/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Let \(A\) denote the \(k \times ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 19

Let \(A\) denote the \(k \times k\) matrix $$ \left(\begin{array}{ccccc} 0 & 0 & \cdots & 0 & -a_{0} \\ 1 & 0 & \cdots & 0 & -a_{1} \\ 0 & 1 & \cdots & 0 & -a_{2} \\ \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{k-1} \end{array}\right), $$ where \(a_{0}, a_{1}, \ldots, a_{k-1}\) are arbitrary scalars. Prove that the characteristic polynomial of \(A\) is $$ (-1)^{k}\left(a_{0}+a_{1} t+\cdots+a_{k-1} t^{k-1}+t^{k}\right) $$ Hint: Use mathematical induction on \(k\), computing the determinant by cofactor expansion along the first row.

Short Answer

Expert verified
We use mathematical induction on the matrix $A$. 1. Base case (k = 1): $\det(A - tI) = -a_0 - t$ which matches the characteristic polynomial for k = 1. 2. Inductive Hypothesis: Assume the characteristic polynomial holds for a $(k-1) \times (k-1)$ matrix $B$ with coefficients $a_0, a_1, \ldots, a_{k-2}$. 3. Inductive Step: We prove that the characteristic polynomial holds for a $k \times k$ matrix $A$. Compute $\det(A - tI)$ using cofactor expansion along the first row to get $-t(-t)^{k-1} - a_0 (-1)^{k-1}\left(a_{1}+a_{2} t+\cdots+a_{k-1} t^{k-2}+t^{k-1}\right)$. 4. Simplify the expression to obtain the characteristic polynomial, $(-1)^{k}\left(a_{0}+a_{1} t+\cdots+a_{k-1} t^{k-1}+t^{k}\right)$, proving the problem statement by induction.

Step by step solution

01

Base case (k = 1)

In this case, matrix A is: \[ A = \begin{pmatrix} -a_0 \end{pmatrix} \] Subtracting t times the identity matrix, we get: \[ A - tI = \begin{pmatrix} -a_0 - t \end{pmatrix} \] The determinant of this matrix is simply the single entry in the matrix: \[ \det(A-tI) = -a_0 - t \] This matches the given expression for the characteristic polynomial with k = 1.
02

Inductive hypothesis

Assume that the characteristic polynomial of a (k-1) x (k-1) matrix B, with coefficients \(a_0, a_1, \ldots, a_{k-2}\), is as the problem specifies: \[ (-1)^{k-1}\left(a_{0}+a_{1} t+\cdots+a_{k-2} t^{k-2}+t^{k-1}\right) \]
03

Inductive step

We need to prove that the characteristic polynomial of the k x k matrix A is as given. We do this by computing the determinant of the matrix obtained by subtracting t times the identity matrix from A, using the cofactor expansion along the first row. \[ A - tI = \left(\begin{array}{ccccc} -t & 0 & \cdots & 0 & -a_{0} \\ 1 & -t & \cdots & 0 & -a_{1} \\ 0 & 1 & \cdots & 0 & -a_{2} \\ \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & -t & -a_{k-1} \end{array}\right) \] Cofactor expansion along the first row: \[ \det(A-tI) = (-t)\det(M_{1,1}) - 0\det(M_{1,2}) - \cdots - 0\det(M_{1,k-1}) - a_0\det(M_{1,k}) \] where \(M_{i,j}\) is the (k-1) x (k-1) matrix obtained from A-tI by deleting the i-th row and j-th column.
04

Compute determinants of submatrices

Firstly, we have: \[ M_{1,1} = \left(\begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & -t \end{array}\right) \] Clearly, \(\det(M_{1,1}) = (-t)^{k-1}\). Secondly, we have: \[ M_{1,k} = \left(\begin{array}{ccccc} 1 & -t & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & -t \end{array}\right) \] This is the same matrix as in the inductive hypothesis, let's call this matrix B. Therefore, we already know its determinant: \[ \det(M_{1,k}) = (-1)^{k-1}\left(a_{1}+a_{2} t+\cdots+a_{k-1} t^{k-2}+t^{k-1}\right) \]
05

Compute the determinant of A-tI

We can now substitute the determinant values of the submatrices into the cofactor expansion expression: \[ \det(A-tI) = -t(-t)^{k-1} - a_0 (-1)^{k-1}\left(a_{1}+a_{2} t+\cdots+a_{k-1} t^{k-2}+t^{k-1}\right) \] Simplify the expression: \[ \det(A-tI) = (-1)^k t^k + (-1)^{k} a_{0}\left(a_{1}+a_{2} t+\cdots+a_{k-1} t^{k-2}+t^{k-1}\right) \] The characteristic polynomial is then: \[ (-1)^{k}\left(a_{0}+a_{1} t+\cdots+a_{k-1} t^{k-1}+t^{k}\right) \] This matches the given expression for the characteristic polynomial, thus proving the problem statement by induction.

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Most popular questions from this chapter

For each of the following linear operators \(T\) on the vector space \(V\), determine whether the given subspace \(W\) is a \(T\)-invariant subspace of V. (a) \(\mathrm{V}=\mathrm{P}_{3}(R), \mathrm{T}(f(x))=f^{\prime}(x)\), and \(\mathrm{W}=\mathrm{P}_{2}(R)\) (b) \(\mathrm{V}=\mathrm{P}(R), \mathrm{T}(f(x))=x f(x)\), and \(\mathrm{W}=\mathrm{P}_{2}(R)\) (c) \(\mathrm{V}=\mathrm{R}^{3}, \mathrm{~T}(a, b, c)=(a+b+c, a+b+c, a+b+c)\), and \(\mathrm{W}=\\{(t, t, t): t \in R\\}\) (d) $\mathrm{V}=\mathrm{C}([0,1]), \mathrm{T}(f(t))=\left[\int_{0}^{1} f(x) d x\right] t$, and \(\mathrm{W}=\\{f \in \mathrm{V}: f(t)=a t+b\) for some \(a\) and \(b\\}\) (e) $\mathrm{V}=\mathrm{M}_{2 \times 2}(R), \mathrm{T}(A)=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right) A$, and \(\mathrm{W}=\left\\{A \in \mathrm{V}: A^{t}=A\right\\}\)

Prove that if both \(T_{w}\) and \(\bar{T}\) are diagonalizable and have no common eigenvalues, then \(T\) is diagonalizable. The results of Theorem \(5.21\) and Exercise 28 are useful in devising methods for computing characteristic polynomials without the use of determinants. This is illustrated in the next exercise.

Let \(V\) be of finite dimension and let \(T\) be a linear operator on \(V\) for which \(T R=I\), for some operator \(R\) on \(V\). (We call \(R\) a right inverse of \(T\).) (a) Show that \(T\) is invertible. (b) Show that \(R=T^{-1}\) (c) Give an example showing that the above need not hold if \(V\) is of infinite dimension.

Give an example of a nonlinear map \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) such that \(F^{-1}(0)=\\{0\\}\) but \(F\) is not one-to-one.

Let \(T\) be the linear operator on \(\mathbf{R}^{2}\) defined by \(T(x, y)=(x+2 y, 3 x+4 y) .\) Find a formula for \(f(T),\) where (a) \(f(t)=t^{2}+2 t-3\) (b) \(f(t)=t^{2}-5 t-2\)
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