9Ddependence


 * Linear Dependence**

Two or more vectors are called Linearly Dependent if __**one**__ of them can be written as a linear combination of the others.

If **__no__** vector can be written as a linear combination of the others then they are Linearly Independent

Alternately, if we can add scalar multiples of the vectors together and get 0, they are Linearly Dependent.

math .\qquad \text{Three vectors : } v_1, \; v_2, \; v_3 \text{ are linearly dependent if } \qquad. \\ . \qquad k_1 v_1 +k_2 v_2 + k_3 v_3 = 0 \text{ for non zero values of some of the constants, k} \qquad. math

math . \qquad \text{ If the only way for } k_1 v_1 +k_2 v_2 + k_3 v_3 \text{ to equal zero is} \qquad. \\ . \qquad \text {when ALL k values are zero, then they are linearly independent.} \qquad. math

Two Vectors Two vectors can only be linearly dependent if one is a multiple of the other. (ie they are parallel)

Example 1 ... ... ... **__v__** = 3**__i__** + 4**__j__** and ... ... ... **__u__** = 6**__i__** + 8**__j__**

are linearly __**dependent**__ because **__u__** = 2**__v__**

... ... ... **__v__** = 3**__i__** + 4**__j__** and ... ... ... **__u__** = 3**__i__** – 4**__j__**

are linearly __**independent**__ because we can not write **__u__** as a multiple of **__v__**.

Three Vectors Three vectors that are in the same plane (coplanar) __must__ be linearly dependent

Example 2

... ... ... **__v__** = 3**__i__** + 4**__k__** and ... ... ... **__u__** = **__i__** + 3**__k__** and ... ... ... **__w__** = 5**__i__** + 5**__k__**

are coplanar (all in the x-z plane) so they are linearly dependent

In this case **__w__** = 2**__v__** – **__u__**

Example 3

Show that w can be written as a linear combination of u and v, where math . \qquad \underline{u}=2\underline{i}-\underline{j} \\ \text{ and} \qquad .\\ . \qquad \underline{v}=\underline{i}-\underline{j} \\ \text{ and} \qquad. \\ . \qquad \underline{w}=4\underline{i}+5\underline{j} math

math \\ . \qquad \text{Let } \underline{w}=a\underline{u}+b\underline{v} \qquad .\\. \\ . \qquad \underline{w}=a \big( 2\underline{i}-\underline{j} \big)+b\big( \underline{i}-\underline{j} \big) \qquad. math . math \\ . \qquad \underline{w}=2a\underline{i}-a\underline{j}+b\underline{i}-b\underline{j} \qquad .\\. \\ . \qquad \underline{w}= \big( 2a+b \big) \underline{i} + \big( -a-b \big) \underline{j} math

Thus, by equating coefficients, we get:

math . \qquad 2a+b=4 \\ \text{ and } \\ . \qquad -a-b=5 \qquad. math Solving simultaneously, we get:

math . \qquad a=9 \\ \text{ and } \\ . \qquad b=-14 \qquad. math

so

math . \quad \underline{w}=9\underline{u}-14\underline{v} \qquad. math

Hence, since w can be written as a linear combination of u and v, the three vectors are linearly dependent.

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