Linear Algebra II

November 7, 2014

Lecture 23

Filed under: 2014 Fall — Y.K. Lau @ 3:50 PM

Some remarks to supplement Friday’s lectures:

  • We mentioned that {\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}} has only 1 eigenvalue {0} and its geometric multiplicity is 1 (strictly less than its algebraic multiplicity 2), so {\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}} is not diagonalizable.

    Indeed for this simple case, we may check that {\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}} is not diagonalizable by brute-force, as follows:

    Suppose {P^{-1}\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} P=\begin{pmatrix} \lambda & 0 \\ 0 & \mu\end{pmatrix}} for some nonsingular {P}.

    Let us write {P=\begin{pmatrix} a & b \\ c& d\end{pmatrix}}. Then,

    \displaystyle  \begin{array}{rcl}  \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \begin{pmatrix} a & b \\ c& d\end{pmatrix} &=& \begin{pmatrix} a & b \\ c& d\end{pmatrix}\begin{pmatrix} \lambda & 0 \\ 0 & \mu\end{pmatrix}\\ \Rightarrow \qquad \qquad \qquad\qquad\begin{pmatrix} c & d \\ 0& 0\end{pmatrix} &=& \begin{pmatrix} \lambda a & \mu b \\ \lambda c& \mu d\end{pmatrix} \end{array}

    Case 1) {\lambda=0} and {\mu=0}, then {c=d=0}.

    Case 2) {\lambda\neq 0} and {\mu =0}, then {c=a=0}.

    Case 3) {\lambda= 0} and {\mu \neq 0}, then {d=b=0}.

    Case 4) {\lambda\neq 0} and {\mu \neq 0}, then {c=d=0}.

    All cases are impossible as {\det \begin{pmatrix} a & b \\ c & d\end{pmatrix} =\det P \neq 0}.

  • Cayley-Hamilton’s Theorem says that for an {n\times n} matrix {A}, if its characteristic polynomial is {c_A(x)=x^n + a_{n-1}x^n +\cdots +a_1 x +a_0}, then

    \displaystyle  A^n + a_{n-1}A^n +\cdots +a_1 A +a_0 I = 0 \quad (zero matrix).

  • Some classmates asked if the characteristic equation {c_A(x)=0} has no real solution, does it mean {A} has no eigenvalue? The answer is NO: it only means {A} has no real eigenvalues! If we use complex numbers, then {A} has (complex) eigenvalues and the theorem of Jordan Canonical Form holds for the case of complex eigenvalues.

 

After Class Exercises: Ex 9.3 Qn 7, 31; Ex 11.1 Qn 2; Ex 11.2 Qn 3 (b). (Solution — To be available.)

Exercises. Find a Jordan Canonical Form of the following matrices:

{\begin{pmatrix} -3 & -1 & 0\\ 4 & -1 & 3\\ 4 & -2 & 4 \end{pmatrix}} ,     {\begin{pmatrix} -3 & 6& 3 & 2\\ -2 & 3 & 2 & 2\\ -1 & 3 & 0 & 1\\ -1 & 1 & 2 & 0 \end{pmatrix}}.

 

Next week we shall turn to the new topic — inner product spaces. Hence let us have a round up with the following chart which indicates how/what are developed in our study of linear transformations.

 

 

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