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Mathematical Expression Editor
Challenge Problems for Chapter 8
Suppose is an matrix consisting entirely of real entries but is a complex eigenvalue
having the eigenvector, Here and are real vectors. Show that then is also an
eigenvalue with the eigenvector, .
You should remember that the conjugate of a product of complex numbers equals
the product of the conjugates. Here is a complex number whose conjugate equals
Click the arrow to see the answer.
. Now take conjugates of both sides. Since is real,
If is and diagonalizable, show that has dimension or .
If , show that is in if and
only if is in .
Let be with distinct real eigenvalues. If , show that is diagonalizable.
Given a polynomial and a square matrix , the matrix is called the evaluation of at
. Let . Show that for all polynomials .
If is diagonalizable and is a polynomial such that for all eigenvalues of , show that
(here, the final O is the zero matrix the same size as ).
The characteristic polynomial of (see Definition def:char_poly_complex) certainly satisfies the requirement
that for all eigenvalues of . In solving this problem you have proved a special case of
the Cayley-Hamilton theorem, see Theorem th:Cayley_Hamilton. In fact, if is the characteristic
polynomial of , then whether or not is diagonalizable.
Let be an invertible matrix. If is any matrix, write . Verify that:
(a)
(b)
(c)
(d)
(e)
for
(f)
If is invertible, .
(g)
If is invertible, .
Let and .
(a)
Show that has real roots by considering .
(b)
Show that by considering .
Assume the matrix is similar to an upper triangular matrix. If , show that is equal
to the zero matrix.
Show that is similar to for all matrices .
Let . If treat the cases and separately.
If , reduce to the case using Exercise prob:5_5_12prob:5_5_12d.
Suppose is an matrix and let be an eigenvector such that . Also suppose the
characteristic polynomial of is \begin{equation*} \det \left ( z I-A\right ) =z ^{n}+a_{n-1} z ^{n-1}+\cdots +a_{1}z +a_{0} \end{equation*}
Explain why \begin{equation*} \left ( A^{n}+a_{n-1}A^{n-1}+\cdots +a_{1}A+a_{0}I\right ) \vec{v}=0 \end{equation*}
Use this to prove that the Cayley-Hamilton theorem holds for any diagonalizable
matrix . (The Cayley-Hamilton theorem says that satisfies its characteristic
equation, i.e., \begin{equation*} A^{n}+a_{n-1}A^{n-1}+\cdots +a_{1}A+a_{0}I=0 \end{equation*}
. (For a proof of the general case, see Theorem th:Cayley_Hamilton)
Suppose the characteristic polynomial of an matrix is . Find where is an
integer.
Click the arrow to see answer.
The eigenvalues are distinct because they are the roots of . Hence if is a given vector
with then so .