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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 \(\lambda_{2}=3 ?\)

Short Answer

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

Step by step solution

01

Find eigenvectors corresponding to eigenvalues

To find the eigenvectors corresponding to the eigenvalue 位鈧=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 位鈧: \[ \left[\begin{array}{rrr} 2 - 1 & -2 & 3 \\ 1 & -1 -1 & 3 \\ 1 & -2 & 4 - 1 \end{array}\right]\left[\begin{array}{r} x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{r} 0\\0\\0\end{array}\right] \]
02

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: \[ \left[\begin{array}{rrr|r} 1 & -2 & 3 & 0 \\ \end{array}\right] \] x鈧 - 2x鈧 + 3x鈧 = 0. Solving for x鈧: x鈧 = 2x鈧 - 3x鈧. Therefore, the eigenvectors corresponding to 位鈧=1 are: \[ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} x_2 + \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}x3 \]
03

Apply Gram-Schmidt procedure to eigenvectors

The eigenvectors are linearly independent, so we can form a basis for E鈧 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 u鈧 from the second vector to get u鈧: \[u_2 = \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} - \frac{\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\cdot u_1}{u_1 \cdot u_1}u_1\] After calculation, you find that: \[u_2 = \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}\] Thus, the orthogonal basis for E鈧 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 E鈧 is the set \[ \left\{ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\} \]
04

Check if E鈧 is orthogonal to E鈧

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Use some form of technology to find a complete set of orthonormal eigenvectors for \(A\) and an orthogonal matrix \(S\) and a diagonal matrix \(D\) such that \(S^{-1} A S=D\). $$A=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right].$$

Use some form of technology to determine the eigenvalues and eigenvectors of \(A\) in the following manner: (1) Form the matrix \(A-\lambda I.\) (2) Solve the characteristic equation \(\operatorname{det}(A-\lambda I)=0\) to determine the eigenvalues of \(A.\) (3) For each eigenvalue \(\lambda_{i}\) found in \((2),\) solve the system \(\left(A-\lambda_{i} I\right) \mathbf{v}=\mathbf{0}\) to determine the eigenvectors of \(A.\) $$\diamond A=\left[\begin{array}{lll}5 & 34 & -41 \\\4 & 17 & -23 \\\5 & 24 & -31 \end{array}\right]$$

Use some form of technology to determine the eigenvalues and eigenvectors of \(A\) in the following manner: (1) Form the matrix \(A-\lambda I.\) (2) Solve the characteristic equation \(\operatorname{det}(A-\lambda I)=0\) to determine the eigenvalues of \(A.\) (3) For each eigenvalue \(\lambda_{i}\) found in \((2),\) solve the system \(\left(A-\lambda_{i} I\right) \mathbf{v}=\mathbf{0}\) to determine the eigenvectors of \(A.\) $$\diamond A=\left[\begin{array}{ll}3 & 1 \\\2 & 4\end{array}\right]$$

Use some form of technology to find a complete set of orthonormal eigenvectors for \(A\) and an orthogonal matrix \(S\) and a diagonal matrix \(D\) such that \(S^{-1} A S=D\). $$A=\left[\begin{array}{rrr} 0 & -1 & 4 \\ -1 & 5 & 2 \\ 4 & 2 & 2 \end{array}\right].$$

Use Jordan canonical forms to determine whether the given pair of matrices are similar. \(A=\left[\begin{array}{rrr}3 & 0 & 4 \\ 0 & 2 & 0 \\ -4 & 0 & -5\end{array}\right] ; B=\left[\begin{array}{rrr}-1 & -1 & 3 \\ 0 & -1 & 1 \\\ 0 & 0 & 2\end{array}\right]\).

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