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Determine the invariant subspace of \(A=\left[\begin{array}{ll}2 & -4 \\ 5 & -2\end{array}\right]\) viewed as a linear operator on (a) \(\mathbf{R}^{2},(\mathrm{b}) \mathbf{C}^{2}\).

Short Answer

Expert verified
For the linear operator A, there are no invariant subspaces in 鈩澛 since the eigenvectors are not real. However, in 鈩偮, there are two invariant subspaces: 1. The line spanned by the eigenvector associated with eigenvalue 位鈧, given by x鈧 = \((2 - 2\sqrt{2})x鈧乗) / 4. 2. The line spanned by the eigenvector associated with eigenvalue 位鈧, given by x鈧 = \((2 + 2\sqrt{2})x鈧乗) / 4.

Step by step solution

01

Find the eigenvalues of A

To find the eigenvalues of A, we need to solve the characteristic equation, given by det(A - 位I) = 0, where 位 is an eigenvalue and I is the identity matrix. For A = \(\begin{bmatrix} 2 & -4\\ 5 & -2 \end{bmatrix}\), we get: \(|A - 位I| = \begin{vmatrix} 2 - 位 & -4 \\ 5 & -2 - 位 \end{vmatrix} = (2 - 位)(-2 - 位) - (-4)(5) = 位虏 - 8\)
02

Solve for the eigenvalues 位

To determine the eigenvalues, we solve the equation 位虏 - 8 = 0. 位虏 - 8 = (位 + 2\(\sqrt{2}\))(位 - 2\(\sqrt{2}\)) = 0 Therefore, we have two eigenvalues: 位鈧 = 2\(\sqrt{2}\) and 位鈧 = -2\(\sqrt{2}\).
03

Find the eigenvectors of A

For each eigenvalue, we want to solve (A - 位I)x = 0, where x is an eigenvector. For 位鈧 = 2\(\sqrt{2}\): \( (A - 位鈧両)x = \begin{bmatrix} 2 - 2\sqrt{2} & -4 \\ 5 & -2 + 2\sqrt{2} \end{bmatrix} \begin{bmatrix} x鈧 \\ x鈧 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) The system of equations can be simplified to: \( (2 - 2\sqrt{2})x鈧 - 4x鈧 = 0 \) \( 5x鈧 + (2\sqrt{2} - 2)x鈧 = 0 \) Any non-zero solution to this system is an eigenvector, observing that, x鈧 = \((2 - 2\sqrt{2})x鈧乗) / 4 is a valid eigenvector. For 位鈧 = -2\(\sqrt{2}\): \( (A - 位鈧侷)x = \begin{bmatrix} 2 + 2\sqrt{2} & -4 \\ 5 & -2 - 2\sqrt{2} \end{bmatrix} \begin{bmatrix} x鈧 \\ x鈧 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) This system of equations also leads to x鈧 = \((2 + 2\sqrt{2})x鈧乗) / 4 being a valid eigenvector.
04

Determine the invariant subspaces

The eigenvectors form the invariant subspaces for the linear operator A. (a) When A acts on 鈩澛: The eigenvectors corresponding to 位鈧 and 位鈧 are not real, so there are no invariant subspaces in 鈩澛. (b) When A acts on 鈩偮: The eigenvectors form two distinct lines: Invariant subspace 1: The line spans by the eigenvector associated with eigenvalue 位鈧, given by x鈧 = \((2 - 2\sqrt{2})x鈧乗) / 4. Invariant subspace 2: The line spans by the eigenvector associated with eigenvalue 位鈧, given by x鈧 = \((2 + 2\sqrt{2})x鈧乗) / 4. These two lines are the invariant subspaces of A acting as a linear operator on 鈩偮.

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Key Concepts

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

Eigenvalues
When we speak about eigenvalues, we're referring to special numbers associated with a matrix or a linear transformation鈥攊n this case, the matrix \(A\). These numbers give us insight into the behavior of the transformation. To find them, we solve the characteristic equation \(\text{det}(A - \lambda I) = 0\), where \(I\) represents the identity matrix and \(\lambda\) the eigenvalues we seek.
In our example, the characteristic equation simplifies to \(\lambda^2 - 8 = 0\). Solving gives us the eigenvalues \(\lambda_1 = 2\sqrt{2}\) and \(\lambda_2 = -2\sqrt{2}\). These hint at the directions along which our transformation affects the vector space most distinctly. Understanding eigenvalues is crucial as they relate closely to invariant subspaces, helping in predicting the transformation's effects.
Eigenvectors
Eigenvectors accompany eigenvalues and represent the directions in which the linear transformation stretches or compresses the space. Calculating them involves solving \((A - \lambda I)\mathbf{x} = 0\), where \(\mathbf{x}\) is the eigenvector of interest.
For our matrix \(A\) with eigenvalues \(\lambda_1 = 2\sqrt{2}\) and \(\lambda_2 = -2\sqrt{2}\), we solve separately for each \(\lambda\). The solutions \(x_2 = \frac{(2 - 2\sqrt{2})x_1}{4}\) and \(x_2 = \frac{(2 + 2\sqrt{2})x_1}{4}\) describe the eigenvectors.
These vectors pinpoint directions that remain unchanged in direction under transformation by \(A\), highlighting the shape and orientation of invariant subspaces.
Linear Operator
A linear operator can be thought of as a function mapping vectors to vectors in a linear manner. It means if you apply the operator to a combination of vectors, the result is the same as applying the operator to each vector individually and combining the results. In mathematical terms, if \(T\) is our linear operator, then \(T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v})\) for any scalars \(a, b\) and vectors \(\mathbf{u}, \mathbf{v}\).
Our matrix \(A = \begin{bmatrix} 2 & -4 \ 5 & -2 \end{bmatrix}\) acts as a linear operator, transforming input vectors. The way it transforms vectors is encapsulated by its eigenvalues and eigenvectors, and profoundly affects the concept of invariant subspaces. This transformation is explored in both real and complex vector spaces, revealing more structure in the latter due to the nature of complex numbers.
Complex Vector Space
A complex vector space extends the idea of real vector spaces by allowing vectors to have complex numbers as components. This flexibility is fundamental in finding solutions to certain equations, like differential equations, that might not have real-number solutions.
In our exercise, the matrix \(A\) exhibited no real invariant subspaces when acting on \(\mathbb{R}^2\), due to its complex eigenvalues and eigenvectors. However, when considered in \(\mathbb{C}^2\), it shows two invariant subspaces corresponding to its complex eigenvectors.
  • Invariant Subspace 1: Defined by the eigenvector related to \(\lambda_1 = 2\sqrt{2}\), with the relational equation \(x_2 = \frac{(2 - 2\sqrt{2})x_1}{4}\).
  • Invariant Subspace 2: Defined by the eigenvector linked to \(\lambda_2 = -2\sqrt{2}\), with \(x_2 = \frac{(2 + 2\sqrt{2})x_1}{4}\).

Understanding the complex vector space's role is crucial for delving deeper into linear algebra, facilitating solutions that real numbers alone can't provide.

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Most popular questions from this chapter

Show that two nilpotent matrices of order 3 are similar if and only if they have the same index of nilpotency. Show by example that the statement is not true for nilpotent matrices of order 4.

Find the rational canonical form for the four-square Jordan block with \(\lambda^{\prime}\) s on the diagonal.

Determine all possible Jordan canonical forms for a linear operator \(T: V \rightarrow V\) whose characteristic polynomial \(\Delta(t)=(t-2)^{3}(t-5)^{2} .\) In each case, find the minimal polynomial \(m(t)\) Because \(t-2\) has exponent 3 in \(\Delta(t), 2\) must appear three times on the diagonal. Similarly, 5 must appear twice. Thus, there are six possibilities: (a) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right], \quad\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)\) (b) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right], \quad[5], \quad[5]\right)\) (c) \(\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5 & 1 \\ 5\end{array}\right]\right)\) (d) \(\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\\ 2\end{array}\right],\left[\begin{array}{ll}2\end{array}\right],\left[\begin{array}{ll}5\end{array}\right], \quad[5]\right)\) (e) \(\operatorname{diag}\left([2],[2],[2],\left[\begin{array}{ll}5 & 1 \\\ 5\end{array}\right]\right)\) (f) \(\operatorname{diag}([2],[2],[2],[5],[5])\) The exponent in the minimal polynomial \(m(t)\) is equal to the size of the largest block. Thus, (a) \(m(t)=(t-2)^{3}(t-5)^{2}\) (b) \(m(t)=(t-2)^{3}(t-5)\) (c) \(\quad m(t)=(t-2)^{2}(t-5)^{2}\) (d) \(m(t)=(t-2)^{2}(t-5)\) (e) \(m(t)=(t-2)(t-5)^{2}\) \((\mathrm{f}) \quad m(t)=(t-2)(t-5)\)

Suppose \(T: V \rightarrow V\) is linear. Show that each of the following is invariant under \(T\) (a) \\{0\\} (b) \(V\) (c) kernel of \(T\) (d) image of \(T\) (a) We have \(T(0)=0 \in\\{0\\} ;\) hence, \\{0\\} is invariant under \(T\) (b) For every \(v \in V, T(v) \in V ;\) hence, \(V\) is invariant under \(T\) (c) Let \(u \in\) Ker \(T\). Then \(T(u)=0 \in\) Ker \(T\) because the kemel of \(T\) is a subspace of \(V\). Thus, Ker \(T\) is invariant under \(T\) (d) Because \(T(v) \in \operatorname{Im} T\) for every \(v \in V\), it is certainly true when \(v \in \operatorname{Im} T .\) Hence, the image of \(T\) is invariant under \(T\)

Show that every subspace of \(V\) is invariant under \(I\) and 0 , the identity and zero operators.

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