Discover Linear Algebra

(a)

Do you remember what $\Span$ means? Explain why the vector $\uvec{x} = 3\uvec{v}_1 + 2\uvec{v}_2 - \uvec{v}_3$ is in $\Span \{\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\}\text{.}$

(b)

Actually, $\uvec{v}_2$ can be expressed as a linear combination of $\uvec{v}_1$ and $\uvec{v}_3$ — do you see how?

Use this and the expression for $\uvec{x}$ in Task a to express $\uvec{x}$ as a linear combination of just $\uvec{v}_1$ and $\uvec{v}_3\text{.}$

(c)

Task b shows that $\uvec{x}$ is in $\Span \{\uvec{v}_1,\uvec{v}_3\}\text{.}$ Do you think that similar calculations and the same reasoning can be carried out for every vector in $\Span \{\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\}\text{?}$ What does this say about $\Span \{\uvec{v}_1,\uvec{v}_2,\uvec{v}_3\}$ versus $\Span \{\uvec{v}_1,\uvec{v}_3\}\text{?}$

(a)

Use the test to verify that $\uvec{v}_1,\uvec{v}_2,\uvec{v}_3$ from Discovery 18.1 are linearly dependent.

(b)

Use the test to verify that $\uvec{v}_1,\uvec{v}_3$ from Discovery 18.1 are linearly independent.

(a)

Consider abstract vectors $\uvec{u}_1,\uvec{u}_2,\uvec{u}_3\text{,}$ and suppose the vector equation

has a nontrivial solution. This means that there are values for the scalars $k_1,k_2,k_3\text{,}$ at least one of which is not zero, so that equation ($\star\star$) is true.

Use some algebra to manipulate equation ($\star\star$) to demonstrate that one of the vectors can be expressed as a linear combination of the others (and hence, by definition, the vectors $\uvec{u}_1,\uvec{u}_2,\uvec{u}_3$ are linearly dependent).

(b)

Consider abstract vectors $\uvec{w}_1,\uvec{w}_2,\uvec{w}_3\text{,}$ and suppose the vector equation

has only the trivial solution. We would like to see why this means that $\uvec{w}_1,\uvec{w}_2,\uvec{w}_3$ are linearly independent.

Suppose they weren't: for example, suppose $\uvec{w}_3 = c_1\uvec{w}_1 + c_2\uvec{w}_2$ were true for some scalars $c_1,c_2\text{.}$ Manipulate this expression for $\uvec{w}_3$ until is says something about equation ($\star\star\star$). Do you see now why $\uvec{w}_1,\uvec{w}_2,\uvec{w}_3$ cannot satisfy the definition of linearly dependence, and hence must be linearly independent?

(a)

$V = \matrixring_2(\R)\text{,}$ $S= \left\{\; \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}, \;\; \begin{bmatrix} 0 \amp 1\\1 \amp 0 \end{bmatrix}, \;\; \begin{bmatrix} 0 \amp 0\\0 \amp 1 \end{bmatrix} \; \right\} \text{.}$

(b)

$V = \matrixring_2(\R)\text{,}$ $S= \left\{\; \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix}, \;\; \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}, \;\; \begin{bmatrix} 3 \amp 0 \\ 0 \amp -2 \end{bmatrix} \; \right\} \text{.}$

(c)

$V = \poly(\R)\text{,}$ $S = \{ 1+x, 1+x^2, 2 - x + 3x^2 \}\text{.}$

(d)

(a)

Do you think it's possible to have a set of three linearly independent vectors in $\R^2\text{?}$ Why or why not?

(b)

Do you think it's possible to have a set of four linearly independent vectors in $\R^3\text{?}$ Why or why not?

(a)

What does the definition of linear dependence say in the case of just two vectors?

(b)

If the test for linear dependence/independence is to remain true in the case of a “set” of vectors consisting of just one vector, how should we define linear dependence/independence for such a set?