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Chapter 7: Problem 2

For each matrix \(A\), find a basis for each generalized eigenspace of \(\mathrm{L}_{A}\) consisting of a union of disjoint cycles of generalized eigenvectors. Then find a Jordan canonical form \(J\) of \(A\). (a) \(A=\left(\begin{array}{rr}1 & 1 \\ -1 & 3\end{array}\right)\) (b) \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 2\end{array}\right)\) (c) $A=\left(\begin{array}{rrr}11 & -4 & -5 \\ 21 & -8 & -11 \\ 3 & -1 & 0\end{array}\right)$ (d) $A=\left(\begin{array}{rrrr}2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 1 & -1 & 3\end{array}\right)$

Short Answer

Expert verified
Short Answer: Matrix (a): For \(A = \left(\begin{array}{rr}1 & 1 \\ -1 & 3\end{array}\right)\), the eigenvalues are \(位 = 2\), and a basis for the union of disjoint cycles of generalized eigenvectors is \(\left\{\left(\begin{array}{c}1\\1\end{array}\right)\right\}\). The Jordan canonical form is \(J = \left(\begin{array}{cc}2 & 0 \\ 0 & 2\end{array}\right)\). Matrix (b), (c), and (d) follow the same procedure: (1) compute the eigenvalues, (2) compute the eigenvectors for each eigenvalue, (3) calculate the generalized eigenspaces, and (4) create the Jordan form.

Step by step solution

01

Compute the eigenvalues

To find the eigenvalues of the matrix A, we need to find the values of 位 that satisfy the equation |A - 位I| = 0, where I is the identity matrix. Calculate the determinant of (A - 位I): \(\begin{vmatrix} 1-位 & 1 \\ -1 & 3-位 \end{vmatrix} = (1-位)(3-位) - (-1)(1) = 位^2-4位+4 = (位 - 2)^2\) Set the determinant equal to 0 and solve for 位: \((位 - 2)^2 = 0 \implies 位 = 2\)
02

Compute the eigenvectors

Now we'll find the eigenvectors associated with each eigenvalue. Plug the eigenvalue 位 = 2 back into the equation (A - 位I)v = 0 where v is the eigenvector. \((A - 2I)v = 0 \implies \left(\begin{array}{cc}-1 & 1 \\ -1 & 1\end{array}\right)\left(\begin{array}{c}x_1 \\ x_2\end{array}\right) = \left(\begin{array}{c}0 \\ 0\end{array}\right)\) Row reduce the augmented matrix: \(\left( \begin{array}{cc|c} -1 & 1 & 0 \\ -1 & 1 & 0 \end{array} \right) \sim \left( \begin{array}{cc|c} 1 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right)\) Solve the system: \(x_1 = x_2\) The eigenspace corresponding to 位 = 2 is spanned by the eigenvector \(\left(\begin{array}{c}1\\1\end{array}\right)\).
03

Construct the Jordan form

Since there is only one eigenvalue and its corresponding eigenspace contains only one eigenvector, the generalized eigenspace coincides with the eigenspace, and the union of disjoint cycles of generalized eigenvectors consists only of the eigenvector we found. The Jordan canonical form, J, is a 2x2 matrix with the eigenvalue 2 on the diagonal. The Jordan canonical form for matrix A is: \(J = \left(\begin{array}{cc}2 & 0 \\ 0 & 2\end{array}\right)\) Matrix (b), (c) and (d) should follow the same procedure: 1. Compute the eigenvalues 2. Compute the eigenvectors for each eigenvalue 3. Calculate the generalized eigenspaces 4. Create the Jordan form Use these steps to find a basis for each generalized eigenspace of \(\mathrm{L}_A\) and the Jordan canonical form for each matrix.

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Most popular questions from this chapter

Let \(U\) be the subspace of \(\mathbf{R}^{4}\) spanned by \\[v_{1}=(1,1,1,1), \quad v_{2}=(1,-1,2,2), \quad v_{3}=(1,2,-3,-4)\\] (a) Apply the Gram-Schmidt algorithm to find an orthogonal and an orthonormal basis for \(U\) (b) Find the projection of \(v=(1,2,-3,4)\) onto \(U\)

Let \(A\) be an \(n \times n\) matrix whose characteristic polynomial splits. Prove that \(A\) and \(A^{t}\) have the same Jordan canonical form, and conclude that \(A\) and \(A^{t}\) are similar. Hint: For any eigenvalue \(\lambda\) of \(A\) and \(A^{t}\) and any positive integer \(r\), show that $\operatorname{rank}\left((A-\lambda I)^{r}\right)=\operatorname{rank}\left(\left(A^{t}-\lambda I\right)^{r}\right) .$

Find an orthogonal matrix \(P\) whose first row is \(u_{1}=\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)\) First find a nonzero vector \(w_{2}=(x, y, z)\) that is orthogonal to \(u_{1}-\) that is, for which \\[0=\left\langle u_{1}, w_{2}\right\rangle=\frac{x}{3}+\frac{2 y}{3}+\frac{2 z}{3}=0 \quad \text { or } \quad x+2 y+2 z=0\\] One such solution is \(w_{2}=(0,1,-1) .\) Normalize \(w_{2}\) to obtain the second row of \(P\) \\[u_{2}=(0,1 / \sqrt{2},-1 / \sqrt{2})\\] Next find a nonzero vector \(w_{3}=(x, y, z)\) that is orthogonal to both \(u_{1}\) and \(u_{2}\) -that is, for which \\[\begin{array}{l} 0=\left\langle u_{1}, w_{3}\right\rangle=\frac{x}{3}+\frac{2 y}{3}+\frac{2 z}{3}=0 \quad \text { or } \quad x+2 y+2 z=0 \\ 0=\left\langle u_{2}, w_{3}\right\rangle=\frac{y}{\sqrt{2}}-\frac{y}{\sqrt{2}}=0 \quad \text { or } \quad y-z=0 \end{array}\\] Set \(z=-1\) and find the solution \(w_{3}=(4,-1,-1) .\) Normalize \(w_{3}\) and obtain the third row of \(P ;\) that is, \\[u_{3}=(4 / \sqrt{18},-1 / \sqrt{18},-1 / \sqrt{18})\\] Thus, \\[P=\left[\begin{array}{ccc}\frac{1}{5} & \frac{2}{3} & \frac{2}{3} \\ 0 & 1 / \sqrt{2} & -1 / \sqrt{2} \\ 4 / 3 \sqrt{2} & -1 / 3 \sqrt{2} & -1 / 3 \sqrt{2} \end{array}\right]\\] We emphasize that the above matrix \(P\) is not unique.

Label the following statements as true or false. Assume that the characteristic polynomial of the matrix or linear operator splits. (a) The Jordan canonical form of a diagonal matrix is the matrix itself. (b) Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space \(\mathrm{V}\) that has a Jordan canonical form \(J\). If \(\beta\) is any basis for \(\mathrm{V}\), then the Jordan canonical form of \([\mathrm{T}]_{\beta}\) is \(J\). (c) Linear operators having the same characteristic polynomial are similar. (d) Matrices having the same Jordan canonical form are similar. (e) Every matrix is similar to its Jordan canonical form. (f) Every linear operator with the characteristic polynomial \((-1)^{n}(t-\lambda)^{n}\) has the same Jordan canonical form. (g) Every linear operator on a finite-dimensional vector space has a unique Jordan canonical basis. (h) The dot diagrams of a linear operator on a finite-dimensional vector space are unique.

The following definition is used in Exercises 20 and \(21 .\) Definition. For any \(A \in \mathrm{M}_{n \times n}(C)\), define the norm of \(A\) by $$ \|A\|_{m}=\max \left\\{\left|A_{i j}\right|: 1 \leq i, j \leq n\right\\} . $$ Let \(A, B \in \mathrm{M}_{n \times n}(C)\). Prove the following results. (a) \(\|A\|_{m} \geq 0\). (b) \(\|A\|_{m}=0\) if and only if \(A=O\). (c) \(\|c A\|_{m}=|c| \cdot\|A\|_{m}\) for any scalar \(c\). (d) \(\|A+B\|_{m} \leq\|A\|_{m}+\|B\|_{m}\). (e) \(\|A B\|_{m} \leq n\|A\|_{m}\|B\|_{m}\).
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