Determine the \(2 \times 2\) matrix \(A\) such that \(A \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ -6 \end{bmatrix}\) and \(A \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\). Prove that the matrix \(A\) is not diagonalizable.
Solution
1. \(A = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix}\).
1.1. The second column of \(A\) is \(A e_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).
\(e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\).
1.2. The first column of \(A\) is \(A e_1 = A\begin{bmatrix} 1 \\ 2 \end{bmatrix} - 2A e_2 = \begin{bmatrix} 5 \\ -6 \end{bmatrix} - 2\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 3 \\ -4 \end{bmatrix}\).
\(\begin{bmatrix} 1 \\ 2 \end{bmatrix} = e_1 + 2e_2\), so \(A e_1 = A\begin{bmatrix} 1 \\ 2 \end{bmatrix} - 2A e_2\).
1.3. Hence \(A = \begin{bmatrix} 3 & 1 \\ -4 & -1 \end{bmatrix}\).
1 and 2 give the two columns.
2. The characteristic polynomial of \(A\) is \((t-1)^2\).
\(\det(tI - A) = \det\begin{bmatrix} t-3 & -1 \\ 4 & t+1 \end{bmatrix} = (t-3)(t+1) + 4 = t^2 - 2t + 1 = (t-1)^2\).
3. The eigenspace for \(\lambda = 1\) has dimension \(1\).
3.1. \(A - I = \begin{bmatrix} 2 & 1 \\ -4 & -2 \end{bmatrix}\).
subtract \(I\).
3.2. \(\operatorname{rank}(A - I) = 1\).
the two rows are scalar multiples of each other.
3.3. Hence \(\dim \ker(A - I) = 2 - 1 = 1\).
rank–nullity theorem.
4. \(A\) is not diagonalizable.
4.1. \(A\) has a single eigenvalue \(\lambda = 1\) of algebraic multiplicity \(2\).
2.
4.2. The geometric multiplicity of \(\lambda = 1\) is \(1\).
3.
4.3. A matrix is diagonalizable iff for each eigenvalue the geometric multiplicity equals the algebraic multiplicity.
standard criterion.
4.4. Hence \(A\) is not diagonalizable.
<2>1–<2>3, since \(1 \neq 2\).
5. Q.E.D.
1 and 4.