Title: Vector Spaces
Course: Demonstrating STACK (v4.3.7)

Determine the rank of the matrix \( {\bf A} = [value]\)
rank(\({\bf A}\)) =
Feedback: The rank of a matrix is the number of non-zero rows in the Echelon form of the matrix. A possible Echelon form of \[ [value] \] is \[ [value] \] Hence, the rank of this matrix is [value]

Determine the dimension of the subspace spanned by the set \[ \left\{ [value],\ [value],\ [value] \right \}.\]
dimension =
Feedback: If we place the vectors in the spanning set \( \left\{ [value],\ [value],\ [value] \right \}\) in the rows of a matrix, i.e., \[ [value], \] then the dimension of the subspace is the same as the dimension of the row space of this matrix, which is the rank of this matrix. The rank of a matrix is the number of non-zero rows in the Echelon form of the matrix. A possible Echelon form of this matrix is \[ [value] \] Hence, the rank of this matrix and the dimension of the subspace is [value]

Evaluate the folowing expression:
\[ \left( ([value]-[value])[value] - [value] [value]\right) + [value] [value].\]
Feedback: \[\left( ([value]-[value])[value] - [value] [value]\right) + [value] [value] = \left([value][value] - [value] [value]\right) + [value] [value] \] \[ =\left( [value] - [value]\right) + [value] = [value]\]

Let \({\bf v_{1}}\) and \({\bf v_2}\) be the following vectors:
\[ {\bf v_{1}} = [value], {\bf v_{2}} = [value].\]
Compute \[ [value] {\bf v_1} + [value] {\bf v_2} .\]
Feedback: Given \[ {\bf v_{1}} = [value], {\bf v_{2}} = [value].\] \[ [value] {\bf v_1} + [value] {\bf v_2} = [value][value]+[value][value]\] \[ = [value] + [value] = [value]\]

Let \({\bf v_{1}}\), \({\bf v_2}\), \({\bf v_3}\) be the following vectors:
\[ {\bf v_{1}} = [value],\quad {\bf v_{2}} = [value],\quad {\bf v_{3}} = [value]. \]
Find the values for \(\lambda_1\) and \(\lambda_2\) such that
\[{\bf v_3} = \lambda_1 {\bf v_1} + \lambda_2 {\bf v_2} .\]
If such values do not exist, enter zero for both entries.
\(\lambda_1\) = \(\lambda_2\) =
Feedback: If \[{\bf v_3} = \lambda_{1} {\bf v_1} + \lambda_{2} {\bf v_2} \] then solving \[[value] = [value]*\lambda_1 + [value]*\lambda_2,\] and \[[value] = [value]*\lambda_1 + [value]*\lambda_2,\] gives \[ \lambda_1 = [value] , \lambda_2 = [value]\]

Let \({\bf v_{1}}\), \({\bf v_2}\), \({\bf v_3}\) be the following vectors:
\[ {\bf v_{1}} = [value],\quad {\bf v_{2}} = [value],\quad {\bf v_{3}} = [value]. \]
Find the values for \(\lambda_1\) and \(\lambda_2\) such that
\[{\bf v_3} = \lambda_1 {\bf v_1} + \lambda_2 {\bf v_2} .\]
If such values do not exist, enter zero for both entries. \(\lambda_1\) = \(\lambda_2\) =
Feedback: If \[{\bf v_3} = \lambda_1 {\bf v_1} + \lambda_2 {\bf v_2},\] we have 3 equations in 2 unknowns. For there to be a solution for \(\lambda_1\) and \(\lambda_2\) the vectors \({\bf v_1},{\bf v_2}\) and \({\bf v_3}\) must be linearly dependent. This requires the determinant of the matrix of vectors to be zero. The matrix is \[[value]\] with a determinant of \( [value]\) Therefore the vectors are not linearly dependent and there is no solution for \(\lambda_1\) and \(\lambda_2\)

Given is a spanning set \(S\) of a subspace \(U \subset R^4\):
\[ S = \{ [value],\ [value],\ [value],\ [value],\ [value] \}. \]
Find a basis of \(U\).
Enter the vectors in the basis as rows in the matrix below, adding zero rows (at the bottom!) when there are less then 5 vectors in the basis.
Feedback: A basis is most easily found by using the vectors in the spanning set as the rows in a matrix: \[ [value]. \] The subspace spanned by them is then the same as the rowspace of this matrix. Reduction to Echelon form will then reveal both the rank (dimension of the subspace) and a basis (the non-zero rows). An Echelon form of the matrix above is \[ [value].\] A basis is then given by the non-zero rows: \[ [value].\] (Reminder: any \([value]\) linear independent vectors that can be written as a combination of the given vectors will form a basis.)

Find a basis for the rowspace of the matrix \[ {\bf A} = [value]. \] Enter the vectors in the basis as rows in the matrix below, adding zero rows when there are less then 5 vectors in the basis.
Feedback: The rank of the matrix [value] is [value]. So we look for a set of [value] rows from the matrix to form the basis set. Check the rank of the matrix formed by the chosen rows and if this is equal to [value] you have found a solution.

Find \(\alpha\) such that the set of vectors \[ { [value],\ [value],\ [value] } \] is linearly dependent.
Feedback: For the vectors to be linearly dependent the determinant of the matrix of vectors must equal zero. So for the vectors \[[value]\] The determinant is \[[value] = 0.\] Hence \[[value] =0,\] and therefore \[\alpha = [value]\]

Given are the vectors: \[ [value],\quad [value]. \]
Find a third vector which is linearly independent from the two given vectors.
Feedback: The two given vectors \( [value],[value] \) and the third vector should form a matrix whose rowspace has dimension 3 (or whose rank is dimension 3). So for example, \[[value]\] has rank 3 since after reduction to Echelon form, we have three non-zero rows.

Calculate the vector product \(\underline{u} \times \underline{v}\) where \(\underline{u}\) has components [value] and \(\underline{v}\) has components [value].

\(\underline{u}\times\underline{v} =\)

Calculate the dot produxct \(\underline{u}\cdot \underline{v}\) where \(\underline{u}\) has components [value] and \(\underline{v}\) has components [value].

\(\underline{u}\cdot\underline{v} =\)

Determine the equation of the plane through the point \(P\) with coordinates [value] and with normal vector \(\underline{n}\) with components [value].

Equation is: