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Chapter 6: Problem 16

Let \(A\) be a \(2 \times 2\) matrix and let \(p(\lambda)=\lambda^{2}+b \lambda+c\) be the characteristic polynomial of \(A .\) Show that \(b=-\operatorname{tr}(A)\) and \(c=\operatorname{det}(A)\)

Short Answer

Expert verified
In order to show that b = -tr(A) and c = det(A), we first find the characteristic polynomial of a general 2x2 matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) and its determinant (A - 位I). The characteristic polynomial is given by \(p(\lambda) = \lambda^2 - (a+d)\lambda + (ad-bc)\). By comparing the coefficients of the given polynomial (\(p(\lambda) = \lambda^2 + b\lambda + c\)) and that of the determinant calculation, we can observe that \(b = -(a+d) = -tr(A)\) and \(c = ad-bc = \operatorname{det}(A)\).

Step by step solution

01

Calculate the characteristic polynomial of a 2x2 matrix A.

Let A be a 2x2 matrix given by: \[A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\] We are asked to find the characteristic polynomial of A, which is given by the determinant of (A - 位I), where I is the identity matrix and 位 is an eigenvalue of A: \[ p(\lambda) = \operatorname{det}(A - \lambda I) \] The matrix (A - 位I) can be represented as: \[A - \lambda I = \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix}\] Now, we can compute the determinant of (A - 位I)
02

Compute the determinant of (A - 位I).

The determinant of a 2x2 matrix, given by: \[\operatorname{det}(A - \lambda I) = \begin{vmatrix} a - \lambda & b \\ c & d - \lambda \end{vmatrix} = (a - \lambda)(d - \lambda) - bc\] Expanding this expression, we get: \[p(\lambda) = \lambda^2 - (a+d)\lambda + (ad-bc)\]
03

Compare the coefficients of the characteristic polynomial.

According to the given polynomial, p(位) = 位^2 + b位 + c, and from the determinant calculation, we have: \[p(\lambda) = \lambda^2 - (a+d)\lambda + (ad-bc)\] By comparing the coefficients, we can observe that: \[b = -(a+d)\] \[c = ad-bc\] Since the trace (tr) of a matrix is the sum of its diagonal elements, it follows that: \[tr(A) = a + d\] And the determinant (det) of a 2x2 matrix is given by: \[\operatorname{det}(A) = ad-bc\] Therefore, we can conclude that:
04

Prove the requested equalities.

From the comparison of coefficients, we have: \[b = -(a+d) = -tr(A)\] \[c = ad - bc = \operatorname{det}(A)\] Hence, b = -tr(A) and c = det(A).

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Most popular questions from this chapter

Which of the matrices that follow are Hermitian? Normal? (a) \(\left(\begin{array}{cc}1-i & 2 \\ 2 & 3\end{array}\right)\) (b) \(\left(\begin{array}{cc}1 & 2-i \\ 2+i & -1\end{array}\right)\) \((\mathrm{c})\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)\) (d) \(\left(\begin{array}{cc}\frac{1}{\sqrt{2}} i & \frac{1}{\sqrt{2}} \\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} i\end{array}\right)\) (e) \(\left(\begin{array}{ccc}0 & i & 1 \\ i & 0 & -2+i \\ -1 & 2+i & 0\end{array}\right)\) (f) \(\left(\begin{array}{ccc}3 & 1+i & i \\ 1-i & 1 & 3 \\ -i & 3 & 1\end{array}\right)\)

Let \(U\) be a unitary matrix. Prove that (a) \(U\) is normal. (b) \(\|U \mathbf{x}\|=\|\mathbf{x}\|\) for all \(\mathbf{x} \in \mathbb{C}^{n}\) (c) if \(\lambda\) is an eigenvalue of \(U,\) then \(|\lambda|=1\)

Let \(B\) be an \(m \times n\) matrix of rank \(n .\) Show that \(B^{T} B\) is positive definite.

Let \\[ A=\left(\begin{array}{rrrr} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array}\right) \\] (a) Compute the \(L U\) factorization of \(A\) (b) Explain why \(A\) must be positive definite.

Which of the matrices that follow are reducible? For each reducible matrix, find a permutation matrix \(P\) such that \(P A P^{T}\) is of the form \\[ \left(\begin{array}{l|l} B & O \\ X & C \end{array}\right) \\] where \(B\) and \(C\) are square matrices. (a) \(\left(\begin{array}{llll}1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1\end{array}\right)\) (b) \(\left(\begin{array}{llll}1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1\end{array}\right)\) (c) \(\left(\begin{array}{ccccc}1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1\end{array}\right)\) (d) \(\left(\begin{array}{lllll}1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1\end{array}\right)\)
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