Comprehensive Exercises on Eigenvalues and Eigenvectors

Approach: Follow the steps for finding eigenvalues and eigenvectors.

Steps:

1. The characteristic equation: $\lvert A - \lambda I\rvert = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} = 0$
2. Expanding: $(2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0$
3. Solving: $\lambda = 1$ or $\lambda = 3$
4. For $\lambda = 1$:
   • Solve the system $(A - I)\vec{x} = \vec{0}$
   • $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
   • This gives the eigenvector $\vec{x} = (1, -1)$
5. For $\lambda = 3$:
   • Solve the system $(A - 3I)\vec{x} = \vec{0}$
   • $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
   • This gives the eigenvector $\vec{x} = (1, 1)$

Answer: the eigenvalues are $\lambda_1 = 1$, $\lambda_2 = 3$; the corresponding eigenvectors are $\vec{x}_1 = (1, -1)$, $\vec{x}_2 = (1, 1)$

Approach: Check whether there are enough linearly independent eigenvectors.

Steps:

1. The characteristic equation: $\lvert A - \lambda I\rvert = (1-\lambda)^2 = 0$
2. The eigenvalue: $\lambda = 1$ (a double root)
3. Find the eigenvectors: solve $(A - I)\vec{x} = \vec{0}$
4. $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
5. This gives the eigenvector $\vec{x} = (1, 0)$
6. There is only one linearly independent eigenvector, fewer than the order of the matrix, which is 2

Answer: it is not diagonalisable

Approach: Use the properties of the eigenvalues.

Steps:

1. The product of the eigenvalues equals the determinant: $\lvert A\rvert = 1 \times 2 \times 3 = 6$
2. The sum of the eigenvalues equals the trace: $\operatorname{tr}A = 1 + 2 + 3 = 6$

Answer: $\lvert A\rvert = 6$, $\operatorname{tr}A = 6$

Approach: Check the symmetry of the matrix and its eigenvalues.

Steps:

1. Check symmetry: $A^T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = A$, so it is a real symmetric matrix
2. Find the eigenvalues: $\lvert A - \lambda I\rvert = \lambda^2 - 1 = 0$
3. The eigenvalues: $\lambda = 1$ or $\lambda = -1$, both real

Answer: it is a real symmetric matrix, and all its eigenvalues are real

Approach: Use the definition of diagonalisation.

Steps:

1. If $A$ is diagonalisable, then there exist an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^{-1}$
2. Here the columns of $P$ are linearly independent eigenvectors of $A$
3. The diagonal entries of $D$ are the eigenvalues of $A$

Answer: $A = PDP^{-1}$, where $D$ is a diagonal matrix and $P$ is the eigenvector matrix

Approach: Follow the steps for finding eigenvalues and eigenvectors.

Steps:

1. The characteristic equation: $\lvert A - \lambda I\rvert = \begin{vmatrix} 3-\lambda & 1 \\ 0 & 2-\lambda \end{vmatrix} = 0$
2. Expanding: $(3-\lambda)(2-\lambda) = 0$
3. Solving: $\lambda = 2$ or $\lambda = 3$
4. For $\lambda = 2$:
   • Solve the system $(A - 2I)\vec{x} = \vec{0}$
   • $\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
   • This gives the eigenvector $\vec{x} = (1, -1)$
5. For $\lambda = 3$:
   • Solve the system $(A - 3I)\vec{x} = \vec{0}$
   • $\begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
   • This gives the eigenvector $\vec{x} = (1, 0)$

Answer: the eigenvalues are $\lambda_1 = 2$, $\lambda_2 = 3$; the corresponding eigenvectors are $\vec{x}_1 = (1, -1)$, $\vec{x}_2 = (1, 0)$

Approach: Follow the steps of diagonalisation.

Steps:

1. The characteristic equation: $\lvert A - \lambda I\rvert = \lambda^2 - 4\lambda + 3 = 0$
2. The eigenvalues: $\lambda_1 = 1$, $\lambda_2 = 3$
3. The eigenvectors:
   • $\lambda_1 = 1$: $\vec{x}_1 = (1, -1)$
   • $\lambda_2 = 3$: $\vec{x}_2 = (1, 1)$
4. Construct the matrices:
   • $P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$
   • $D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$
5. The diagonalisation: $A = PDP^{-1}$

Answer: $A = PDP^{-1}$, where $P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ and $D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$

Approach: Follow the steps of orthogonal diagonalisation.

Steps:

1. The characteristic equation: $\lvert A - \lambda I\rvert = (3-\lambda)^2 - 1 = \lambda^2 - 6\lambda + 8 = 0$
2. The eigenvalues: $\lambda_1 = 2$, $\lambda_2 = 4$
3. The eigenvectors:
   • $\lambda_1 = 2$: $\vec{x}_1 = (1, -1)$
   • $\lambda_2 = 4$: $\vec{x}_2 = (1, 1)$
4. Normalisation:
   • $\vec{q}_1 = \frac{1}{\sqrt{2}}(1, -1)$
   • $\vec{q}_2 = \frac{1}{\sqrt{2}}(1, 1)$
5. Construct the matrices:
   • $Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$
   • $D = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$
6. The orthogonal diagonalisation: $A = QDQ^T$

Answer: $A = QDQ^T$, where $Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ and $D = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$

Approach: Check the eigenvalues or the leading principal minors.

Steps:

1. Method 1: find the eigenvalues
   • The characteristic equation: $\lvert A - \lambda I\rvert = \lambda^2 - 4\lambda + 3 = 0$
   • The eigenvalues: $\lambda_1 = 1$, $\lambda_2 = 3$, both positive
2. Method 2: check the leading principal minors
   • $\lvert A_1\rvert = 2 > 0$
   • $\lvert A_2\rvert = 3 > 0$

Answer: it is a positive definite matrix

Approach: Use the property that similar matrices have the same characteristic polynomial.

Steps:

1. Let $A \sim B$, that is, $B = P^{-1}AP$
2. Similar matrices have the same characteristic polynomial: $\lvert A - \lambda I\rvert = \lvert B - \lambda I\rvert$
3. The eigenvalues are the roots of the characteristic polynomial
4. Therefore $A$ and $B$ have the same eigenvalues

Answer: the proof is complete

Approach: Use the property of powers of a diagonalised matrix.

Steps:

1. $A = PDP^{-1}$
2. $A^n = (PDP^{-1})^n = PD^nP^{-1}$
3. $D^n = \operatorname{diag}(\lambda_1^n, \lambda_2^n, \dots, \lambda_n^n)$

Answer: $A^n = PD^nP^{-1}$, where $D^n = \operatorname{diag}(\lambda_1^n, \lambda_2^n, \dots, \lambda_n^n)$

Approach: Use a proof by contradiction.

Steps:

1. Suppose that the eigenvectors $\vec{x}_1, \vec{x}_2, \dots, \vec{x}_k$ corresponding to the distinct eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_k$ are linearly dependent
2. Then there exist constants $c_1, c_2, \dots, c_k$, not all zero, such that $c_1\vec{x}_1 + c_2\vec{x}_2 + \dots + c_k\vec{x}_k = \vec{0}$
3. Taking $c_1 \neq 0$, we get $\vec{x}_1 = -\frac{c_2}{c_1}\vec{x}_2 - \dots - \frac{c_k}{c_1}\vec{x}_k$
4. Left-multiplying by $A$: $A\vec{x}_1 = -\frac{c_2}{c_1}A\vec{x}_2 - \dots - \frac{c_k}{c_1}A\vec{x}_k$
5. That is, $\lambda_1\vec{x}_1 = -\frac{c_2}{c_1}\lambda_2\vec{x}_2 - \dots - \frac{c_k}{c_1}\lambda_k\vec{x}_k$
6. This contradicts the fact that the eigenvalues are distinct

Answer: the proof is complete

Approach: Follow the steps for finding eigenvalues and eigenvectors.

Steps:

1. The characteristic equation: $\lvert A - \lambda I\rvert = \begin{vmatrix} 1-\lambda & 1 \\ 0 & 1-\lambda \end{vmatrix} = 0$
2. Expanding: $(1-\lambda)^2 = 0$
3. Solving: $\lambda = 1$ (a double root)
4. For $\lambda = 1$:
   • Solve the system $(A - I)\vec{x} = \vec{0}$
   • $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
   • This gives the eigenvector $\vec{x} = (1, 0)$

Answer: the eigenvalue is $\lambda = 1$ (a double root); the corresponding eigenvector is $\vec{x} = (1, 0)$

Approach: Use the properties of real symmetric matrices.

Steps:

1. Let $\lambda$ be an eigenvalue of $A$ and $\vec{x}$ the corresponding eigenvector
2. $A\vec{x} = \lambda\vec{x}$
3. Take the conjugate transpose of both sides: $\vec{x}^*A^T = \lambda^*\vec{x}^*$
4. Since $A$ is real symmetric, $A^T = A$, so: $\vec{x}^*A = \lambda^*\vec{x}^*$
5. Right-multiply both sides by $\vec{x}$: $\vec{x}^*A\vec{x} = \lambda^*\vec{x}^*\vec{x}$
6. On the left: $\vec{x}^*A\vec{x} = \vec{x}^*\lambda\vec{x} = \lambda\vec{x}^*\vec{x}$
7. Therefore: $\lambda\vec{x}^*\vec{x} = \lambda^*\vec{x}^*\vec{x}$
8. Since $\vec{x} \neq \vec{0}$, we have $\vec{x}^*\vec{x} > 0$, so $\lambda = \lambda^*$, that is, $\lambda$ is real

Answer: the proof is complete

Approach: Use the properties of real symmetric matrices.

Steps:

1. All the eigenvalues of the real symmetric matrix $A$ are real, so it has a complete set of eigenvalues over the real numbers.
2. The eigenvectors of a real symmetric matrix belonging to distinct eigenvalues are necessarily mutually orthogonal; those belonging to the same eigenvalue can also be made orthogonal by Gram-Schmidt orthogonalisation.
3. Therefore $A$ has $n$ linearly independent orthogonal eigenvectors, which arranged as columns form an orthogonal matrix $Q$.
4. Then $Q^{-1}AQ$ is a diagonal matrix, which is exactly what orthogonal diagonalisation means.

Answer: the proof is complete