Question10. (a) Show that the vectors $(1,0,0),(0,1,0),(0,0,1)$ span $\mathbf{R}^{3}$ and that they are also linearly independent. (b) Show that the vectors $(1,0,0),(0,1,0),(0,0,1)$ , $(0,1,1)$ span $\mathbf{R}^{3}$ . Demonstrate that it is not an efficient spanning set by showing that an arbitrary vector in $\mathbf{R}^{3}$ can be expressed in more than one way as a linear combination of these vectors. We can think of $(0,1,1)$ as being a redundant vector. (c) Show that $\{(1,0,0),(0,1,0),(0,0,1),(0,1,1)\}$ is linearly dependent and is thus not a basis for $\mathbf{R}^{2}$ . A basis consists of a set of vectors, all of which are needed.

Studdy Solution

STEP 1

1. We are working in the vector space $\mathbf{R}^3$ .
2. A set of vectors spans $\mathbf{R}^3$ if any vector in $\mathbf{R}^3$ can be expressed as a linear combination of these vectors.
3. A set of vectors is linearly independent if no vector in the set can be expressed as a linear combination of the others.
4. A set of vectors is linearly dependent if at least one vector in the set can be expressed as a linear combination of the others.

STEP 2

1. Show that the vectors $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ span $\mathbf{R}^3$ and are linearly independent.
2. Show that the vectors $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ , $(0,1,1)$ span $\mathbf{R}^3$ and demonstrate redundancy.
3. Show that the set $\{(1,0,0),(0,1,0),(0,0,1),(0,1,1)\}$ is linearly dependent.

STEP 3

To show that $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ span $\mathbf{R}^3$ , we need to express an arbitrary vector $(x,y,z)$ in $\mathbf{R}^3$ as a linear combination of these vectors:
$(x, y, z) = x(1,0,0) + y(0,1,0) + z(0,0,1)$
This shows that any vector $(x, y, z)$ can be expressed as a linear combination of the given vectors, hence they span $\mathbf{R}^3$ .

STEP 4

To show linear independence, we need to check if the equation:
$a(1,0,0) + b(0,1,0) + c(0,0,1) = (0,0,0)$
implies that $a = b = c = 0$ .
Solving this gives:
$a = 0, \quad b = 0, \quad c = 0$
Thus, the vectors are linearly independent.

STEP 5

To show that $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ , $(0,1,1)$ span $\mathbf{R}^3$ , we note that the first three vectors already span $\mathbf{R}^3$ . Adding $(0,1,1)$ does not change the span, but introduces redundancy.
To demonstrate redundancy, express $(x,y,z)$ as:
$(x, y, z) = x(1,0,0) + (y-z)(0,1,0) + z(0,1,1)$
and also as:
$(x, y, z) = x(1,0,0) + y(0,1,0) + z(0,0,1)$
This shows multiple representations, indicating redundancy.

STEP 6

To show linear dependence, consider the set $\{(1,0,0),(0,1,0),(0,0,1),(0,1,1)\}$ .
Check if there exist scalars, not all zero, such that:
$a(1,0,0) + b(0,1,0) + c(0,0,1) + d(0,1,1) = (0,0,0)$
Choose $b = -d$ and $d = 1$ , then:
$a(1,0,0) + (-1)(0,1,0) + c(0,0,1) + 1(0,1,1) = (0,0,0)$
This simplifies to:
$a = 0, \quad c = 0, \quad b = -1, \quad d = 1$
Thus, the set is linearly dependent.

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