Definition: An \( n \times n \) matrix \( A \) is said to be diagonalizable if there exists an invertible matrix \( P \) (having an inverse) and a diagonal matrix \( D \) such that :
$$ A = P D P^{-1} $$Theorem: An \( n \times n \) square matrix \( A \) is diagonalizable if and only if it has \( n \) linearly independent eigenvectors. Matrix \( P \) consists of the \( n \) eigenvectors as its columns, and matrix \( D \) is a diagonal matrix whose entries are the corresponding eigenvalues of \( A \) .
Let \( A = \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} \) .
a) Find the eigenvalues of \( A \) and their corresponding eigenvectors .
b) Show that the eigenvectors form a basis for \( \mathbb{R}^2 \) .
c) Diagonalize matrix \( A \) if possible .
a) Eigenvalues and Eigenvectors
Find the eigenvalues using the characteristic polynomial \( \det(A - \lambda I) = 0 \) :
$$ \det \left( \begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = \det \begin{bmatrix} -1-\lambda & 2 \\ 0 & 1-\lambda \end{bmatrix} = (-1-\lambda)(1-\lambda) = 0 $$Solving gives the eigenvalues: \( \lambda_1 = -1 \) and \( \lambda_2 = 1 \) .
Eigenvector for \( \lambda_1 = -1 \): Solving \( (A + I)\mathbf{x} = 0 \):
$$ \begin{bmatrix} 0 & 2 \\ 0 & 2 \end{bmatrix} \sim \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \implies x_1 \text{ is free, } x_2 = 0 \implies \mathbf{x}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$Eigenvector for \( \lambda_2 = 1 \): Solving \( (A - I)\mathbf{x} = 0 \):
$$ \begin{bmatrix} -2 & 2 \\ 0 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} \implies x_2 \text{ is free, } x_1 = x_2 \implies \mathbf{x}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$b) Basis for \( \mathbb{R}^2 \)
To show the eigenvectors \( \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\} \) form a basis, we test for linear independence using an augmented matrix :
$$ \begin{bmatrix} 1 & 1 & | & 0 \\ 0 & 1 & | & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & | & 0 \\ 0 & 1 & | & 0 \end{bmatrix} $$The system has only the trivial solution \( \mathbf{0} \), confirming they are linearly independent and thus form a basis for \( \mathbb{R}^2 \) .
c) Diagonalization
Construct \( P \) from the eigenvectors and find \( P^{-1} \) :
$$ P = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \quad P^{-1} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} $$Form matrix \( D \) using the corresponding eigenvalues on the diagonal :
$$ A = P D P^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} $$Let \( A = \begin{bmatrix} 2 & -3 & 0 \\ 0 & -1 & 0 \\ 1 & 3 & 1 \end{bmatrix} \) .
a) Find the eigenvalues of \( A \) and their corresponding eigenvectors .
b) Show that the eigenvectors form a basis for \( \mathbb{R}^3 \) .
c) Diagonalize matrix \( A \) if possible .
a) Eigenvalues and Eigenvectors
Evaluating the characteristic polynomial \( \det(A - \lambda I) = (2-\lambda)(-1-\lambda)(1-\lambda) = 0 \), we find eigenvalues :
\( \lambda_1 = -1, \quad \lambda_2 = 1, \quad \lambda_3 = 2 \)
Solving the respective homogeneous systems for each eigenvalue yields the eigenvectors :
$$ \lambda_1 = -1 \implies \mathbf{x}_1 = \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix}, \quad \lambda_2 = 1 \implies \mathbf{x}_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad \lambda_3 = 2 \implies \mathbf{x}_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $$b) Basis for \( \mathbb{R}^3 \)
Row reducing the augmented matrix formed by the three eigenvectors shows they are linearly independent and form a basis for \( \mathbb{R}^3 \) .
c) Diagonalization
$$ P = \begin{bmatrix} -1 & 0 & 1 \\ -1 & 0 & 0 \\ 2 & 1 & 1 \end{bmatrix}, \quad P^{-1} = \begin{bmatrix} 0 & -1 & 0 \\ -1 & 3 & 1 \\ 1 & -1 & 0 \end{bmatrix} $$ $$ A = P D P^{-1} = \begin{bmatrix} -1 & 0 & 1 \\ -1 & 0 & 0 \\ 2 & 1 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 & 0 \\ -1 & 3 & 1 \\ 1 & -1 & 0 \end{bmatrix} $$Find the eigenvalues and corresponding eigenvectors of matrix \( A = \begin{bmatrix} 2 & 0 & 4 \\ 2 & -1 & 3 \\ -2 & 1 & -3 \end{bmatrix} \) and diagonalize it if possible .
Solving the characteristic polynomial \( -\lambda^3 - 2\lambda^2 = 0 \), we obtain eigenvalues :
\( \lambda_1 = 0 \) (multiplicity 2) and \( \lambda_3 = -2 \)
For \( \lambda = 0 \), row-reducing \( A - 0I \) yields only one linearly independent eigenvector: \( \mathbf{x} = \begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix} \) .
Because the eigenvalue \( \lambda = 0 \) has an algebraic multiplicity of 2 but only 1 associated linearly independent eigenvector, the vectors cannot span \( \mathbb{R}^3 \). Therefore, matrix \( A \) is not diagonalizable .
Knowing that \( 1 \) and \( 3 \) are eigenvalues of matrix \( A = \begin{bmatrix} 2 & 0 & 0 & 1 \\ 0 & 0 & 4 & 0 \\ 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 2 \end{bmatrix} \), find all eigenvalues and eigenvectors, and diagonalize it .
Calculate \( A^{10} \) where \( A \) is the matrix from Example 4 .
Using the property \( A^n = P D^n P^{-1} \), we raise the diagonal matrix entries to the 10th power :
$$ D^{10} = \begin{bmatrix} (-2\sqrt{3})^{10} & 0 & 0 & 0 \\ 0 & 1^{10} & 0 & 0 \\ 0 & 0 & 3^{10} & 0 \\ 0 & 0 & 0 & (2\sqrt{3})^{10} \end{bmatrix} = \begin{bmatrix} 248832 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 59049 & 0 \\ 0 & 0 & 0 & 248832 \end{bmatrix} $$Multiplying \( P D^{10} P^{-1} \) yields :
$$ A^{10} = \begin{bmatrix} 29525 & 0 & 0 & 29524 \\ 0 & 248832 & 0 & 0 \\ 0 & 0 & 248832 & 0 \\ 29524 & 0 & 0 & 29525 \end{bmatrix} $$Note: These exercises demonstrate that matrix invertibility and matrix diagonalization are independent properties.
Part 1: Show that matrix \( A = \begin{bmatrix} 4 & -1 \\ 1 & 2 \end{bmatrix} \) is invertible but not diagonalizable .
Part 2: Show that matrix \( A = \begin{bmatrix} -1 & -2 \\ 0 & 0 \end{bmatrix} \) is not invertible (singular) but diagonalizable .
Part 3: Show that matrix \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \) is neither invertible nor diagonalizable .
Part 4: Show that matrix \( A = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} \) is both invertible and diagonalizable .
Solution to Part 1:
\( \det(A) = 9 \neq 0 \), so \( A \) is invertible . However, eigenvalue \( \lambda = 3 \) has multiplicity 2 but yields only one independent eigenvector, so it is not diagonalizable .
Solution to Part 2:
\( \det(A) = 0 \), making it not invertible . Its eigenvalues are \( \lambda = -1, 0 \) with two independent eigenvectors, making it diagonalizable with \( A = P D P^{-1} \) :
$$ \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} $$Solution to Part 3:
\( \det(A) = 0 \) (not invertible) . Eigenvalue \( \lambda = 0 \) has multiplicity 2 with only one independent eigenvector, so it is not diagonalizable .
Solution to Part 4:
\( \det(A) = -1 \neq 0 \) (invertible) . Eigenvalues \( \lambda = -1, 1 \) yield two independent eigenvectors \( \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \end{bmatrix} \), making it diagonalizable :
$$ A = \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -\frac{1}{2} \\ 0 & \frac{1}{2} \end{bmatrix} $$