Assignment-daixieTM为您提供华盛顿大学University of WashingtonMATH 209 Linear Analysis 线性代数代写代考和辅导服务！
Instructions:
In broad terms, vectors are things you can add and linear functions are
functions of vectors that respect vector addition. The goal of this text is to
teach you to organize information about vector spaces in a way that makes
problems involving linear functions of many variables easy. (Or at least
tractable.)
To get a feel for the general idea of organizing information, of vectors,
and of linear functions this chapter has brief sections on each. We start
here in hopes of putting students in the right mindset for the odyssey that
follows; the latter chapters cover the same material at a slower pace. Please
be prepared to change the way you think about some familiar mathematical
objects and keep a pencil and piece of paper handy!
(a) $A$ and $B$ are any matrices with the same number of rows. What can you say (and explain why it is true) about the comparison of
$$
\operatorname{rank} \text { of } A \quad \operatorname{rank} \text { of the block mat rix }\left[\begin{array}{cc}
A & B
\end{array}\right]
$$
(b) Suppose $B=A^2$. How do those ranks compare ? Explain your reasoning,
(c) If $A$ is $m$ by $n$ of rank $r$, what are the dimensions of these nullspaces?
Nullspace of $A$
Nullspace of $\left[\begin{array}{ll}A & A\end{array}\right]$
(a) All you can say is that $\operatorname{rank} A \leq \operatorname{rank}[A B]$. ( $A$ can have any number $r$ of pivot columns, and these will all be pivot columns for $[A B]$; but there could be more pivot columns among the columns of $B$.)
(b) Now $\operatorname{rank} A=\operatorname{rank}\left[A A^2\right]$. (Every column of $A^2$ is a linear combination of columns of $A$. For instance, if we call $A$ ‘s first column $a_1$, then $A a_1$ is the first column of $A^2$. So there are no new pivot columns in the $A^2$ part of $\left[A A^2\right]$.)
(c) The nullspace $N(A)$ has dimension $n-r$, as always. Since $[A A]$ only has $r$ pivot columns – the $n$ columns we added are all duplicates – $[A A]$ is an $m$-by-2n matrix of rank $r$, and its nullspace $N([A A])$ has dimension $2 n-r$.
Suppose $A$ is a 5 by 3 matrix and $A x$ is never zero (except when $x$ is the zero vector).
(a) What can you say about the columns of $A$ ?
(b) Show that $A^{\mathrm{T}} A x$ is also never zero (except when $x=0$ ) by explaining this key step:
If $A^{\mathrm{T}} A x=0$ then obviously $x^{\mathrm{T}} A^{\mathrm{T}} A x=0$ and then (WHY?) $A x=0$.
(c) We now know that $A^{\mathrm{T}} A$ is invertible. Explain why $B=\left(A^{\mathrm{T}} A\right)^{-1} A^{\mathrm{T}}$ is a one-sided inverse of $A$ (which side of $A$ ?). $B$ is NOT a 2-sided inverse of $A$ (explain why not).
(a) $N(A)=0$ so $A$ has full column rank $r=n=3$ : the columns are linearly independent.
(b) $x^{\mathrm{T}} A^{\mathrm{T}} A x=(A x)^{\mathrm{T}} A x$ is the squared length of $A x$. The only way it can be zero is if $A x$ has zero length (meaning $A x=0$ ).
(c) $B$ is a left inverse of $A$, since $B A=\left(A^{\mathrm{T}} A\right)^{-1} A^{\mathrm{T}} A=I$ is the (3-by-3) identity matrix. $B$ is not a right inverse of $A$, because $A B$ is a 5-by-5 matrix but can only have rank 3 . (In fact, $B A=A\left(A^{\mathrm{T}} A\right)^{-1} A^{\mathrm{T}}$ is the projection onto the (3-dimensional) column space of $A$.)
The sentence in paranthesis can be completed a couple of different ways. One could write “There will be exactly one plane containing $0, a_1, a_2$ unless these three points are colinear”. Another acceptable answer is “There will be exactly one plane containing $0, a_1, a_2$ unless the vectors $a_1$ and $a_2$ are linearly dependent”.
In general, $0, a_1, \ldots, a_{n-1}$ will be contained in an unique hyperplane unless all of the points $0, a_1, \ldots, a_{n-1}$ are contained in an $n-2$ dimensional subspace. Said another way, $0, a_1, \ldots, a_{n-1}$ will be contained in an unique hyperplane unless the vectors $a_1, \ldots, a_{n-1}$ are linearly dependent.
