# 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.