Tutorial note 1 (review of linear algebra and multivariable calculus)

Example 1.
(a) $y'=(e-5)/3$.

(b) $y'=(e-1)/(2e^2-e)$

(c) $y'=-9/(9+e)$

It is not necessary to do implicit differentiation in each part.

In (a) we can make $y$ the subject, namely, \[
    y=\sqrt[3]{\frac{5+e^x-x^4}{x}}.
    \]
    In (b) we can use the formula \[
    \frac{dy}{dx}=\frac{1}{\dis \frac{dx}{dy}}.
    \]
Finally in (c) it is seemingly hopeless to express $y$ as a function of $x$, thus we must assume $y$ is a function of $x$ to do implicit differentiation. The feasibility will be guaranteed by the theorem in later lectures called implicit function theorem. Implicit function theorem guarantees there exists a function $f(x)$ such that $y=f(x)$ is a solution of the equation \[
    x^7+x^2y^2+y^7+e^y=3+e.
    \] Note that although it exists, there is no general method to find $f(x)$. The situation is the same as that we know the maximum of $\dis f:[1,2]\to \R$ given by \[f(x)=\frac{ e^x\sin x}{\sqrt{x}} + \cos e^{e^x}\] must occur somewhere, say $x_0\in [1,2]$, but there is no general method to find $x_0$.

Example 2.
(a) We consider case by case:

Case 1. When $a^2+b>0$, $F_x(a,b)$ exists with $F_x(a,b)=2a\cos (a^2+b)$.

Case 2.  When $a^2+b<0$, $F_x(a,b)$ exists with $F_x(a,b) =-2a\cos(a^2+b)$.

Case 3. If $a^2+b=0$, then we have \begin{align*}

F_x(a,b)&=\lim_{h\to 0}\frac{F(a+h,b)-F(a,b)}{h} \\

&= \lim_{h\to 0}\frac{\sin (|h||2a+h|)}{h}\\

&=\begin{cases}

0,&\text{when }a=0,\\

\text{does not exist},&\text{when }a>0,\\

\text{does not exist},&\text{when }a<0.

\end{cases}

\end{align*} So in this case the only point that has derivative is $(0,0)$, i.e., $F_x(0,0)=0$.

(b) If $a\neq b$, then for $(x,y)$ near $(a,b)$, we still have $x\neq y$, hence $G(x,y)=x+y$, it follows that \[

G_y(x,y)=1\implies G_y(a,b)=1.

\] If $a=b$, then by definition we have \[

G_x(a,b)=\lim_{h\to 0}\frac{G(a,b+h)-G(a,b)}{h}=\lim_{h\to 0}\frac{h-a}{h}=\begin{cases}

1,&a=0,\\

\text{does not exist}, &a\neq 0.

\end{cases}

\]



Example 3.
(a) The equation of the tangent plane is given by \[

\nabla (e^x+y^2z+z\cos x-1) (0,0,0)\cdot  ((x,y,z)-(0,0,0))=0,

\] a simple computation gives $\nabla (e^x+y^2z+z\cos x-1) (0,0,0) = (1,0,1)$, therefore the equation that characterizes the tangent plane is \[

x+z=0.

\] (b) The equation of tangent plane is directly given by the linear approximation of $f$ at $(0,0)$, namely, \[

z=f(0,0) + \nabla f(0,0) ((x,y)-(0,0)) = 0+(1,1)\cdot (x,y)=x+y.

\]

Example 4.
(i) Yes, it is linear since \[

T(x,y)=\matrixx{

4&-3\\

1/2&1\\

-1&-1

}\matrixx{

x\\y},

\] and from linear algebra we know that $x\mapsto Ax$ is always linear when $A$ is a matrix.



(ii) This is just the standard matrix, denoted by $[T]$, i.e., \[

[T]=\matrixx{

4&-3\\

1/2&1\\

-1&-1
}.

\](iii) Recall that the following are equivalent: Let $S:V\to W$ be a linear map between vector spaces.

1. $S$ is injective (or 1-1);
2. $\ker S=\{0\}$;
3. $Sx=0\implies x=0$.

We will use the third condition. Suppose that $T(x,y)=0$, i.e., \begin{align*}

4x-3y&=0\\

\frac{1}{2}x+y&=0\\

-x-y&=0.

\end{align*} The first and third equations imply $x=y=0$, thus $T$ is injective.

(iv) By rank-nullity theorem we have \[

2=\dim\ker T+\dim \range T,

\] as $T$ is injective, $\ker T=\{0\}$, so it has zero dimension, and we have $\dim \range T=2$, hence $\range T\neq \R^3$, i.e., $T$ is not surjective.

(v) Since it is not surjective, it is not bijective.

Example 5.
Let $[T]$ denote its standard matrix, then \[

[T]=\matrixx{a&b\\ 3&ab},\]

and we learn that $T$ is invertible if and only if $[T]$ does, and $[T]$ is invertible iff $\det [T]=b(a^2-3)\neq 0$, thus $b\neq 0$, $a\neq \sqrt{3}$ and $a\neq -\sqrt{3}$.

Example 6.
(a) For clarity all vectors will be column vectors. We can write the integral as \[I=\int_{\R^3}\exp (-\frac{1}{2}x^TAx)\,dV(x),\] where $x\in \R^3$. Since $A$ is symmetric, it is orthogonally diagonalizable, namely, there is an orthogonal $3\times 3$ matrix $P$ (i.e., $P^TP=I$) such that \[

P^TAP=\matrixx{

\lambda_1&0&0\\

0&\lambda_2&0\\

0&0& \lambda_3},

\] for some $\lambda_1,\lambda_2,\lambda_3\in \R$. Therefore by change of variable $x=Py$ we have \[
I:=\int_{\R^3}\exp\brac{-\frac{1}{2}(Py)^TA Py}\underbrace{|\det P|}_{=1} \,dV =\int_{\R^3}e^{\dis -0.5 \lambda_1y_1^2-0.5 \lambda_2y_2^2-0.5 \lambda_3y_3^2}\,dV.
\] By using the fact that volume integral can be computed by writing $dV=dy_1dy_2dy_3$, we have \[
I=\int_\R e^{-\frac{\lambda_1}{2} t^2}\,d t \int_\R e^{-\frac{\lambda_2}{2} t^2}\,dt\int_\R e^{-\frac{\lambda_3}{2} t^2}\,dt.
\] Each of the three integrals are at least positive, and thus we have $I<\infty$ if and only if $\lambda_i >0$ for all $i$ if and only if $A$ is positive definite (given that $A$ is symmetric).