Review exercises

Homework: Let $V$ be the space of infinite sequences $\{x_1,\ldots,x_n,\ldots\}$. Find an infinite dimensional subspace $U$ of $V$ that can be equipped with a non-trivial inner product. Prove.

Example: $A(f) = f-f'$. $A \colon C^1 \rightarrow C^1$. Find the kernel of $A$: $$\begin{array}{} {\rm ker \hspace{3pt}} A &\stackrel { { \rm def } } {=} \{f \colon A(f)=0\} \\ &= \{f \colon f-f'=0 \} \\ &= \{ f \colon f = f' \} \\ &= \{ ce^x \colon c \in {\bf R} \} \end{array}.$$

This is a subspace of dimension 1. Prove that.

Example: $A$ is singular iff 0 is an eigenvalue.

Example: $P^{-1} I P = I$.

Example: $A= \left[ \begin{array}{} 2 & 0 \\ 0 & 3 \end{array} \right] \sim \left[ \begin{array}{} 3 & 0 \\ 0 & 2 \end{array} \right]=B.$$

We need $P$ with $A = P^{-1} B P.$

It's the opposite!

So basis $\{e_1,e_2\} \stackrel{p}{\rightarrow} \{e_2,e_1\}$, therefore $p = \left[ \begin{array}{} 0 & 1 \\ 1 & 0 \end{array} \right].$

Example: Find $A,B$ such that ${\rm tr \hspace{3pt}} A = {\rm tr \hspace{3pt}} B$ but ${\rm det \hspace{3pt}} A \neq {\rm det \hspace{3pt}} B$.

$\left[ \begin{array}{} 1 & 0 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{} 1 & 1 \\ 1 & 1 \end{array} \right]$

Example: Suppose $T \colon {\bf P}_3 \rightarrow {\bf P}_3$ is given by $T(f)(x) = xf'(x)$. Find the matrix of $T$.

This means (not $T(f(x))$!):

$T(f)=g$ and $g(x)=xf'(x)$ for all $x$.

Basis of ${\bf P}_3$ is $\{1, x, x^2, x^3 \}, {\rm dim \hspace{3pt}}{\bf P}^3 = 4$.

$$\begin{array}{} T(x^3) &= x \cdot 3x^2 &= 3x^3 \\ T(x^2) &= x \cdot 2x &= 2x^2 \\ T(x) &= x \cdot 1 &= x \\ T(1) &= x \cdot 0 &= 0 \end{array}$$

Rewrite: $e_1=1$, $e_2=x$, $e_3=x^2$, $e_4=x^3$.

• $T(e_1)=0$,
• $T(e_2)=e_2$,
• $T(e_3)=2e_3$,
• $T(e_4)=3e_4$.

We write these are columns in terms of $e_1,\ldots,e_4$.

Then $$e_2 = \left[ \begin{array}{} 0 \\ 1 \\ 0 \\ 0 \end{array} \right]$$ etc.

$$T = \left[ \begin{array}{} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{array} \right]$$

Example. Compute $$\left[ \begin{array}{} 0 & 0 & a \\ 0 & b & 0 \\ c & 0 & 0 \end{array} \right]^{-1} = ?; a,b,c \neq 0$$

$$\left[ \begin{array}{ccc|ccc} 0 & 0 & a & 1 & 0 & 0 \\ 0 & b & 0 & 0 & 1 & 0 \\ c & 0 & 0 & 0 & 0 & 1 \end{array} \right] \rightarrow R_1 \longleftrightarrow R_3 \left[ \begin{array}{ccc|ccc} 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & b & 0 & 0 & 1 & 0 \\ 0 & 0 & a & 1 & 0 & 0 \end{array} \right]$$

$$\rightarrow (\frac{1}{c}R_1, \frac{1}{b}R_2, \frac{1}{a}R_3) \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{c} \\ 0 & 1 & 0 & 0 & \frac{1}{b} & 0 \\ 0 & 0 & 1 & \frac{1}{a} & 0 & 0 \end{array} \right]$$

Homework: Find the trace ${\rm Tr \hspace{3pt}}A$ from from the characteristic polynomial $\chi_A$.

Hint: Experiment in ${\rm dim \hspace{3pt}} 1,2,3$, conclude something.

Answer: $\chi_A(\lambda) = \lambda^n + {\rm Tr \hspace{3pt}}A \lambda^{n-1}$+lower degree terms

Proof, outline...

Expand along first row

$$\det \left\{ \begin{array}{} a_{11}-\lambda & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22}-\lambda & \ldots & a_{2n} \\ \vdots & & & \\ a_{n1} & a_{n2} & \ldots & a_{nn}-\lambda \end{array} \right\}$$

$$=(a_{11}-\lambda)\det B_{11}-a_{12}\det B_{12} {\rm etc}$$

Two $\lambda$'s are cut out, so degree of $\lambda$ is $\leq n-2$.

By induction, $\det B_{11}=\chi_{{A_{11}}}$, where $A_{11}$ is $A$ with first row, first column cut out.

So the coefficient of $\lambda^{n-2}$ in $\chi_{A_{11}}$ is ${\rm Tr \hspace{3pt}}A_{11}$.

Hence the coefficient of $\lambda^{n-1}$ in $\chi_A$ is ${\rm Tr \hspace{3pt}}A$. Done.

Alternatively, what if $A=D$ is diagonal?

$$\det \left[ \begin{array}{} a_{11}-\lambda & 0 & \ldots & 0 & 0 \\ 0 & a_{21}-\lambda & 0 & \ldots & 0 \\ \vdots & & & & \\ 0 & 0 & \ldots & 0 & a_{nm}-\lambda \end{array} \right]$$

$$=(a_{11}-\lambda) \ldots (a_{nn}-\lambda)$$

Coefficient of $\lambda^{n-1}$ is

$$a_{11}\lambda^{n-1}+a_{22}\lambda^{n-1}+\ldots+a_{nm}\lambda^{n-1} = {\rm Tr \hspace{3pt}}A \cdot \lambda^{n-1}$$

$D=P^{-1}AP$, ten ${\rm Tr \hspace{3pt}}D={\rm Tr \hspace{3pt}}A$ (as one of the properties of ${\rm Tr}$).

All good, but... not every matrix is diagonalizable. Problem?

• $A$ is not diagonalizable over ${\bf R}$ but
• $A$ is diagonalizable over ${\bf C}$!

So $P$ may be complex. But that's OK. Suppose $D=P^{-1}AP$ is diagonal and maybe $D$, but ${\rm Tr \hspace{3pt}} D$ is real $={\rm Tr \hspace{3pt}}A$.