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\title{MATH 351, Fall 2017, Homework \#7}

\author{\underline{\bf Due Friday, October 27, 2017}}



\maketitle

\begin{problem} Show that a non-zero nilpotent matrix is not diagonalizable.
\end{problem}


\begin{problem}[5.36, p. 291] If $A$ is diagonalizable with eigenvalues $\lambda_1,\dots, \lambda_k$ and $q(x) = a_n x^n + a_{n-1}x^{n-1} + \dotsb + a_ 1 x + a_0$ is a polynomial such that $q(\lambda_i)=0$ for all $i=1,\dotsc,k$, show that $$q(A) = a_nA^n + a_{n-1}A^{n-1} + \dotsb + a_1A + a_0 I = 0.$$
\end{problem}



\begin{problem}cf. 5.37] Find all values of $a,b,c$ so that $$\begin{bmatrix*}[r] 2 & a & b\\ 0 & -5 & c\\ 0 & 0 & 2\end{bmatrix*}$$ is diagonalizable.
\end{problem}





\begin{problem}[cf. 6.4 and 6.5, p. 317] Geometrically describe each set.
\end{problem}

\begin{itemize}
\item[a)] For $\vec x = [1,1]^t$, the set $\{\vec y \in \mathbb R : \vec x\cdot\vec  y\leq \|\vec y\|\}$.

\item[b)] For $\vec x = [1,1,1]$, the set $\{\vec y\in \mathbb R^3 : \vec x\cdot\vec y\geq \frac{3}{2}\|\vec y\|\}.$


\end{itemize}

\begin{problem}[cf. 6.11, p. 318] For an orthogonal set of vectors $\{\vec v_1, \vec v_2, \vec v_3\}$ in $\mathbb R^4$, show that there is a vector $\vec v_4$ so that  $\{\vec v_1, \vec v_2, \vec v_3, \vec v_4\}$ forms an orthogonal basis for $\mathbb R^4$.
\end{problem}

\begin{problem}[cf. 6.14, p. 318] Show the Cauchy-Schwart inequality $$- \|\vec x\| \|\vec y\| \leq \vec x\cdot\vec y\leq  \|\vec x\| \|\vec y\|$$ for all $\vec x,\vec y\in \mathbb R^n$.
\end{problem}

Hint: Show that for any real number $R$ $$0\leq \|\vec x + R\vec y\|^2 = \|\vec x\|^2 + 2\|\vec x\| \|\vec y\|R + \|\vec y\|^2 R^2,$$ so this quadratic polynomial in $R$ has at most one root.

\begin{problem}[cf. 6.17, p. 318] Show that any orthogonal set of vectors $\{\vec v_1,\dotsc,\vec v_k\}$ in $\mathbb R^n$ is linearly independent.
\end{problem}





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