Diagonalize a 2 by 2 Symmetric Matrix

Diagonalization Problems and Solutions in Linear Algebra

Problem 629

Diagonalize the $2\times 2$ matrix $A=\begin{bmatrix}
2 & -1\\
-1& 2
\end{bmatrix}$ by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

 
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Solution.

The characteristic polynomial $p(t)$ of the matrix $A$ is
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
2-t & -1\\
-1& 2-1
\end{vmatrix}\\[6pt] &=(2-t)^2-1 =t^2-4t+3\\
&=(t-1)(t-3).
\end{align*}
It follows that the eigenvalues of $A$ are $\lambda=1, 3$ with algebraic multiplicities are both $1$.
Hence, the geometric multiplicities are $1$ and thus any nonzero vector in eahc eigenspace forms a eigenbasis.


Now let us find a eigenbasis for each eigenspace $E_{\lambda}=\calN(A-\lambda I)$.
For the eigenvalue $1$, we have
\[A-I=\begin{bmatrix}
1 & -1\\
-1& 1
\end{bmatrix}\xrightarrow{R_2+R_1} \begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}\] This yields that the eigenvectors corresponding to the eigenvalue $1$ are $x_2\begin{bmatrix}
1 \\
1
\end{bmatrix}$ with $x_2\neq 0$. Hence
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1
\end{bmatrix} \in E_1\] is an eigenbasis for $E_1$.


Similarly, as we have
\[A-3I=\begin{bmatrix}
-1 & -1\\
-1& -1
\end{bmatrix} \xrightarrow{-R_1}\begin{bmatrix}
1 & 1\\
-1& -1
\end{bmatrix} \xrightarrow{R_2+R_1} \begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix},\] we see that
\[\mathbf{v}_2=\begin{bmatrix}
-1 \\
1
\end{bmatrix} \in E_3\] is an eigenbasis for $E_3$.


Let
\[S=\begin{bmatrix}
\mathbf{v}_1 & \mathbf{v}_2
\end{bmatrix}=
\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix} \text{ and } D=\begin{bmatrix}
1 & 0\\
0& 3
\end{bmatrix}.\]

Then the diagonalization procedure yields that $S$ is nonsingular and $S^{-1}AS= D$.


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1 Response

  1. 01/03/2018

    […] 1 end{bmatrix} text{ and } begin{bmatrix} -1 \ 1 end{bmatrix},] respectively. (See the post Diagonalize a 2 by 2 Symmetric Matrix for […]

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