Diagonalization of Matrix

Diagonalization of 2×2 Matrix using Similarity (SAS⁻¹ / S⁻¹AS)

Diagonalization of a 2×2 Matrix using Similarity Transformation

Method: Choose eigenvectors as columns of \(S\), then compute \(D = S^{-1}AS\) (equivalently \(A = SDS^{-1}\)).

Given
\[ A=\begin{pmatrix}5&4\\[2pt]1&2\end{pmatrix} \]

We will diagonalize \(A\) using \(D=S^{-1}AS\), where \(S\) is formed from eigenvectors.

Step 1

Find the eigenvalues

Characteristic equation: \(\det(A-\lambda I)=0\)

\[ \det(A-\lambda I)= \begin{vmatrix} 5-\lambda & 4\\ 1 & 2-\lambda \end{vmatrix} =(5-\lambda)(2-\lambda)-4 \]
\[ =(10-7\lambda+\lambda^2)-4=\lambda^2-7\lambda+6 \]
\[ \lambda^2-7\lambda+6=0 \quad\Rightarrow\quad (\lambda-6)(\lambda-1)=0 \]
\[ \boxed{\lambda_1=6,\qquad \lambda_2=1} \]
Step 2

Find the eigenvectors

For \(\lambda=6\)

\[ A-6I=\begin{pmatrix}-1&4\\[2pt]1&-4\end{pmatrix} \]

Solve \((A-6I)\begin{pmatrix}x\\y\end{pmatrix}=0\):

\[ -x+4y=0\Rightarrow x=4y \]

Choose \(y=1\Rightarrow x=4\):

\[ \boxed{v_1=\begin{pmatrix}4\\[2pt]1\end{pmatrix}} \]

For \(\lambda=1\)

\[ A-I=\begin{pmatrix}4&4\\[2pt]1&1\end{pmatrix} \]

Solve \((A-I)\begin{pmatrix}x\\y\end{pmatrix}=0\):

\[ x+y=0\Rightarrow x=-y \]

Choose \(y=1\Rightarrow x=-1\):

\[ \boxed{v_2=\begin{pmatrix}-1\\[2pt]1\end{pmatrix}} \]
Step 3

Form the similarity matrix \(S\) and find \(S^{-1}\)

Put eigenvectors as columns (order matched with eigenvalues \(6,1\)):

\[ S=\big[v_1\;v_2\big] = \begin{pmatrix} 4 & -1\\[2pt] 1 & 1 \end{pmatrix} \]

Check determinant:

\[ \det(S)=4\cdot1-(-1)\cdot1=5\neq 0 \]

Therefore \(S\) is invertible and diagonalization is possible.

Inverse of a \(2\times2\) matrix:

\[ \begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1} =\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix} \]

So,

\[ S^{-1}=\frac{1}{5}\begin{pmatrix}1&1\\[2pt]-1&4\end{pmatrix} \]
Step 4

Compute the diagonal matrix \(D=S^{-1}AS\)

Since columns of \(S\) are eigenvectors, \(S^{-1}AS\) becomes a diagonal matrix with eigenvalues on the diagonal:

\[ \boxed{ D=S^{-1}AS= \begin{pmatrix}6&0\\[2pt]0&1\end{pmatrix} } \]

Diagonalization statement:

\[ \boxed{ A = S D S^{-1} } \]
\[ A= \begin{pmatrix}4&-1\\[2pt]1&1\end{pmatrix} \begin{pmatrix}6&0\\[2pt]0&1\end{pmatrix} \frac{1}{5}\begin{pmatrix}1&1\\[2pt]-1&4\end{pmatrix} \]

Final answers: eigenvalues \(6\) and \(1\); diagonal matrix \(D=\mathrm{diag}(6,1)\).

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