$\newcommand{\la}{\langle}$ $\newcommand{\ra}{\rangle}$ $\newcommand{\vu}{\mathbf{u}}$ $\newcommand{\vv}{\mathbf{v}}$ $\newcommand{\vw}{\mathbf{w}}$ $\newcommand{\vzz}{\mathbf{z}}$ $\newcommand{\nc}{\newcommand}$ $\nc{\Cc}{\mathbb{C}}$ $\nc{\Rr}{\mathbb{R}}$ $\nc{\Qq}{\mathbb{Q}}$ $\nc{\Nn}{\mathbb{N}}$ $\nc{\cB}{\mathcal{B}}$ $\nc{\cE}{\mathcal{E}}$ $\nc{\cC}{\mathcal{C}}$ $\nc{\cD}{\mathcal{D}}$ $\nc{\mi}{\mathbf{i}}$ $\nc{\sp}[1]{\langle #1 \rangle}$ $\nc{\ol}[1]{\overline{#1} }$ $\nc{\norm}[1]{\left\| #1 \right\|}$ $\nc{\abs}[1]{\left| #1 \right|}$ $\nc{\vz}{\mathbf{0}}$ $\nc{\vo}{\mathbf{1}}$ $\nc{\DMO}{\DeclareMathOperator}$ $\DMO{\tr}{tr}$ $\DMO{\nullsp}{nullsp}$ $\nc{\va}{\mathbf{a}}$ $\nc{\vb}{\mathbf{b}}$ $\nc{\vx}{\mathbf{x}}$ $\nc{\ve}{\mathbf{e}}$ $\nc{\vd}{\mathbf{d}}$ $\nc{\vh}{\mathbf{h}}$ $\nc{\ds}{\displaystyle}$ $\nc{\bm}[1]{\begin{bmatrix} #1 \end{bmatrix}}$ $\nc{\gm}[2]{\bm{\mid & \cdots & \mid \\ #1 & \cdots & #2 \\ \mid & \cdots & \mid}}$ $\nc{\MN}{M_{m \times n}(K)}$ $\nc{\NM}{M_{n \times m}(K)}$ $\nc{\NP}{M_{n \times p}(K)}$ $\nc{\MP}{M_{m \times p}(K)}$ $\nc{\PN}{M_{p \times n}(K)}$ $\nc{\NN}{M_n(K)}$ $\nc{\im}{\mathrm{Im\ }}$ $\nc{\ev}{\mathrm{ev}}$ $\nc{\Hom}{\mathrm{Hom}}$ $\nc{\com}[1]{[\phantom{a}]^{#1}}$ $\nc{\rBD}[1]{ [#1]_{\cB}^{\cD}}$ $\DMO{\id}{id}$ $\DMO{\rk}{rk}$ $\DMO{\nullity}{nullity}$ $\DMO{\End}{End}$ $\DMO{\proj}{proj}$ $\nc{\GL}{\mathrm{GL}}$
Expand the set below to form a basis for $\Rr^3$ by adding vectors to the set. $$\left\{\bm{1 \\ 2 \\1}\right\}$$
Ans: There are many choices of vectors to add to this set to complete a basis of $\Rr^3$. But let us illustrate the idea exemplified in Notes 20 Lemma 3. First, call the only vector in the given set $\vv_1$. Since $\vv_1 \neq \vz$, the set $\{\vv_1\}$ is linearly independent. To expand it to a bigger independent set, according to Lemma 3, we can add to it a vector $\vv_2$ that in not in the span of $\{\vv_1\}$. Since we only have one vector in $\{\vv_1\}$, $\sp{\{\vv_1\}} = \{c \vv_1 \colon c \in \Rr\}$ is simply the set of multiples of $\vv_1$. Thus, we can take $\vv_2$ as $\ve_1 = \bm{1 \\ 0 \\0}$ which is clearly not a multiple of $\vv_1$. Now the set $\{\vv_1, \vv_2\}$ is still, by Lemma 3, linearly independent.
To complete a basis of $\Rr^3$, we need to add one more vector, say $\vv_3$ to $\{\vv_1, \vv_2\}$ so that the resulting set $\{\vv_1,\vv_2,\vv_3\}$ is still linearly independent (Proposition 2 Notes 19). Again, according to Lemma 3 of Notes 20, all we need is to choose a $\vv_3$ that is not in the span of $\{\vv_1,\vv_2\}$. We can pick $\ve_3$ to be $\ve_2 = \bm{0 \\1 \\0}$. To see check that $\{\vv_1, \vv_2, \vv_3\}$ is linearly independent, we form the matrix with these vectors as columns and compute its rref:
%display latex
A = column_matrix([(1,2,1),(1,0,0),(0,0,1)]); A, A.rref()
Since the rref of $A$ (which is $I_3$) has no free columns, according to Proposition 3 in Notes 16, the three columns of $A$ are linearly independent and hence form a basis of $\Rr^3$.
Find a basis of the subspace of $\Rr^4$ consisting of all vectors of the form $$ \bm{x_1 \\ -8x_1 + x_2 \\ -5x_1 +5x_2 \\ -9x_1 -4x_2} $$
Ans. Note that $$ \bm{x_1 \\ -8x_1 + x_2 \\ -5x_1 +5x_2 \\ -9x_1 -4x_2} = x_1\bm{1\\-8\\-5\\-9} + x_2\bm{0 \\1 \\ 5 \\-4} $$
Thus the set of all vectors of the form above is simply the span of $$\left\{ \bm{1\\-8\\-5\\-9}, \bm{0 \\1 \\ 5 \\-4} \right\}$$
Certainly, it is a spanning set of its span (just a tautology). It is also clear that these two vectors are linearly independent (if you cannot spot it by inspection, again you can form a matrix with these vectors as columns and verify that its rref has no free columns). Therefore, the last set on display is a basis of its span.
Accrording to the question, we need to enter the answer as a list of row vectors separated by commas. So it will be <1,-8,-,5,-9>, <0,1,5,-4>
Let A be the following matrix
A = column_matrix([(2,4,-6,8),(4,1,-5,2),(-2,-3,5,-6),(0,-4,4,-8),(-4,-4,9,-9)]); A
(a) Find a basis of the column space of $A$.
Ans. First compute the rref of $A$.
A, A.rref()
The pivotal columns, i.e. 1st, 2nd and the 5th column of $R_A$ form a basis of its column space. Since $A$ and $R_A$ are row equivalent, according to Theorem 1 of Notes 21, the correspond columns, i.e. the 1st column =$\bm{2\\4\\-6\\8}$, the 2nd column = $\bm{4\\1\\-5\\2}$ and the 5th column = $\bm{-4\\-4\\9\\-9}$ of $A$ form a basis of the column space of $A$.
Type <2,4,-6,8>, <4,1,-5,2>, <-4,-4,9,-9> to enter them as a list of row vectors separated by commas.
(b) Thus, the dimension of the column space of $A$ is 3.
As we have seen, this is because three out of five columns in rref($A$) have pivots. Also because the basis we found for column space of $A$ has three vectors.
(c) The column space of $A$ is a subspace of $\Rr^4$ because each column vector is in $\Rr^4$. (In webwork type R^4 for $\Rr^4$).
(d) The geometry of the column space of $A$ is a 3-dimensional subspace of $\Rr^4$.