Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 21E from Chapter 5.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 21E
Chapter:
Problem:
In Exercises 21 and 22, A is an n X n matrix. Mark each statement True or False. Justify each answer...
Step-by-Step Solution
Step 1
Given a Matrix of size $n \times n$ we have to prove or disprove some statements given below:
(a) If $A{\bf{x}} = \lambda {\bf{x}}$ for some vector ${\bf{x}}$, then $\lambda$ is an eigenvalue of A.
(b) A matrix A is not invertible if and only if 0 is an eigenvalue of A.
(c) A number c is an eigenvalue of A if and only if the equation $\left( {A - cI} \right){\bf{x}} = {\bf{0}}$ has a nontrivial solution.
(d) Finding an eigenvector of A may be difficult, but checking whether a given vector is in fact an eigenvector iseasy.
(e) To find the eigenvalues of A, reduce A to echelon form.
Step 2: (a)
An eigenvector of an $n \times n$ matrix A is a nonzero vector x such that $A{\bf{x}} = \lambda {\bf{x}}$ for some scalar $\lambda $. A scalar $\lambda $ is called an eigenvalue of A if there is a nontrivial solution $\bf{x}$ of $A{\bf{x}} = \lambda {\bf{x}}$; such an $\bf{x}$ is called an eigenvector corresponding to $\lambda $.
From the above theorem, the statement (a) is TRUE.
Step 3: (b)
A matrix will have an Eigen value equal to 0 only if the equation has a non-trivial solution. Moreover, a system has a non-trivial solution if and only if A is not invertible.
Thus if $0$ is an eigenvalue of A then A is not invertible.
Therefore, the given statement is TRUE.
Step 4: (c)
By definition, a scalar c is called an eigenvalue of A if there is a nontrivial solution $\bf{x}$ of $A{\bf{x}} = c {\bf{x}}$
\[\begin{array}{l}A{\bf{x}} = c{\bf{x}}\\A{\bf{x}} - c{\bf{x}} = {\bf{0}}\\\left( {A - c} \right){\bf{x}} = {\bf{0}}\end{array}\]Therefore, the given statement is TRUE.
Step 5: (d)
For finding an eigenvector of a matrix, we first need to determine the eigenvalue and then corresponding eigenvector. However,for checking eigenvector we just need to solve following equation. If $A{\bf{x}} = \lambda {\bf{x}}$, then $x$ is an eigenvector, or else it is not.
Hence, the given statement is TRUE.
Step 6: (e)
To find the Eigenvalues of a matrix we have to solve the characteristic equation of the matrix ($\det \left( {A - \lambda I} \right) = 0$). So we can not find eigenvalues just by reducing to echelon form. The echelon forms of give the eigenvectors and not the eigenvalues.Hence, the given statement is FALSE.
ANSWER
(a) TRUE
(b) TRUE
(c) TRUE
(d) TRUE
(e) FALSE
Given a Matrix of size $n \times n$ we have to prove or disprove some statements given below:
(a) If $A{\bf{x}} = \lambda {\bf{x}}$ for some vector ${\bf{x}}$, then $\lambda$ is an eigenvalue of A.
(b) A matrix A is not invertible if and only if 0 is an eigenvalue of A.
(c) A number c is an eigenvalue of A if and only if the equation $\left( {A - cI} \right){\bf{x}} = {\bf{0}}$ has a nontrivial solution.
(d) Finding an eigenvector of A may be difficult, but checking whether a given vector is in fact an eigenvector iseasy.
(e) To find the eigenvalues of A, reduce A to echelon form.
Step 2: (a)
An eigenvector of an $n \times n$ matrix A is a nonzero vector x such that $A{\bf{x}} = \lambda {\bf{x}}$ for some scalar $\lambda $. A scalar $\lambda $ is called an eigenvalue of A if there is a nontrivial solution $\bf{x}$ of $A{\bf{x}} = \lambda {\bf{x}}$; such an $\bf{x}$ is called an eigenvector corresponding to $\lambda $.
From the above theorem, the statement (a) is TRUE.
Step 3: (b)
A matrix will have an Eigen value equal to 0 only if the equation has a non-trivial solution. Moreover, a system has a non-trivial solution if and only if A is not invertible.
Thus if $0$ is an eigenvalue of A then A is not invertible.
Therefore, the given statement is TRUE.
Step 4: (c)
By definition, a scalar c is called an eigenvalue of A if there is a nontrivial solution $\bf{x}$ of $A{\bf{x}} = c {\bf{x}}$
\[\begin{array}{l}A{\bf{x}} = c{\bf{x}}\\A{\bf{x}} - c{\bf{x}} = {\bf{0}}\\\left( {A - c} \right){\bf{x}} = {\bf{0}}\end{array}\]Therefore, the given statement is TRUE.
Step 5: (d)
For finding an eigenvector of a matrix, we first need to determine the eigenvalue and then corresponding eigenvector. However,for checking eigenvector we just need to solve following equation. If $A{\bf{x}} = \lambda {\bf{x}}$, then $x$ is an eigenvector, or else it is not.
Hence, the given statement is TRUE.
Step 6: (e)
To find the Eigenvalues of a matrix we have to solve the characteristic equation of the matrix ($\det \left( {A - \lambda I} \right) = 0$). So we can not find eigenvalues just by reducing to echelon form. The echelon forms of give the eigenvectors and not the eigenvalues.Hence, the given statement is FALSE.
ANSWER
(a) TRUE
(b) TRUE
(c) TRUE
(d) TRUE
(e) FALSE