%display latex
latex.matrix_delimiters(left='[', right=']')

$\newcommand{\la}{\langle}$ $\newcommand{\ra}{\rangle}$ $\newcommand{\vu}{\mathbf{u}}$ $\newcommand{\vv}{\mathbf{v}}$ $\newcommand{\vw}{\mathbf{w}}$ $\newcommand{\vzz}{\mathbf{z}}$ $\newcommand{\nc}{\newcommand}$ $\nc{\Cc}{\mathbb{C}}$ $\nc{\Rr}{\mathbb{R}}$ $\nc{\Qq}{\mathbb{Q}}$ $\nc{\Nn}{\mathbb{N}}$ $\nc{\cB}{\mathcal{B}}$ $\nc{\cE}{\mathcal{E}}$ $\nc{\cC}{\mathcal{C}}$ $\nc{\cD}{\mathcal{D}}$ $\nc{\mi}{\mathbf{i}}$ $\nc{\sp}[1]{\langle #1 \rangle}$ $\nc{\ol}[1]{\overline{#1} }$ $\nc{\norm}[1]{\left\| #1 \right\|}$ $\nc{\abs}[1]{\left| #1 \right|}$ $\nc{\vz}{\mathbf{0}}$ $\nc{\vo}{\mathbf{1}}$ $\nc{\DMO}{\DeclareMathOperator}$ $\DMO{\tr}{tr}$ $\DMO{\nullsp}{nullsp}$ $\nc{\va}{\mathbf{a}}$ $\nc{\vb}{\mathbf{b}}$ $\nc{\vx}{\mathbf{x}}$ $\nc{\ve}{\mathbf{e}}$ $\nc{\vd}{\mathbf{d}}$ $\nc{\vh}{\mathbf{h}}$ $\nc{\ds}{\displaystyle}$ $\nc{\bm}[1]{\begin{bmatrix} #1 \end{bmatrix}}$ $\nc{\gm}[2]{\bm{\mid & \cdots & \mid \\ #1 & \cdots & #2 \\ \mid & \cdots & \mid}}$ $\nc{\MN}{M_{m \times n}(K)}$ $\nc{\NM}{M_{n \times m}(K)}$ $\nc{\NP}{M_{n \times p}(K)}$ $\nc{\MP}{M_{m \times p}(K)}$ $\nc{\PN}{M_{p \times n}(K)}$ $\nc{\NN}{M_n(K)}$ $\nc{\im}{\mathrm{Im\ }}$ $\nc{\ev}{\mathrm{ev}}$ $\nc{\Hom}{\mathrm{Hom}}$ $\nc{\com}[1]{[\phantom{a}]^{#1}}$ $\nc{\rBD}[1]{ [#1]_{\cB}^{\cD}}$ $\DMO{\id}{id}$ $\DMO{\rk}{rk}$ $\DMO{\nullity}{nullity}$ $\DMO{\End}{End}$ $\DMO{\proj}{proj}$ $\nc{\GL}{\mathrm{GL}}$

Linear Dependence (Part III)

Checkpoint 3. Suppose $S=\{\vu,\vv,\vw\}$ is a set of linearly independent vectors.

If $\vx=3\vu+\vv+3\vw$, determine whether $T=\{\vu,\vv,\vx\}$ is a linearly independent set.

Suppose $a \vu + b \vv + c\vx = \vz$ for some scalars $a,b$ and $c$.

So $a\vu + b\vv + c(3 \vu + \vv + 3\vw) = (a + 3c) \vu + (b+ c)\vv + 3c \vw = \vz$.

Since $S$ is, by assumption, linearly independent, we must have $$ \begin{cases} a + 3c &= 0 \\ b + c &= 0 \\ 3c & = 0 \end{cases} $$

By solving the system, we see that these equations imply $a=b=c=0$.

And this shows that $T$ is linearly independent.

Checkpoint 4. Determine that value(s) of $k$ so that the following vectors are linearly independent.

$$ \vu = \left[\begin{array}{c} -1\cr -5\cr -8 \end{array}\right], \ \ \ \ \vv = \left[\begin{array}{c} 3\cr 6\cr -8+k\cr \end{array}\right], \ \ \ \ \vw = \left[\begin{array}{c} 2\cr 1\cr -2 \end{array}\right] $$

Again let's form the matrix with these vectors as columns.

var('k');
AL = [[-1,-5,-8],[3,6,-8+k],[2,1,-2]]
A = column_matrix(AL); A

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{rrr} -1 & 3 & 2 \\ -5 & 6 & 1 \\ -8 & k - 8 & -2 \end{array}\right]$$

The columns of $A$ are independent iff rref($A$) has no free columns (Proposition 3 Notes 16) iff $\det(A) \neq 0$ since $A$ is a square matrix in this case (Theorem 1 Notes 11).

det(A) #compute the determinant of A.

$$\newcommand{\Bold}[1]{\mathbf{#1}}-9 \, k + 126$$

Thus, we conclude that the vectors $\vu, \vv, \vw$ are linearly independent iff $-9k+126 \neq 0$, i.e. $k \neq 14$.

Certainly, we can also use row reduction to arrive to the same conclusion:

A.rescale_row(0,-1); A.add_multiple_of_row(1,0,5); A.add_multiple_of_row(2,0,8);A

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{rrr} 1 & -3 & -2 \\ 0 & -9 & -9 \\ 0 & k - 32 & -18 \end{array}\right]$$

A.rescale_row(1,-1/9); A

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{rrr} 1 & -3 & -2 \\ 0 & 1 & 1 \\ 0 & k - 32 & -18 \end{array}\right]$$

A.add_multiple_of_row(2,1,32-k);A

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{rrr} 1 & -3 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & -k + 14 \end{array}\right]$$

So again, the three columns will form a linearly independent set unless $k = 14$.