From 0d2752001aeadf8cb7900303bed2c3692f637006 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Tue, 15 Oct 2024 13:01:47 -0500 Subject: [PATCH] update ev5 sample --- .../source/02-EV/samples/05.ptx | 214 +++++++----------- 1 file changed, 79 insertions(+), 135 deletions(-) diff --git a/source/linear-algebra/source/02-EV/samples/05.ptx b/source/linear-algebra/source/02-EV/samples/05.ptx index a988c1be..88c83ec3 100644 --- a/source/linear-algebra/source/02-EV/samples/05.ptx +++ b/source/linear-algebra/source/02-EV/samples/05.ptx @@ -1,138 +1,82 @@ EV5 - -
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  1. -

    -Write a statement involving spanning and independence properties -that's equivalent to each claim below. -

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    • -

      -The set of vectors \left\{ \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ --4 -\end{array}\right] , \left[\begin{array}{c} -0 \\ -1 \\ -3 \\ --3 -\end{array}\right] , \left[\begin{array}{c} -3 \\ -11 \\ -18 \\ --18 -\end{array}\right] , \left[\begin{array}{c} --2 \\ --7 \\ --11 \\ -11 -\end{array}\right] \right\} is a basis -of \mathbb{R}^4. -

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      • -
    • -

      -The set of vectors \left\{ \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ --4 -\end{array}\right] , \left[\begin{array}{c} -0 \\ -1 \\ -3 \\ --3 -\end{array}\right] , \left[\begin{array}{c} -3 \\ -11 \\ -18 \\ --18 -\end{array}\right] , \left[\begin{array}{c} --2 \\ --7 \\ --11 \\ -11 -\end{array}\right] \right\} is not a basis -of \mathbb{R}^4. -

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      • -
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    2. -
  2. -

    -Explain how to determine which of these statements is true. -

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    3. -
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The set of vectors \left\{ \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ --4 -\end{array}\right] , \left[\begin{array}{c} -0 \\ -1 \\ -3 \\ --3 -\end{array}\right] , \left[\begin{array}{c} -3 \\ -11 \\ -18 \\ --18 -\end{array}\right] , \left[\begin{array}{c} --2 \\ --7 \\ --11 \\ -11 -\end{array}\right] \right\} is a basis -of \mathbb{R}^4 exactly when it is linearly independent and the set spans \mathbb{R}^4. - If it is either linearly dependent, or the set does not span \mathbb{R}^4, then the set is not a basis. -

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To answer this, we compute - -\mathrm{RREF}\, \left[\begin{array}{cccc} -1 & 0 & 3 & -2 \\ -3 & 1 & 11 & -7 \\ -4 & 3 & 18 & -11 \\ --4 & -3 & -18 & 11 -\end{array}\right] = \left[\begin{array}{cccc} -1 & 0 & 3 & -2 \\ -0 & 1 & 2 & -1 \\ -0 & 0 & 0 & 0 \\ -0 & 0 & 0 & 0 -\end{array}\right] - . -

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We see that this set of vectors is linearly dependent, so therefore -the set of vectors \left\{ \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ --4 -\end{array}\right] , \left[\begin{array}{c} -0 \\ -1 \\ -3 \\ --3 -\end{array}\right] , \left[\begin{array}{c} -3 \\ -11 \\ -18 \\ --18 -\end{array}\right] , \left[\begin{array}{c} --2 \\ --7 \\ --11 \\ -11 -\end{array}\right] \right\} is - -not a basis. -

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Consider the set of vectors \left\{ \left[\begin{array}{c} 1 \\ -2 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -4 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ -6 \\ 0 \\ 3 \end{array}\right] , \left[\begin{array}{c} 5 \\ -5 \\ -2 \\ 4 \end{array}\right] \right\}

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Write a statement involving the solutions of a vector equation that's equivalent to this claim: The set of vectors is a basis for

+ + +

The vector equation + x_1 \left[\begin{array}{c} 1 \\ -2 \\ 0 \\ 1 \end{array}\right] +x_2 \left[\begin{array}{c} 2 \\ -4 \\ 0 \\ 2 \end{array}\right] +x_3 \left[\begin{array}{c} 3 \\ -6 \\ 0 \\ 3 \end{array}\right] +x_4 \left[\begin{array}{c} 5 \\ -5 \\ -2 \\ 4 \end{array}\right] =\vec w + has exactly one solution for every \vec w\in\mathbb R^4.

+ + + + +

Explain and demonstrate how to determine whether or not this statement is true.

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Since \mathrm{RREF}\, \left[\begin{array}{cccc} 1 & 2 & 3 & 5 \\ -2 & -4 & -6 & -5 \\ 0 & 0 & 0 & -2 \\ 1 & 2 & 3 & 4 \end{array}\right] = \left[\begin{array}{cccc} 1 & 2 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right], + we see from the zero row that there are some vectors \vec w for which the equation is not true, so the set fails to span and therefore fails to be a basis.

+ + +

Since \mathrm{RREF}\, \left[\begin{array}{cccc} 1 & 2 & 3 & 5 \\ -2 & -4 & -6 & -5 \\ 0 & 0 & 0 & -2 \\ 1 & 2 & 3 & 4 \end{array}\right] = \left[\begin{array}{cccc} 1 & 2 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right], + we see from the non-pivot column that there are some vectors \vec w for which the equation has infinitely-many solutions, so the set is linearly dependent and therefore fails to be a basis.

+ + + + + +

Consider the set of vectors \left\{ \left[\begin{array}{c} 1 \\ 3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -3 \end{array}\right] \right\}

+ + + +

Write a statement involving the solutions of a vector equation that's equivalent to this claim: The set of vectors is a basis for

+ + +

The vector equation + x_1 \left[\begin{array}{c} 1 \\ 3 \\ 4 \\ -4 \end{array}\right] +x_2 \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ 4 \end{array}\right] +x_3 \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -3 \end{array}\right] =\vec w + has exactly one solution for every \vec w\in\mathbb R^4.

+ + + + +

Explain and demonstrate how to determine whether or not this statement is true.

+ + +

Since \mathrm{RREF}\, \left[\begin{array}{ccc} 1 & -1 & 0 \\ 3 & -3 & 1 \\ 4 & -4 & 3 \\ -4 & 4 & -3 \end{array}\right] = \left[\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] + we see from the zero row that there are some vectors \vec w for which the equation is not true, so the set fails to span and therefore fails to be a basis.

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The set has only three vectors, so the set cannot span and there must be vectors for which the equation has no solutions. Therefore the set is not a basis.

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Consider the set of vectors \left\{ \left[\begin{array}{c} 3 \\ 2 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ -1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ -1 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 0 \\ -2 \end{array}\right] \right\}

+ + + +

Write a statement involving the solutions of a vector equation that's equivalent to this claim: The set of vectors is a basis for

+ + +

The vector equation + x_1 \left[\begin{array}{c} 3 \\ 2 \\ -1 \\ 0 \end{array}\right] +x_2 \left[\begin{array}{c} -2 \\ -1 \\ 0 \\ -1 \end{array}\right] +x_3 \left[\begin{array}{c} -2 \\ -1 \\ 1 \\ 0 \end{array}\right] + x_4\left[\begin{array}{c} -4 \\ -1 \\ 0 \\ -2 \end{array}\right] =\vec w + has exactly one solution for every \vec w\in\mathbb R^4.

+ + + + +

Explain and demonstrate how to determine whether or not this statement is true.

+ + +

Since \mathrm{RREF}\, \left[\begin{array}{cccc} 3 & -2 & -2 & -4 \\ 2 & -1 & -1 & -1 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & -2 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] + we see the equation always has exactly one solution (each row and column has a pivot). Therefore the set is spanning and linearly independent, and therefore the set is a basis.

+ + +