Problem 29 - Differential Equations and Linear Algebra

Question

The matrix $$ A=\left[\begin{array}{lll} 2 & -2 & 3 \\ 1 & -1 & 3 \\ 1 & -2 & 4 \end{array}\right] $$ has eigenvalues \(\lambda_{1}=1\) and \(\lambda_{2}=3\) (a) Determine a basis for the eigenspace \(E_{1}\) corresponding to \(\lambda_{1}=1\) and then use the GramSchmidt procedure to obtain an orthogonal basis for \(E_{1}\) (b) Are the vectors in \(E_{1}\) orthogonal to the vectors in \(E_{2},\) the eigenspace corresponding to \(\lambda_{2}=3 ?\)

Step-by-Step Solution

Answer

The orthogonal basis for eigenspace E₁ is the set: \[ \left\{ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\} \] The vectors in eigenspace E₁ are not orthogonal to the vectors in eigenspace E₂.

Step 1: Find eigenvectors corresponding to eigenvalues

To find the eigenvectors corresponding to the eigenvalue λ₁=1, we need to solve the following equation for the vector x: \[ (A - \lambda_{1}I)x = 0 \] with A being the given matrix and I being the identity matrix. Plug in the values for A and λ₁: \[ \left[\begin{array}{rrr} 2 - 1 & -2 & 3 \\ 1 & -1 -1 & 3 \\ 1 & -2 & 4 - 1 \end{array}\right]\left[\begin{array}{r} x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{r} 0\\0\\0\end{array}\right] \]

Step 2: Solve for eigenvectors

Now, we need to row reduce the augmented matrix. The matrix is: \[ \left[\begin{array}{rrr|r} 1 & -2 & 3 & 0 \\ 1 & -2 & 3 & 0 \\ 1 & -2 & 3 & 0 \end{array}\right] \] Since all rows are the same, we can remove rows 2 and 3 and still have an equivalent matrix: \[ \left[\begin{array}{rrr|r} 1 & -2 & 3 & 0 \\ \end{array}\right] \] x₁ - 2x₂ + 3x₃ = 0. Solving for x₁: x₁ = 2x₂ - 3x₃. Therefore, the eigenvectors corresponding to λ₁=1 are: \[ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} x_2 + \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}x3 \]

Step 3: Apply Gram-Schmidt procedure to eigenvectors

The eigenvectors are linearly independent, so we can form a basis for E₁ using these vectors: \[E_1 = \left\{ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\}\] Now, apply GramSchmidt to these vectors. Since the first vector is already non-zero, we can use it as it is: \[u_1 = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}\] Next, subtract its projection onto u₁ from the second vector to get u₂: \[u_2 = \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} - \frac{\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\cdot u_1}{u_1 \cdot u_1}u_1\] After calculation, you find that: \[u_2 = \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\] Thus, the orthogonal basis for E₁ using the Gram-Schmidt procedure is: \[ \left\{ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\} \] (a) answer: The orthogonal basis for eigenspace E₁ is the set \[ \left\{ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\} \]

Step 4: Check if E₁ is orthogonal to E₂

First, we need to find the eigenvectors for λ₂=3. Similar to step 1, plug in the values for A and λ₂: \[ \left[\begin{array}{rrr} 2 - 3 & -2 & 3 \\ 1 & -1 -( 3) & 3 \\ 1 & -2 & 4 - 3 \end{array}\right]\left[\begin{array}{r} x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{r} 0\\0\\0\end{array}\right] \] Solving the equation, we obtain the eigenvector corresponding to λ₂=3: \[ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \] Since E₂ has only one vector, we can check if it is orthogonal to the vectors in E₁: \[ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} = 3 \neq 0 \] So the vectors in E₁ are not orthogonal to the vectors in E₂. (b) answer: The vectors in eigenspace E₁ are not orthogonal to the vectors in eigenspace E₂.

Key Concepts

Eigenspace

Eigenspaces are a fundamental concept when dealing with matrices and linear transformations. They consist of all eigenvectors corresponding to a given eigenvalue, along with the zero vector. This space thus represents all directions where the transformation induced by the matrix acts like mere scaling by the eigenvalue. If a matrix has an eigenvalue, the associated eigenspace is a vector space itself. For our matrix example, the eigenspace corresponding to eigenvalue \( \lambda_1 = 1 \) is denoted as \( E_1 \), and is determined by solving the eigenvector equation \( (A - \lambda_1 I)x = 0 \), where \( A \) is the matrix, \( \lambda_1 \) is the eigenvalue, and \( I \) is the identity matrix. The solutions, in this case, form the set of all possible linear combinations of the found eigenvectors for \( \lambda_1 \). This means they span the eigenspace \( E_1 \). Understanding eigenspaces is crucial as they can reveal important structural properties of transformations represented by the matrix.

Orthogonal Basis

An orthogonal basis for a vector space is a set of vectors that are mutually perpendicular to each other, simplifying many calculations, including projections and expansions of vectors. When constructing an orthogonal basis for an eigenspace, like \( E_1 \) with eigenvectors \( \begin{bmatrix} 2 & 1 & 0 \end{bmatrix} \) and \( \begin{bmatrix} -3 & 0 & 1 \end{bmatrix} \), our goal is to develop a set of vectors that maintains span while ensuring mutual orthogonality. This process allows property advantageous for decomposing vectors and simplifying the matrix-representation of transformations. Because orthogonal vectors essentially operate independently, calculations such as inner products and projections become trivial tasks, greatly aiding in solving complex linear algebra systems.

Gram-Schmidt Procedure

The Gram-Schmidt Procedure is a method used to transform a set of vectors into an orthogonal set while maintaining their span within the space. It is particularly useful in linear algebra for ensuring that a basis is orthogonal, which simplifies many calculations. The process involves iteratively subtracting the projections of a vector onto other vectors in your recursively forming orthogonal set.

Given a set of linearly independent vectors, such as \( \begin{bmatrix} 2 & 1 & 0 \end{bmatrix} \) and \( \begin{bmatrix} -3 & 0 & 1 \end{bmatrix} \), the first vector \( u_1 \) remains unchanged. For the second vector, \( u_2 \), subtract its projection onto \( u_1 \) to make it orthogonal to \( u_1 \). The projection is calculated as the dot product of the second vector with \( u_1 \), divided by the dot product of \( u_1 \) with itself, and then multiplied by \( u_1 \). This forms a new orthogonal vector \( u_2 \), if calculated correctly. This method provides an orthogonal basis that makes vector operations much more manageable. The result is a set of vectors ready for extensive computational applications while maintaining the structural properties of the original space.