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Course Notes for Linear Algebra

Remkes Kooistra

Section 5.4 Week 5 Activity

Subsection 5.4.1 Dimensions and Bases of Spans

Activity 5.4.1.

Calculate the dimension of this span and give a basis.
\begin{equation*} \Span \left\{ \left( \begin{matrix} -3 \\ -6 \\ 1 \end{matrix} \right), \left( \begin{matrix} 2 \\ 5 \\ -3 \end{matrix} \right), \left( \begin{matrix} 1 \\ 4 \\ -5 \end{matrix} \right) \right\} \end{equation*}
Solution.
I need to determine if there are any redundancies in the span. To do this, I put the vectors in the span as rows of a matrix.
\begin{equation*} \left( \begin{matrix} -3 \amp -6 \amp 1 \\ 2 \amp 5 \amp -3 \\ 1 \amp 4 \amp -5 \end{matrix} \right) \end{equation*}
I row reduce this matrix. I will either row reduce without interchanging rows or keeping track of row exchanges. (If done by computer, as is the usual practice, I’ll have to be careful choosing a basis. Because of this, I’ll recheck the bases after I find them.)
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp \frac{13}{3} \\ 0 \amp 1 \amp \frac{-7}{3} \\ 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}
Every row that has a leading one is a row that is necessary for the span; every row without a leading one is redundant. Here, there is one vector that is redundant, since there is a row of zeros. There are two leading ones, so the dimension of this span is 2. I can take the vectors that corresponded to rows with leading ones as a basis.
\begin{equation*} \left\{ \left( \begin{matrix} -3 \\ -6 \\ 1 \end{matrix} \right), \left( \begin{matrix} 2 \\ 5 \\ -3 \end{matrix} \right) \right\} \end{equation*}
I can check with a matrix that these four are independent using the same steps as before (I’ve not shown that work). I need this check to make sure rows weren’t exchanged during row reduction.

Activity 5.4.2.

Calculate the dimension of this span and give a basis.
\begin{equation*} \Span \left\{ \left( \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right), \left( \begin{matrix} 3 \\ 0 \\ -4 \end{matrix} \right), \left( \begin{matrix} 0 \\ 0 \\ 6 \end{matrix} \right), \left( \begin{matrix} -1 \\ -1 \\ -1 \end{matrix} \right) \right\} \end{equation*}
Solution.
I need to determine if there are any redundancies in the span. To do this, I put the vectors in the span as rows of a matrix.
\begin{equation*} \left( \begin{matrix} 1 \amp 1 \amp 0 \\ 3 \amp 0 \amp -4 \\ 0 \amp 0 \amp 6 \\ -1 \amp -1 \amp -1 \end{matrix} \right) \end{equation*}
I row reduce this matrix. I will either row reduce without interchanging rows or keeping track of row exchanges. (If done by computer, as is the usual practice, I’ll have to be careful choosing a basis. Because of this, I’ll recheck the bases after I find them.)
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}
Every row that has a leading one is a row that is necessary for the span; every row without a leading one is redundant. Here, the fourth vector is all zeros, so there is a redundant vector. There are three leading ones, so the dimension of this span is 3. I can take the vectors that corresponded to rows with leading ones as a basis.
\begin{equation*} \left\{ \left( \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right), \left( \begin{matrix} 3 \\ 0 \\ -4 \end{matrix} \right), \left( \begin{matrix} 0 \\ 0 \\ 6 \end{matrix} \right) \right\} \end{equation*}
I can check with a matrix that these four are independent using the same steps as before (I’ve not shown that work). I need this check to make sure rows weren’t exchanged during row reduction.

Activity 5.4.3.

Calculate the dimension of this span and give a basis.
\begin{equation*} \Span \left\{ \left( \begin{matrix} 7 \\ 0 \\ -5 \\ -5 \end{matrix} \right), \left( \begin{matrix} 0 \\ -3 \\ -3 \\ 4 \end{matrix} \right), \left( \begin{matrix} 7 \\ -3 \\ -8 \\ -1 \end{matrix} \right) \right\} \end{equation*}
Solution.
I need to determine if there are any redundancies in the span. To do this, I put the vectors in the span as rows of a matrix.
\begin{equation*} \left( \begin{matrix} 7 \amp 0 \amp -5 \amp -5 \\ 0 \amp -3 \amp -3 \amp 4 \\ 7 \amp -3 \amp -8 \amp -1 \end{matrix} \right) \end{equation*}
I row reduce this matrix. I will either row reduce without interchanging rows or keeping track of row exchanges. (If done by computer, as is the usual practice, I’ll have to be careful choosing a basis. Because of this, I’ll recheck the bases after I find them.)
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp \frac{-5}{7} \amp \frac{-5}{7} \\ 0 \amp 1 \amp 1 \amp \frac{-4}{3} \\ 0 \amp 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}
Every row that has a leading one is a row that is necessary for the span; every row without a leading one is redundant. Here I see that there is a row of zeros, so there is a redundant vector. There are two leading ones, to the dimension of this span is I can take the vectors that corresponded to rows with leading ones as a basis.
\begin{equation*} \left\{ \left( \begin{matrix} 7 \\ 0 \\ -5 \\ -5 \end{matrix} \right), \left( \begin{matrix} 0 \\ -3 \\ -3 \\ 4 \end{matrix} \right) \right\} \end{equation*}
I can check with a matrix that these four are independent using the same steps as before (I’ve not shown that work). I need this check to make sure rows weren’t exchanged during row reduction.

Activity 5.4.4.

Calculate the dimension of this span and give a basis.
\begin{equation*} \Span \left\{ \left( \begin{matrix} 6 \\ -1 \\ -1 \\ 4 \\ 7 \end{matrix} \right), \left( \begin{matrix} 0 \\ 4 \\ 3 \\ 0 \\ -5 \end{matrix} \right), \left( \begin{matrix} -4 \\ 10 \\ 2 \\ -5 \\ -8 \end{matrix} \right), \left( \begin{matrix} -8 \\ -6 \\ 3 \\ 4 \\ 0 \end{matrix} \right), \left( \begin{matrix} -6 \\ 7 \\ 7 \\ 3 \\ -6 \end{matrix} \right) \right\} \end{equation*}
Solution.
I need to determine if there are any redundancies in the span. To do this, I put the vectors in the span as rows of a matrix.
\begin{equation*} \left( \begin{matrix} 6 \amp -1 \amp -1 \amp 4 \amp 7 \\ 0 \amp 4 \amp 3 \amp 0 \amp -5 \\ -4 \amp 10 \amp 2 \amp -5 \amp -8 \\ -8 \amp -6 \amp 3 \amp 4 \amp 0 \\ -6 \amp 7 \amp 7 \amp 3 \amp -6 \end{matrix} \right) \end{equation*}
I row reduce this matrix. I will either row reduce without interchanging rows or keeping track of row exchanges. (If done by computer, as is the usual practice, I’ll have to be careful choosing a basis. Because of this, I’ll recheck the bases after I find them.)
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 0 \amp 0 \amp \frac{-1}{4} \\ 0 \amp 1 \amp 0 \amp 0 \amp \frac{23}{62} \\ 0 \amp 0 \amp 1 \amp 0 \amp \frac{-67}{31} \\ 0 \amp 0 \amp 0 \amp 1 \amp \frac{52}{31} \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}
Every row that has a leading one is a row that is necessary for the span; every row without a leading one is redundant. Here I see that the fifth row is a row of zeros, so one vector is redundant. There are four leading ones, so the dimension of this span is 4. I can take the vectors that corresponded to rows with leading ones as a basis.
\begin{equation*} \left\{ \left( \begin{matrix} 6 \\ -1 \\ -1 \\ 4 \\ 7 \end{matrix} \right), \left( \begin{matrix} 0 \\ 4 \\ 3 \\ 0 \\ -5 \\ \end{matrix} \right), \left( \begin{matrix} -4 \\ 10 \\ 2 \\ -5 \\ -8 \\ \end{matrix} \right), \left( \begin{matrix} -8 \\ -6 \\ 3 \\ 4 \\ 0 \end{matrix} \right) \right\} \end{equation*}
I can check with a matrix that these four are independent using the same steps as before (I’ve not shown that work). I need this check to make sure rows weren’t exchanged during row reduction.

Activity 5.4.5.

Calculate the dimension of this span and give a basis.
\begin{equation*} \Span \left\{ \left( \begin{matrix} 4 \\ 0 \\ -3 \\ -3 \end{matrix} \right), \left( \begin{matrix} 1 \\ 2 \\ 2 \\ 0 \end{matrix} \right), \left( \begin{matrix} -3 \\ 2 \\ 5 \\ 3 \end{matrix} \right), \left( \begin{matrix} 8 \\ 0 \\ -7 \\ 1 \end{matrix} \right) \right\} \end{equation*}
Solution.
I need to determine if there are any redundancies in the span. To do this, I put the vectors in the span as rows of a matrix.
\begin{equation*} \left( \begin{matrix} 4 \amp 0 \amp -3 \amp -3 \\ 1 \amp 2 \amp 2 \amp 0 \\ -3 \amp 2 \amp 5 \amp 3 \\ 8 \amp 0 \amp -7 \amp 1 \end{matrix} \right) \end{equation*}
I row reduce this matrix. I will either row reduce without interchanging rows or keeping track of row exchanges. (If done by computer, as is the usual practice, I’ll have to be careful choosing a basis. Because of this, I’ll recheck the bases after I find them.)
\begin{equation*} \left( \begin{matrix} 1 \amp 0 \amp 0 \amp -6 \\ 0 \amp 1 \amp 0 \amp 10 \\ 0 \amp 0 \amp 1 \amp -7 \\ 0 \amp 0 \amp 0 \amp 0 \end{matrix} \right) \end{equation*}
Every row that has a leading one is a row that is necessary for the span; every row without a leading one is redundant. I see that there is a row of zeros, so one vector is redundant. There are three leading ones, so the dimension of this span is 3. I can take the vectors that corresponded to rows with leading ones as a basis.
\begin{equation*} \left\{ \left( \begin{matrix} 4 \\ 0 \\ -3 \\ -3 \end{matrix} \right), \left( \begin{matrix} 1 \\ 2 \\ 2 \\ 0 \end{matrix} \right), \left( \begin{matrix} -3 \\ 2 \\ 5 \\ 3 \end{matrix} \right) \right\} \end{equation*}
If I check, however, I find that this is still redundant. This is not a basis, so one of these rows must have been redundant and the row reduction must have involved some exchange of rows. If I take the fourth vector instead of the third, I get this basis.
\begin{equation*} \left\{ \left( \begin{matrix} 4 \\ 0 \\ -3 \\ -3 \end{matrix} \right), \left( \begin{matrix} 1 \\ 2 \\ 2 \\ 0 \end{matrix} \right), \left( \begin{matrix} 8 \\ 0 \\ -7 \\ 1 \end{matrix} \right) \right\} \end{equation*}
If I check, I will find that there are no redundancies now, so this is a basis.

Subsection 5.4.2 Dimensions and Bases of Loci

Activity 5.4.6.

Calculate the dimension of this locus.
\begin{align*} 4x - y + 7z \amp = 3\\ 2x + z \amp = 0 \\ 2x - y + 6z \amp = 3 \end{align*}
Solution.
I change the locus into a matrix.
\begin{equation*} \left( \begin{array}{ccc|c} 4 \amp -1 \amp 7 \amp 3 \\ 2 \amp 0 \amp 1 \amp 0 \\ 2 \amp -1 \amp 6 \amp 3 \end{array} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp \frac{1}{2} \amp 0 \\ 0 \amp 1 \amp -5 \amp -3 \\ 0 \amp 0 \amp 0 \amp 0 \end{array} \right) \end{equation*}
Then the dimension is determined by the number of free variables, as long as there isn’t a row equivalent to \(0=1\text{.}\) Here, there is one free variable, so the locus has dimension 1.

Activity 5.4.7.

Calculate the dimension of this locus.
\begin{align*} x - y - z \amp = 7\\ 2x + 5y - 3z \amp = 2 \\ 3x + 4y - 4z \amp = 4 \end{align*}
Solution.
I change the locus into a matrix.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp -1 \amp -1 \amp 7 \\ 2 \amp 5 \amp -3 \amp 2 \\ 3 \amp 4 \amp -4 \amp 4 \end{array} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp \frac{-8}{7} \amp 0 \\ 0 \amp 1 \amp \frac{-1}{7} \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{array} \right) \end{equation*}
Then the dimension is determined by the number of free variables, as long as there isn’t a row equivalent to \(0=1\text{.}\) Here, the last row is such a row, so this locus is empty and has no dimension. (Note this is not dimension 0, which is the dimension of a point. The empty set simply doesn’t have any dimension.)

Activity 5.4.8.

Calculate the dimension of this locus. (In this course, the order of the variables in \(\RR^4\) is \(w,x,y,z\text{.}\))
\begin{align*} 9w - 3x + 5y - z \amp = 0 \\ 3w - 12x + 4y \amp = 0 \\ 3w + 21x - 3y - z \amp = 0 \end{align*}
Solution.
I change the locus into a matrix.
\begin{equation*} \left( \begin{array}{cccc|c} 9 \amp -3 \amp 5 \amp -1 \amp 0 \\ 3 \amp -12 \amp 4 \amp 0 \amp 0 \\ 3 \amp 21 \amp -3 \amp -1 \amp 0 \end{array} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp 0 \amp \frac{16}{33} \amp \frac{-4}{33} \amp 0 \\ 0 \amp 1 \amp \frac{-7}{33} \amp \frac{-1}{33} \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{array} \right) \end{equation*}
Then the dimension is determined by the number of free variables, as long as there isn’t a row equivalent to \(0=1\text{.}\) Here, there are two free variables, so the locus has dimension 2.

Activity 5.4.9.

Calculate the dimension of this locus. (In this course, the order of the variables in \(\RR^4\) is \(w,x,y,z\text{.}\))
\begin{align*} w - 2x + y - 3z \amp = 0 \\ 2w - 4x + 2y - 6z \amp = 0 \\ 3w - 6x + 3y - 9z \amp = 0 \end{align*}
Solution.
I change the locus into a matrix.
\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp -2 \amp 1 \amp -3 \amp 0 \\ 2 \amp -4 \amp 2 \amp -6 \amp 0 \\ 3 \amp -6 \amp 3 \amp -9 \amp 0 \end{array} \right) \end{equation*}
I row reduce this matrix.
\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp -2 \amp 1 \amp -3 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right) \end{equation*}
Then the dimension is determined by the number of free variables, as long as there isn’t a row equivalent to \(0=1\text{.}\) Here, there are three free variables, so the dimension of the locus is three.

Subsection 5.4.3 Calculating Linear Combinations

Activity 5.4.10.

Write the first vector as a linear combination of the vectors in the span.
\begin{align*} \amp \left( \begin{matrix} 2 \\ -2 \end{matrix} \right) \amp \amp \Span \left\{ \left( \begin{matrix} 4 \\ 9 \end{matrix} \right), \left( \begin{matrix} -3 \\ 12 \end{matrix} \right) \right\} \end{align*}
Solution.
I can convert into an equation with unknowns by writing out the definition of a linear combination.
\begin{equation*} \left( \begin{matrix} 2 \\ -2 \end{matrix} \right) = a \left( \begin{matrix} 4 \\ 9 \end{matrix} \right) + b \left( \begin{matrix} -3 \\ 12 \end{matrix} \right) \end{equation*}
Then I can translate this into a system of equation in the unknowns \(a,b\text{.}\)
\begin{align*} 2 \amp = 4a - 3b \\ -2 \amp = 9a + 12 b \end{align*}
Then I can translate this into a matrix.
\begin{equation*} \left( \begin{array}{cc|c} 4 \amp -3 \amp 2 \\ 9 \amp 12 \amp -2 \end{array} \right) \end{equation*}
Then I row reduce this matrix.
\begin{equation*} \left( \begin{array}{cc|c} 1 \amp 0 \amp \frac{6}{25} \\ 0 \amp 1 \amp \frac{-26}{75} \end{array} \right) \end{equation*}
This has a unique solution, which gives me the coefficient of the linear combination. Finally, I can write the original vector as a linear combination of the spanning vectors.
\begin{equation*} \left( \begin{matrix} 2 \\ -2 \end{matrix} \right) = \frac{6}{25} \left( \begin{matrix} 4 \\ 9 \end{matrix} \right) + \frac{-26}{75} \left( \begin{matrix} -3 \\ 12 \end{matrix} \right) \end{equation*}

Activity 5.4.11.

Write the first vector as a linear combination of the vectors in the span.
\begin{align*} \amp \left( \begin{matrix} 1 \\ 0 \\ 4 \end{matrix} \right) \amp \amp \Span \left\{ \left( \begin{matrix} 4 \\ 0 \\ 13 \end{matrix} \right), \left( \begin{matrix} -1 \\ -1 \\ 3 \end{matrix} \right), \left( \begin{matrix} 0 \\ -5 \\ 4 \end{matrix} \right) \right\} \end{align*}
Solution.
I can convert into an equation with unknown by writing out the definition of a linear combination.
\begin{equation*} \left( \begin{matrix} 1 \\ 0 \\ 4 \end{matrix} \right) = a \left( \begin{matrix} 4 \\ 0 \\ 13 \end{matrix} \right) + b \left( \begin{matrix} -1 \\ -1 \\ 3 \end{matrix} \right) + c \left( \begin{matrix} 0 \\ -5 \\ 4 \end{matrix} \right) \end{equation*}
Then I can translate this into a system of equation in the unknowns \(a,b\text{.}\)
\begin{align*} 1 \amp = 4a - b \\ 0 \amp = -b - 5c\\ 4 \amp = 13a + 3b + 4c \end{align*}
Then I can translate this into a matrix.
\begin{equation*} \left( \begin{array}{ccc|c} 4 \amp -1 \amp 0 \amp 1 \\ 0 \amp -1 \amp -5 \amp 0 \\ 13 \amp 3 \amp 4 \amp 4 \end{array} \right) \end{equation*}
Then I row reduce this matrix.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp \frac{31}{109} \\ 0 \amp 1 \amp 0 \amp \frac{15}{109} \\ 0 \amp 0 \amp 1 \amp \frac{-3}{109} \end{array} \right) \end{equation*}
This has a unique solution, which gives me the coefficient of the linear combination. Finally, I can write the original vector as a linear combination of the spanning vectors.
\begin{equation*} \left( \begin{matrix} 1 \\ 0 \\ 4 \end{matrix} \right) = \frac{31}{109} \left( \begin{matrix} 4 \\ 0 \\ 13 \end{matrix} \right) + \frac{15}{109} \left( \begin{matrix} -1 \\ -1 \\ 3 \end{matrix} \right) + \frac{-3}{109} \left( \begin{matrix} 0 \\ -5 \\ 4 \end{matrix} \right) \end{equation*}

Activity 5.4.12.

Write the first vector as a linear combination of the vectors in the span.
\begin{align*} \amp \left( \begin{matrix} -6 \\ 9 \\ -25 \\ 16 \end{matrix} \right) \amp \amp \Span \left\{ \left( \begin{matrix} 0 \\ 5 \\ -3 \\ 8 \end{matrix} \right), \left( \begin{matrix} -1 \\ -1 \\ 4 \\ 3 \end{matrix} \right), \left( \begin{matrix} -5 \\ -5 \\ 0 \\ 2 \end{matrix} \right) \right\} \end{align*}
Solution.
I can convert into an equation with unknown by writing out the definition of a linear combination.
\begin{equation*} \left( \begin{matrix} -6 \\ 9 \\ -25 \\ 16 \end{matrix} \right) = a \left( \begin{matrix} 0 \\ 5 \\ -3 \\ 8 \end{matrix} \right) + b \left( \begin{matrix} -1 \\ -1 \\ 4 \\ 3 \end{matrix} \right) c \left( \begin{matrix} -5 \\ -5 \\ 0 \\ 2 \end{matrix} \right) \end{equation*}
Then I can translate this into a system of equation in the unknowns \(a,b\text{.}\)
\begin{align*} -6 \amp = -b - 5c \\ 9 \amp = 5a - b - 5c \\ -25 \amp = -3a + 4b \\ 16 \amp = 8a + 3b + 2c \end{align*}
Then I can translate this into a matrix.
\begin{equation*} \left( \begin{array}{ccc|c} 0 \amp -1 \amp -5 \amp -6 \\ 5 \amp -1 \amp -5 \amp 9 \\ -3 \amp 4 \amp 0 \amp -25 \\ 8 \amp 3 \amp 2 \amp 16 \end{array} \right) \end{equation*}
Then I row reduce this matrix.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp 3 \\ 0 \amp 1 \amp 0 \amp -4 \\ 0 \amp 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \end{array} \right) \end{equation*}
This has a unique solution, which gives me the coefficient of the linear combination. Finally, I can write the original vector as a linear combination of the spanning vectors.
\begin{equation*} \left( \begin{matrix} -6 \\ 9 \\ -25 \\ 16 \end{matrix} \right) = 3 \left( \begin{matrix} 0 \\ 5 \\ -3 \\ 8 \end{matrix} \right) - 4 \left( \begin{matrix} -1 \\ -1 \\ 4 \\ 3 \end{matrix} \right) + 2 \left( \begin{matrix} -5 \\ -5 \\ 0 \\ 2 \end{matrix} \right) \end{equation*}

Subsection 5.4.4 Conceptual Review Questions

  • What are dimensions?
  • How are spans “built up” but loci are “restricted down”?
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