Linear Algebra Final Exam

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Feel free to use any source you want when completing this exam. If you use any sources besides class notes or the text, please give appropriate references. You may use technology for row reduction. All other computations must be done by hand. Good Luck!

1. 1. Let $A$ be a $4\times 4$ strictly upper-triangular matrix; i.e.,
      $$A=\begin{pmatrix} 0&a&b&c\\ 0&0&d&e\\ 0&0&0&f\\ 0&0&0&0\\ \end{pmatrix}.$$
      Show that $A^k=0_M$ for some $k\ge0$. That is, show that $A$ is nilpotent.
      
      Note:This result is true in general. That is, any $n\times n$ strictly upper-triangular matrix $A$ is nilpotent. Moreover, its nilpotency class is less than or equal to $n$ - that is, $A^n=0_M$.
      

   2. Let $Nt$ be the $n\times n$ matrix with $t$ on the first superdiagonal and zeros elsewhere.
      $$Nt=\begin{pmatrix} 0&t&0&\cdots&0\\ 0&0&t&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&t\\ 0&0&0&0&0\\ \end{pmatrix}.$$

      (i)

      If $n=4$, determine the form of the matrix $(Nt)^k$, for each $1\le k\le 4$.
      

      (ii)

      The above pattern for $(Nt)^k$ holds for any $n$. Use your answer from (i) to deduce the form of the matrix $(Nt)^k$, for each $1\le k\le n$.
      

      (iii)

      Recall the power series for $e^x$.

      $$e^x=\displaystyle \sum^{\infty}_{i=0}\displaystyle \frac{x^i}{i!}= 1+x+\displaystyle \frac{x^2}{2!}+\displaystyle \frac{x^3}{3!}+\cdots$$

      
      For any $n\times n$ matrix $A$, we define $e^A$ to be the power series for $e^x$ evaluated at the matrix $A$. That is,

      $$e^A=\displaystyle \sum^{\infty}_{i=0}\displaystyle \frac{A^i}{i!}= I+A+\displaystyle \frac{A^2}{2!}+\displaystyle \frac{A^3}{3!}+\cdots$$

      
      Using your answer in (ii), show that

      $$e^{Nt}=\begin{pmatrix} 1\;&\;t&\;\displaystyle \frac{t^2}{2!}&\cd... ...&\ddots&t&\displaystyle \frac{t^2}{2!}\\ &&&&1&t\\ &&&&&1\\ \end{pmatrix}$$

      Note: Remember that $Nt$ is a nilpotent matrix.
      

2. Consider the matrix
   $$A=\begin{pmatrix} 1/2&1/4&0&-1/4\\ 0&3/8&0&1/8\\ 0&0&1/2&0\\ 0&1/8&0&3/8\\ \end{pmatrix}$$

   (i)

   Determine the characteristic polynomial of $A$, and find all of its roots. You may use technology to factor this polynomial.
   

   (ii)

   Give a $PDP^{-1}$ factorization of $A$.
   

   (iii)

   Use (ii) to convince yourself that $$\lim_{n\to\infty}A^n=0_M$$.
   

   (iv)

   Prove that the matrix $(I-A)$ is invertible and determine $(I-A)^{-1}$.
   

   (v)

   Recall the algebraic identity

   $$(1-x)(1+x+x^2+\cdots+x^n)=1-x^{n+1}.$$

   Then since $$\lim_{n\to\infty}A^n=0_M$$, as $n$ gets large, we get

   $$(1-A)(1+A+A^2+\cdots+A^n)\approx I.$$

   Therefore, we can approximate $(I-A)^{-1}$ as

   $$(1-A)^{-1}\approx(1+A+A^2+\cdots+A^n)$$

   Use the first five terms in this formula and the $PDP^{-1}$ factorization of $A$ to approximate $(I-A)^{-1}$. What would you need to define in order to decide if this is a ``close'' approximation?
   

3. Consider a polynomial $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. We define the companion matrix of $f(x)$ to be
   $$C_f=\begin{pmatrix} 0&0&0&\cdots&0&-a_0\\ 1&0&0&\cdots&0&-a_1\\... ...vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&1&-a_{n-1}\\ \end{pmatrix}$$
   
   For this problem, suppose $f(x)=x^4-2x^3+x+1$. Verify that the characteristic polynomial of $C_f$ is $f(x)$.
   
   Note: Companion matrices are used when forming the Rational Canonical Form of a matrix.
   

4. The following result is one of the most important theorem in linear algebra.
   Theorem 1 (Cayley-Hamilton)   Let $p(x)$ be the characteristic polynomial of an $n\times n$ matrix $A$. Then $p(A)=0_M$.
   

   Verify the Cayley-Hamilton theorem for the matrix

   $$A=\begin{pmatrix} 1&2&1\\ 2&0&2\\ 1&2&3\\ \end{pmatrix}$$

5. Suppose we have a linear system of differential equations
   

   $$y'_1(t) = a_{11}y_1(t)+a_{12}y_2(t)+\cdots+a_{1n}y_{n}(t)$$

   $$y'_2(t) = a_{21}y_1(t)+a_{22}y_2(t)+\cdots+a_{2n}y_{n}(t)$$

   $$y'_3(t) = a_{31}y_1(t)+a_{32}y_2(t)+\cdots+a_{3n}y_{n}(t)$$

   $$\vdots = \vdots$$

   $$y'_n(t) = a_{n1}y_1(t)+a_{n2}y_2(t)+\cdots+a_{nn}y_{n}(t)$$

   
   
   Form the vectors
   $$\bar{y}'=\begin{pmatrix} y'_1(t)\\ y'_2(t)\\ y'_3(t)\\ \vdots\... ...begin{pmatrix} y_1(t)\\ y_2(t)\\ y_3(t)\\ \vdots\\ y_n(t)\\ \end{pmatrix}$$
   and the matrix
   $$A=\begin{pmatrix} a_{11}&a_{22}&\cdots&a_{1n}\\ a_{21}&a_{22}&\... ... \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\\ \end{pmatrix}.$$
   
   Then we can write this system of equations as the matrix-vector equation $\bar{y}'=A\bar{y}$.
   

   In particular, consider the system
   

   $$y'_1(t) = 2y_1(t)+2y_2(t)+y_{3}(t)$$

   $$y'_2(t) = y_1(t)+3y_2(t)+y_{3}(t)$$

   $$y'_3(t) = y_1(t)+2y_2(t)+2y_{3}(t)$$

   
   

   (i)

   Form the matrix $A$ and find a $PDP^{-1}$ factorization of $A$.
   
   Using the matrix $P$ in (i), we can define a new vector $\bar{x}=P^{-1}\bar{y}$ so that $\bar{y}=P\bar{x}$. Then $(P\bar{x})'=\bar{y}'=A\bar{y}=AP\bar{x}$. But $P$ is a constant matrix, and so $(P\bar{x})'=P\bar{x}'$. It follows that $\bar{x}'=P^{-1}\bar{y}'=P^{-1}A\bar{y}=P^{-1}AP\bar{x}=D\bar{x}$.
   

   (iv)

   Rewrite the new linear system of differential equations corresponding to $\bar{x}'=D\bar{x}$, and show that

   $$\bar{x}=\begin{pmatrix} c_1e^{5t}\\ c_2e^{t}\\ c_3e^{t}\\ \end{pmatrix}$$

   (v)

   Determine $\bar{y}$ using $\bar{y}=P\bar{x}$, and show that all solutions $\bar{y}$ to the system of differential equations will be some linear combination of the set

   $$\left\{\begin{pmatrix} e^{5t}\\ e^{5t}\\ e^{5t}\\ \end{pmatri... ...^t\\ \end{pmatrix},\begin{pmatrix} e^t\\ -e^t\\ 0\\ \end{pmatrix}\right\}$$

   
   A theorem of differential equations guarantees that these vectors form a basis for all solutions to the system of differential equations.
   

   Note:It is worth noting that the columns of the matrix $e^{At}$ (as defined in problem 1.iii.) will contain a (possibly different) basis for all solutions to the system of equations. That is, all solutions to $\bar{y}'=A\bar{y}$ are spanned by the columns of the matrix $e^{At}$. Notice the similarity to the solution, $y=e^{at}$, of the single differential equation $y'(t)=ay(t)$, where here $a$ is a scalar.
   

6. Consider $V=C[0,2\pi]$, the vector space of all continuous functions on the interval $[0,2\pi]$. For functions $p(t)$and $q(t)$, define the inner product
   $$\langle p,q\rangle=\displaystyle \int^{2\pi}_{0}p(t)q(t)\;dt$$

   (i)

   Show that $\sin(t)$ and $\cos(t)$ are orthogonal with respect to this inner product.
   

   (ii)

   Let $W$ be a subspace of $V$ with orthogonal basis $\{w_1(t),\dots,w_p(t)\}$. Write the formula for an orthogonal projection of some function $f(t)$ onto $W$.
   

   For any $n$, the set $\{1,\sin(t),\sin(2t),\dots,\sin(nt),\cos(t),\cos(2t),\dots,\cos(nt)\}$ is an orthogonal set in $C[0,2\pi]$. Let

   $$W=span\{1,\sin(t),\sin(2t),\sin(3t),\cos(t),\cos(2t),\cos(3t)\}$$

   (iii)

   Find the projection of $f(t)=1-t$ onto $W$. You should use technology to evaluate any of the integrals.
   

   (iv)

   On the interval $[0,2\pi]$, graph the function $f(t)=1-t$ and its orthogonal projection from (iii).
   

   The projection in (iii) is the third order Fourier approximation of $f(t)=1-t$.
   

   Note:Fourier series and Fourier approximations play a major role when solving partial differential equations.
   

7. We first supply some background for this problem. For functions of a single variable (that is functions $f:\mathbb{R}\rightarrow\mathbb{R}$), the derivative is the slope of the tangent line (which is, recall, the best linear approximation to the graph of the original function) and can be used to find the equation for this tangent line. Suppose we are now working with vector-valued functions; i.e., $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$, where $n,m\ge1$. In a similar fashion, we want the derivative of a vector valued function to help us find a linear approximation to the function. That is, we want an $m\times n$ matrix A so that
   $$\displaystyle \lim_{\bar{x}\mapsto\bar{a}}\displaystyle \frac{\ve... ...}}\displaystyle \frac{\vert A(\bar{x}-\bar{a})\vert}{\vert\bar{x}-\bar{a}\vert}$$
   This is usually stated in the following way.
   
   Definition 1   A function $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is differentiable at $\bar{a}\in\mathbb{R}^n$ if there is an $m\times n$ matrix $A:\mathbb{R}^n\rightarrow\mathbb{R}^m$ such that

   $$\displaystyle \lim_{\bar{x}\mapsto\bar{a}}\displaystyle \frac{\vert f(\bar{x})-f(\bar{a})-A(\bar{x}-\bar{a})\vert}{\vert\bar{x}-\bar{a}\vert}=0$$

   If such a limit exists, the matrix $A$ is denoted by $Df(\bar{a})$ and is called the Jacobian matrix.
   
   A vector valued function $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ will often be given by $m$ vector valued functions $f_1(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n)$, and written in the form
   $$f(x_1,\dots,x_n)=\begin{pmatrix} f_1(x_1,\dots,x_n)\\ \vdots\\ f_m(x_1,\dots,x_m)\\ \end{pmatrix}$$
   
   It is easy to show that if $f$ is differentiable, then each of $f_i$ is differentiable, and moreover, that the Jacobian matrix is
   $$Df(\bar{x})=\begin{pmatrix} \displaystyle \frac{\partial f_1}{\pa... ... x_2}& \cdots&\displaystyle \frac{\partial f_m}{\partial x_n}\\ \end{pmatrix}$$

   We now consider the following theorem.
   

   Theorem 2 (Inverse Function Theorem)   Let $f$ be a continuously differentaible, vector-valued function mapping the open set $E\subset\mathbb{R}^n$ to $\mathbb{R}^n$. If for some point $\bar{a}\in E$, the determinant of the Jacobian matrix, $\vert Df(\bar{a})\vert$, is non-zero, then $f$ is invertible on some open neighborhood of $\bar{a}$.
   

   Note:Although possibly confusing, the determinant of the Jacobian matrix $, \vert Df(\bar{a})\vert$, is simply called the Jacobian.
   

   For this problem, consider the vector-valued function $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ defined as

   $$f(x_1,x_2,x_3)=\begin{pmatrix} f_1(x_1,x_2,x_3)\\ f_2(x_1,x_2,x_... ...atrix}=\begin{pmatrix} \cos(x_1)+x_2\\ 2x_3-x_1\\ \sin(x_2)\\ \end{pmatrix}$$
   Note: $f$ is NOT a linear function.
   

   Then the Jacobian matrix is

   $$Df(\bar{x})=\begin{pmatrix} \displaystyle \frac{\partial f_1}{\pa... ...}=\begin{pmatrix} -\sin(x_1)&1&0\\ -1&0&2\\ 0&\cos(x_2)&0\\ \end{pmatrix}$$
   Use the Theorem 2 to decide at what points $\bar{a}=(a_1,a_2,a_3)\in\mathbb{R}^3$ the function $f$ is NOT invertible on any neighborhood of $\bar{a}$.