_1=\calN(A-I)=\Span \left\{\, \begin{bmatrix}

1 \\

1

\end{bmatrix} \,\right\},

\end{align*}

and its dimension is $1$, which is less than the algebraic multiplicity.

Thus, when $a\neq 0$, the matrix $A$ is not diagonalizable.

If $a=0$, then the matrix $A=\begin{bmatrix}

1 & 0\\

0& 1

\end{bmatrix}$ is already diagonal, hence it is diagonalizable.

In conclusion, the matrix $A$ is diagonalizable if and only if $a=0$.

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