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$\newcommand{\la}{\langle}$ $\newcommand{\ra}{\rangle}$ $\newcommand{\vu}{\mathbf{u}}$ $\newcommand{\vv}{\mathbf{v}}$ $\newcommand{\vw}{\mathbf{w}}$ $\newcommand{\vzz}{\mathbf{z}}$ $\newcommand{\nc}{\newcommand}$ $\nc{\Cc}{\mathbb{C}}$ $\nc{\Rr}{\mathbb{R}}$ $\nc{\Qq}{\mathbb{Q}}$ $\nc{\Nn}{\mathbb{N}}$ $\nc{\cB}{\mathcal{B}}$ $\nc{\cE}{\mathcal{E}}$ $\nc{\cC}{\mathcal{C}}$ $\nc{\cD}{\mathcal{D}}$ $\nc{\mi}{\mathbf{i}}$ $\nc{\sp}[1]{\langle #1 \rangle}$ $\nc{\ol}[1]{\overline{#1} }$ $\nc{\norm}[1]{\left\| #1 \right\|}$ $\nc{\abs}[1]{\left| #1 \right|}$ $\nc{\vz}{\mathbf{0}}$ $\nc{\vo}{\mathbf{1}}$ $\nc{\DMO}{\DeclareMathOperator}$ $\DMO{\tr}{tr}$ $\DMO{\nullsp}{nullsp}$ $\nc{\va}{\mathbf{a}}$ $\nc{\vb}{\mathbf{b}}$ $\nc{\vx}{\mathbf{x}}$ $\nc{\ve}{\mathbf{e}}$ $\nc{\vd}{\mathbf{d}}$ $\nc{\vh}{\mathbf{h}}$ $\nc{\ds}{\displaystyle}$ $\nc{\bm}[1]{\begin{bmatrix} #1 \end{bmatrix}}$ $\nc{\gm}[2]{\bm{\mid & \cdots & \mid \\ #1 & \cdots & #2 \\ \mid & \cdots & \mid}}$ $\nc{\MN}{M_{m \times n}(K)}$ $\nc{\NM}{M_{n \times m}(K)}$ $\nc{\NP}{M_{n \times p}(K)}$ $\nc{\MP}{M_{m \times p}(K)}$ $\nc{\PN}{M_{p \times n}(K)}$ $\nc{\NN}{M_n(K)}$ $\nc{\im}{\mathrm{Im\ }}$ $\nc{\ev}{\mathrm{ev}}$ $\nc{\Hom}{\mathrm{Hom}}$ $\nc{\com}[1]{[\phantom{a}]^{#1}}$ $\nc{\rBD}[1]{ [#1]_{\cB}^{\cD}}$ $\DMO{\id}{id}$ $\DMO{\rk}{rk}$ $\DMO{\nullity}{nullity}$ $\DMO{\End}{End}$ $\DMO{\proj}{proj}$ $\nc{\GL}{\mathrm{GL}}$
Let us see how to apply Proposition 3 in Part I to decide if a give set of vectors is linearly dependent or not.
Checkpoint 1. Is the following set $A$ of vectors in $K^2$ linearly dependent or not? $$ \left\{\left[\begin{array}{r} -4 \\ -1 \end{array}\right], \left[\begin{array}{r} -3 \\ -3 \end{array}\right], \left[\begin{array}{r} -3 \\ 6 \end{array}\right]\right\} $$ If it is linearly dependent, express $\vz$ as a nontrivial linear combination of vectors in $A$.
Ans. A linearly independent subset of $K^2$ has size at most $2$, hence $A$ must be linearly dependent.
AL = {(-4,-1),(-3,-3),(-3,6)}; latex([column_matrix(aa) for aa in AL])
To express $\vz$ as a nontrivial linear combination of the vectors in $A$ amongs to finding a nontrivial solution to the system $A\vx = \vz$.
A = column_matrix(AL); A,A.rref()
The corresponding system is $$ \begin{cases} x_1 + 3x_3 & = 0 \\ x_2 - 3x_3 &= 0 \end{cases} $$
So $x_3$ is the free variable here and let's rename it to $s_1$. Therefore, the system is $$ x_1 + 3s_1 = 0, x_2 - 3s_1 =0, x_3 = s_1 $$ And by setting the free variable $s_1$ to $1$ we get
So $\bm{-3 \\ 3 \\ 1}$ is a nontrivial solution of system. In other words, $$ (-3)\bm{-4 \\ -1} + (3)\bm{-3 \\-3} + (1)\bm{-3 \\ 6} = \vz $$
Checkpoint 2. Determine whether the following set of $2 \times 2$ real matrices is linearly dependent or not
$$ S= \left\{ \begin{bmatrix} 1&-2\cr 2&3 \end{bmatrix},\, \begin{bmatrix} -2&1\cr 1&2 \end{bmatrix},\, \begin{bmatrix} -2&2\cr 1&0 \end{bmatrix} \right\} $$As far as being a vector space is concerned, $M_{n \times m}(K)$ are $\mathbb{K}^{mn}$ are the same. (The technical way of saying this is that they are isomorphic).
In particular $M_2(\Rr)$ and $\Rr^4$ are isomorphic as $\Rr$-vector space via the identification (Example 2 in Notes 15).
$$ \bm{a & b \\ c & d} \longleftrightarrow \bm{a \\ b \\ c \\d} $$Thus, the three matrices in $S$ correspond to the following three vectors in $\Rr^4$:
AL = [[1,-2,2,3],[-2,1,1,2],[-2,2,1,0]]; [column_matrix(aa) for aa in AL]
A = column_matrix(AL); A, A.rref() #form the matrix A with the those vectors as columns then row reduced it.
Since rref($A$) has no free columns, the three columns are independent and hence $S$ is linearly independent.