diff --git a/source/linear-algebra/source/02-EV/06.ptx b/source/linear-algebra/source/02-EV/06.ptx index 9a9c2d06d..51e83ee9c 100644 --- a/source/linear-algebra/source/02-EV/06.ptx +++ b/source/linear-algebra/source/02-EV/06.ptx @@ -43,19 +43,20 @@ Class Activities +

-Recall from section that a subspace of a vector space is -the result of spanning a set of vectors from that vector space. +In we saw an example of +two linearly independent vectors spanning a planar subspace of \IR^3.

-Recall also that a linearly dependent set contains redundant vectors. For example, -only two of the three vectors in are needed to span -the planar subspace. +Because these independent vectors fail to span \IR^3, they are not +a basis for \IR^3. However, they still span a subspace of +\IR^3...

- +

Consider the subspace of \IR^4 given by W=\vspan\left\{ @@ -68,33 +69,61 @@ the planar subspace.

-

- Mark the column of \RREF\left[\begin{array}{cccc} - 2&2&2&1\\ - 3&0&-3&5\\ - 0&1&2&-1\\ - 1&-1&-3&0 - \end{array}\right] that shows that W's spanning set - is linearly dependent. -

+ +

+Which feature of \RREF\left[\begin{array}{cccc} +2&2&2&1\\ +3&0&-3&5\\ +0&1&2&-1\\ +1&-1&-3&0 +\end{array}\right]= +\left[\begin{array}{cccc} +1&0&-1&0\\ +0&1& 2&0\\ +0&0& 0&1\\ +0&0& 0&0 +\end{array}\right] shows that W's spanning set +is linearly dependent? +

    +
  1. The third column.
    2. +
  2. The fourth column.
    3. +
  3. The third row.
    4. +
  4. The fourth row.
    5. +
+

+ + +

A.

+

+The third columns lacks a pivot, introducing a free variable that +prevents uniqueness of linear combinations. +

+ -

-What would be the result of removing the vector that gave us this column? -

    -
  1. The set still spans W, and remains linearly dependent.
    2. -
  2. The set still spans W, but is now also linearly independent.
    3. -
  3. The set no longer spans W, and remains linearly dependent.
    4. -
  4. The set no longer spans W, but is now linearly independent.
    5. -
-

- + +

+If we removed the vector that causes this issue, +what could we say about that set of three vectors? +

    +
  1. The set spans the vector space \IR^4, but remains linearly dependent.
    2. +
  2. The set spans the subspace W of \IR^4, but remains linearly dependent.
    3. +
  3. The set spans the subspace W of \IR^4, and is now linearly independent.
    4. +
  4. The set no longer spans the subspace W of \IR^4, but is now linearly independent.
    5. +
+

+ + +

C.

+

+Because the removed vector was already a linear combination of the others, +we still span W. Now that all vectors yield pivot columns, the set +is now independent. +

+ + - -rref([2,2,2,1; 3,0,-3,5; 0,1,2,-1; 1,-1,-3,0]) - -

@@ -145,55 +174,6 @@ What would be the result of removing the vector that gave us this column?

- - - - - - - -

@@ -248,19 +228,22 @@ T=\left\{ .

-Thus the basis for a subspace is not unique in general. +Thus a given basis for a subspace need not be unique.

- - +

Any non-trivial real vector space has infinitely-many different bases, but all the bases for a given vector space are exactly the same size. + So we say the dimension of a vector space or subspace is equal to the + size of any basis for the vector space.

- For example, + As you'd expect, \IR^n has dimension n. + For example, \IR^3 has dimension 3 because any basis for \IR^3 + such as \setList{\vec e_1,\vec e_2,\vec e_3} \text{ and } @@ -276,37 +259,23 @@ Thus the basis for a subspace is not unique in general. \left[\begin{array}{c}3\\-2\\5\end{array}\right] } - are all valid bases for \IR^3, and they all contain three vectors. -

- - - - - -

- The dimension of a vector space or subspace is equal to the size - of any basis for the vector space. + contains exactly three vectors.

- As you'd expect, \IR^n has dimension n. - For example, \IR^3 has dimension 3 because any basis for \IR^3 - such as + Likewise, the planar subspace with the following two bases - \setList{\vec e_1,\vec e_2,\vec e_3} - \text{ and } \setList{ - \left[\begin{array}{c}1\\0\\0\end{array}\right], - \left[\begin{array}{c}0\\1\\0\end{array}\right], - \left[\begin{array}{c}1\\1\\1\end{array}\right] + \left[\begin{array}{c}1\\2\\3\end{array}\right], + \left[\begin{array}{c}-2\\0\\5\end{array}\right] } \text{ and } \setList{ - \left[\begin{array}{c}1\\0\\-3\end{array}\right], - \left[\begin{array}{c}2\\-2\\1\end{array}\right], - \left[\begin{array}{c}3\\-2\\5\end{array}\right] + \left[\begin{array}{c}-1\\2\\8\end{array}\right], + \left[\begin{array}{c}0\\4\\11\end{array}\right] } - contains exactly three vectors. + has dimension 2 because any basis for the subspace + will have exactly two vectors.