For the following parts, use one (or more) of the equivalent properties (from class) that define positive-definiteness. (There are multiple possible approaches.)
(a) If $A$ and $B$ are positive-definite $m \times m$ matrices, why must their sum $A + B$ be positive definite?
(b) If $B$ is a positive-definite $m \times m$ matrix and $C$ is an $m \times n$ matrix with full column rank, why must $C^H B C$ be positive-definite?
(a) Recall the following definition of positive-definiteness: "a hermitian $m\times m$ matrix $M$ is positive-definite if for any non-zero complex $m$-vector $v$ we have $v^HMv>0$".
If $A, B$ are positive-definite $m\times m$ matrices, then for any non-zero vector $v\in\mathbb C^m$ we have $v^HAv>0$ and $v^HBv>0$. Hence $v^H(A+B)v=v^HAv+v^HBv>0$. Moreover $A+B$ is hermitian: $(A+B)^H=A^H+B^H=A+B$. Thus $A+B$ is positive definite.
(b) We can use the same definition as in part (a). $C^HBC$ is hermitian because $(C^HBC)^H=C^HB^HC=C^HBC$. Now assume that $v$ is a non-zero $n$-vector and consider $v^HC^HBCv$. Note that we can rewrite $$ v^HC^HBCv=(Cv)^HB(Cv). $$ Since $C$ is full-column rank the nullspace $N(C)$ consists only of the vector $0$, so $Cv\neq 0$. Hence, using that $B$ is positive-definite, we get $v^HC^HBCv=(Cv)^HB(Cv)>0$.
Alternative solution of (b) Recall another definition: "a hermitian $m\times m$ matrix $M$ is positive-definite if $M=A^HA$ for some matrix $A$ with full column rank". Since $B$ is positive-definite $B=A^HA$ for some full column rank matrix $A$. Consequently we have $C^HBC=C^HA^HAC=(AC)^H(AC)$ and to show that $C^HBC$ is positive-definite it is enough to show that $AC$ is of full column rank. For the latter fact recall that being of full column rank is equivalent to having $0$ nullspace and we should show that $N(AC)=0$. But this follows from the fact that $N(A)=0$ and $N(C)=0$ ($A$ and $C$ are both of full column rank): if $ACy=0$ then $Cy=0$ since $N(A)=0$, which implies $y=0$ since $N(C)=0$.
In class, we analyzed a system with $n$ masses and $n+1$ springs and showed that it satisfied an equation $$ m \frac{d^2x}{dt^2} = -D^T K D x $$ for the vector $x \in \mathbb{R}^n$ of displacements, and we showed that $-D^T (K/m) D $ was negative-definite, and that this led to oscillating solutions.
Suppose that the masses are not identical, and let $M$ be the diagonal $n \times n$ matrix of masses $m_1,\ldots,m_n > 0$. If we define $y = \sqrt{M} x$, then show that $\frac{d^2y}{dt^2} = By$ where $B$ is negative-definite, and hence we still have oscillating solutions.
Recall that in the equation above the right-hand side $-D^TKDx=\begin{pmatrix}F_1(x)\\ \dots \\ F_n(x)\end{pmatrix}$ corresponds to the forces generated by the springs (which do not depend on the masses) and the equation above is an application of the Newton's law to each particle: $$ m\frac{d^2x}{dt^2}=\begin{pmatrix} m\ddot{x}_1\\ \dots\\ m\ddot{x}_n\end{pmatrix}=\begin{pmatrix} F_1\\ \dots\\ F_n\end{pmatrix}=-D^TKDx. $$ Now, if we make masses not identical we need to change $m$ in the left-hand side to the corresponding masses, which can be expressed in the following way: $$ M\frac{d^2x}{dt^2}=\begin{pmatrix} m_1\ddot{x}_1\\ \dots\\ m_n\ddot{x}_n\end{pmatrix}=-D^TKDx, $$ where $M$ is the diagonal matrix of masses as in the statement of the problem. Note also that $\sqrt{M}$ is simply the diagonal matrix of the (real & positive) square roots of the masses.
So, the problem asks to rewrite the equation $$ M\frac{d^2x}{dt^2}=-D^TKDx $$ as $\frac{d^2y}{dt^2}=By$, where $B$ is negative-definite. As suggested by the problem, set $\sqrt{M}^{-1}y=x$ and substitute it in our equation with different masses: $$ M\frac{d^2(\sqrt{M}^{-1}y)}{dt^2}=-D^TKD\sqrt{M}^{-1}y. $$ Since the derivative commutes with multiplying by a constant matrix, we have $\frac{d^2(\sqrt{M}^{-1}y)}{dt^2}=\sqrt{M}^{-1}\frac{d^2y}{dt^2}$. Hence our equation is equivalent to $$ \sqrt{M}\frac{d^2y}{dt^2}=-D^TKD\sqrt{M}^{-1}y, $$ which is exactly $\frac{d^2y}{dt^2}=-By$ for the matrix $\boxed{B=-\sqrt{M}^{-1}D^TKD\sqrt{M}^{-1}}$.
Finally, we need to show that $B$ is negative-definite. This is equivalent to showing that $-B=\sqrt{M}^{-1}D^TKD\sqrt{M}^{-1}$ is positive definite. But we already know that $D^TKD$ is positive definite, and $\sqrt{M}^{-1} = (\sqrt{M}^{-1})^H$ is a diagonal real matrix (hence Hermitian) with full column rank. Hence, applying part (b) of Problem 1 and using that $M^H=M$, we get that $-B$ is positive-definite.
The nullspace $N(A)$ of the real matrix $A$ is spanned by the vector $v = \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix}$.
(a) Give an eigenvector and eigenvalue of the matrix $B = (3I - A^T A)(3I + A^T A)^{-1}$.
(b) Aside from the eigenvalue identified in (a), all other eigenvalues $\lambda$ of $B$ must be which of the following must be true (list all that apply):
Justify your answer.
(c) Give a good approximate formula for $B^n \begin{pmatrix} 0 \\ -1 \\ 0 \\ 8 \end{pmatrix}$ for a large $n$ (give an explicit numerical vector).
(a) Recall that $N(A)=N(A^TA)$, so the nullspace of $A^TA$ is also spanned by $v$. In other words, $0$ is an eigenvalue of $A^TA$ and all eigenvectors of $A^TA$ with eigenvalue $0$ are proportional to $v$. In particular, we have $$ (3I-A^TA)v=3Iv=3v, \qquad (3I+A^TA)v=3Iv=3v $$ and, consequently, we get $(3I+A^TA)^{-1}v=\frac{1}{3}v$ and $$ \boxed{Bv=(3I-A^TA)(3I+A^TA)^{-1}v=\frac{1}{3}(3I-A^TA)v=v} $$ In other words, $v$ is an eigenvector of $B$ with eigenvalue $1$.
(b) The argument we have used in part (a) can be repeated for any eigenvector $w$ of $A^TA$. Namely, if $w$ is an eigenvector of $A^TA$ with eigenvalue $\mu$, we have $$ (3I-A^TA)w=(3-\mu)w, \qquad (3I+A^TA)^{-1}w=(3+\mu)^{-1}w,\qquad (3I-A^TA)(3I+A^TA)^{-1}w=\frac{3-\mu}{3+\mu}w. $$ Moreover, this argument gives all eigenvalues of $B$: since $A^TA$ is symemtric real it is diagonalizable $A^TA=X\Lambda X^{-1}$, hence $$ B=(3I-X\Lambda X^{-1})(3I+X\Lambda X^{-1})^{-1}=(3XX^{-1}-X\Lambda X^{-1})(3XX^{-1}+X\Lambda X^{-1})^{-1}=X(3I-\Lambda)X^{-1}(X(3I+\Lambda)X^{-1})^{-1}=X(3I-\Lambda)(3I+\Lambda)^{-1}X^{-1} $$ so $B$ is diagonalizable with exaclty the same matrix $X$ and the diagonal matrix $(3I-\Lambda)(3I+\Lambda)^{-1}$.
So, all eigenvalues of $B$ are of the form $\lambda=\frac{3-\mu}{3+\mu}$ where $\mu$ is an eigenvalue of $A^TA$. Since $A^TA$ is real symmetric all $\mu$ are real, hence all eigenvalues of $B$ are real.
Since $A^TA$ is non-negative definite (always true for any $A$), we know that all eigenvalues $\mu$ of $A^TA$ are non-negative. Moreover, since we are asked to describe eigenvalues $\lambda$ different from part (a) (which corresponds to $\mu=0$), we can assume $\mu>0$. But in this case $$ \frac{3-\mu}{3+\mu}<\frac{3}{3+\mu}<1, \qquad \frac{3-\mu}{3+\mu}>\frac{-3-\mu}{3+\mu}=-1. $$ So $\boxed{|\lambda|<1}$ and of course that also means that $\boxed{|\lambda|\leq 1}$.
All other options are incorrect: since $A^TA$ can have arbitrary eigenvalues $\mu\geq 0$, we can have $\mu=1$ or $\mu=4$ as eigenvalues of $A^TA$. These correspond to possible eigenvalues of $B$ equal to $\lambda=0.5$ or $\lambda=-\frac{1}{7}$, giving counter-examples for the other options.
(c) Recall that to describe $B^n\begin{pmatrix}0\\-1\\0\\8\end{pmatrix}$ we want to write $\begin{pmatrix}0\\-1\\0\\8\end{pmatrix}=c_1x_1+c_2x_2+c_3x_3+c_4x_4$ and use that $$ B^n\begin{pmatrix}0\\-1\\0\\8\end{pmatrix}=\lambda_1^nc_1x_1+\lambda_2^nc_2x_2+\lambda_3^nc_3x_3+\lambda_4^nc_4x_4, $$ where $x_1, x_2, x_3, x_4$ and $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ are the eigenvectors and eigenvalues of $B$.
Part (a) shows that one of these eigenvalues (say $\lambda_1$) is equal to $1$, with the corresponding eigenvector $x_1=v$. In part (b) we have showed that $|\lambda_2|<1, |\lambda_3|<1, |\lambda_4|<1$. Hence, for large $n$ the powers $\lambda_2^n, \lambda_3^n,\lambda_4^n$ are small and we can give an approximation $$ B^n\begin{pmatrix}0\\-1\\0\\8\end{pmatrix}\approx c_1 \begin{pmatrix}1\\2\\3\\4\end{pmatrix}. $$
We only need to compute $c_1$. To do it note that the eigenvectors $x_1, x_2, x_3, x_4$ can be chosen to be orthogonal to each other, since matrix $B$ is real symmetric (to show it note that $3I+A^TA$ and $3I-A^TA$ are symmetric matrices which commute with each other. Also in part (b) we noted that $A^TA$ has the same eigenvectors as $B$).
Hence $c_1 v$ is just the projection of $u=\begin{pmatrix}0\\-1\\0\\8\end{pmatrix}$ onto $v$, and so $$ c_1 v= \frac{u^Tv}{v^Tv}v=\frac{30}{30}\begin{pmatrix}1\\2\\3\\4\end{pmatrix}=\begin{pmatrix}1\\2\\3\\4\end{pmatrix}. $$
So, $\boxed{B^n\begin{pmatrix}0\\-1\\0\\8\end{pmatrix}\approx\begin{pmatrix}1\\2\\3\\4\end{pmatrix}}$.