Linear Algebra

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

Nagoya University Linear Algebra Exam Problems and Solutions

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*}

Note

If you put $b=1-t$, then $(1-a-t)(1+a-t)=(b-a)(b+a)=b^2-a^2=(1-t)^2-a^2$.

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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