MATH347: Linear algebra for applications

This assignment is a worksheet of exercises intended as preparation for the Final Examination. You should:

1. Review Lessons 1 to 12

2. Set aside 60 minutes to solve these exercises. Each exercise is meant to be solved within 3 minutes. If you cannot find a solution within 3 minutes, skip to the next one.

3. Check your answers in Matlab. Revisit theory for skipped or incorrectly answered exercies.

4. Turn in a PDF with your brief handwritten answers that specify your motivation, approach, calculations, answer. It is good practice to start all answers by briefly recounting the applicable definitions.

When constructing a solution follow these steps:

1. Ask yourself: “what course concept is being verified?”

2. Identify relevant definitions and include them in your answer.

3. Briefly describe your approach

4. Carry out calculations

5. Present final answer

1Vector operations

1. Find the linear combination of vectors \(\boldsymbol{u}= \left[ \begin{array}{lll} 1 & 1 & 1 \end{array} \right]\), \(\boldsymbol{v}= \left[ \begin{array}{lll} 1 & 2 & 3 \end{array} \right]\) with scaling coefficients \(\alpha = 2\), \(\beta = 1\).

   Solution. By definition of linear combination

   \(\displaystyle \boldsymbol{w}= \alpha \boldsymbol{u}+ \beta \boldsymbol{v}= 2 \cdot \left[ \begin{array}{lll} 1 & 1 & 1 \end{array} \right] + 1 \cdot \left[ \begin{array}{lll} 1 & 2 & 3 \end{array} \right] = \left[ \begin{array}{lll} 3 & 4 & 5 \end{array} \right] .\)

2. Express the above linear combination \(\boldsymbol{b}\) as a matrix-vector product \(\boldsymbol{b}=\boldsymbol{A}\boldsymbol{x}\). Define \(\boldsymbol{x}\) and the column vectors of \(\boldsymbol{A}= \left[ \begin{array}{ll} \boldsymbol{a}_1 & \boldsymbol{a}_2 \end{array} \right]\).

   Solution. Vectors that enter the linear combination are matrix \(\boldsymbol{A}\) columns, scaling coefficients are components of \(\boldsymbol{x}\)

   \(\displaystyle \boldsymbol{b}=\boldsymbol{A}\boldsymbol{x}= \left[ \begin{array}{ll} \boldsymbol{a}_1 & \boldsymbol{a}_2 \end{array} \right] \left[ \begin{array}{l} x_1\\ x_2 \end{array} \right] = \left[ \begin{array}{ll} 1 & 1\\ 1 & 2\\ 1 & 3 \end{array} \right] \left[ \begin{array}{l} 2\\ 1 \end{array} \right] = \left[ \begin{array}{l} 3\\ 4\\ 5 \end{array} \right] .\)

3. Consider \(\boldsymbol{u}= \left[ \begin{array}{lll} 1 & 1 & 0 \end{array} \right]^T\), \(\boldsymbol{v}= \left[ \begin{array}{lll} 1 & 1 & 1 \end{array} \right]^T\). Compute the 2-norms of \(\boldsymbol{u}, \boldsymbol{v}\). Determine the angle between \(\boldsymbol{u}, \boldsymbol{v}\).

   Solution. By definition the 2-norms are

   \(\displaystyle \| \boldsymbol{u} \|_2 = \left( \sum_{i = 1}^3 u_i^2 \right)^{1 / 2} = \sqrt{2}, \| \boldsymbol{v} \|_2 = \left( \sum_{i = 1}^3 v_i^2 \right)^{1 / 2} = \sqrt{3} .\)

   The cosine of the angle \(\theta\) between \(\boldsymbol{u}, \boldsymbol{v}\) is defined as

   \(\displaystyle \cos \theta = \frac{\boldsymbol{u}^T \boldsymbol{v}}{\| \boldsymbol{u} \|_2 \| \boldsymbol{v} \|_2} = \frac{2}{\sqrt{6}} .\)

4. Consider \(\boldsymbol{u}= \left[ \begin{array}{lll} 1 & 1 & 0 \end{array} \right]^T\), \(\boldsymbol{v}= \left[ \begin{array}{lll} 1 & 1 & 1 \end{array} \right]^T\). Define vector \(\boldsymbol{w}\) such that \(\boldsymbol{v}+\boldsymbol{w}\) is orthogonal to \(\boldsymbol{u}\). Write the equation to determine \(\boldsymbol{w}\), and then compute \(\boldsymbol{w}\).

   Solution. Orthogonal vectors satisfy

   \(\displaystyle \boldsymbol{u}^T (\boldsymbol{v}+\boldsymbol{w}) = 0,\)

   whence

   \(\displaystyle \boldsymbol{u}^T \boldsymbol{w}= \left[ \begin{array}{lll} 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{l} w_1\\ w_2\\ w_3 \end{array} \right] = -\boldsymbol{u}^T \boldsymbol{v}= - \left[ \begin{array}{lll} 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{l} 1\\ 1\\ 1 \end{array} \right] = - 2 \Rightarrow w_1 + w_2 = - 2.\)

   The above equation has an infinity of solutions, say

   \(\displaystyle \boldsymbol{w}= \left[ \begin{array}{l} - 2\\ 0\\ 0 \end{array} \right] .\)

   Verify:

   \(\displaystyle \boldsymbol{v}+\boldsymbol{w}= \left[ \begin{array}{l} 1\\ 1\\ 1 \end{array} \right] + \left[ \begin{array}{l} - 2\\ 0\\ 0 \end{array} \right] = \left[ \begin{array}{l} - 1\\ 1\\ 1 \end{array} \right], \boldsymbol{u}^T (\boldsymbol{v}+\boldsymbol{w}) = \left[ \begin{array}{lll} 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{l} - 1\\ 1\\ 1 \end{array} \right] = 0. \checked\)

5. Determine \(\boldsymbol{q}_1, \boldsymbol{q}_2\) to be of unit norm and in the direction of vectors \(\boldsymbol{u}, \boldsymbol{v}+\boldsymbol{w}\) from Ex. 4. Form \(\hat{\boldsymbol{Q}} = \left[ \begin{array}{ll} \boldsymbol{q}_1 & \boldsymbol{q}_2 \end{array} \right]\). Compute \(\hat{\boldsymbol{Q}} \hat{\boldsymbol{Q}}^T\) and \(\widehat{\boldsymbol{Q} }^T \hat{\boldsymbol{Q}}\).

   Solution. By definition of a unit norm vector

   \(\displaystyle \boldsymbol{q}_1 = \frac{\boldsymbol{u}}{\| \boldsymbol{u} \|_2} = \frac{1}{\sqrt{2}} \left[ \begin{array}{l} 1\\ 1\\ 0 \end{array} \right], \boldsymbol{q}_2 = \frac{\boldsymbol{v}+\boldsymbol{w}}{\| \boldsymbol{v}+\boldsymbol{w} \|_2} = \frac{1}{\sqrt{3}} \left[ \begin{array}{l} - 1\\ 1\\ 1 \end{array} \right],\)

   leading to

   \(\displaystyle \hat{\boldsymbol{Q}} = \left[ \begin{array}{ll} 1 / \sqrt{2} & - 1 / \sqrt{3}\\ 1 / \sqrt{2} & 1 / \sqrt{3}\\ 0 & 1 / \sqrt{3} \end{array} \right] .\)

   Compute products

   \(\displaystyle \hat{\boldsymbol{Q}} \hat{\boldsymbol{Q}}^T = \left[ \begin{array}{ll} 1 / \sqrt{2} & - 1 / \sqrt{3}\\ 1 / \sqrt{2} & 1 / \sqrt{3}\\ 0 & 1 / \sqrt{3} \end{array} \right] \left[ \begin{array}{lll} 1 / \sqrt{2} & 1 / \sqrt{2} & 0\\ - 1 / \sqrt{3} & 1 / \sqrt{3} & 1 / \sqrt{3} \end{array} \right] = \left[ \begin{array}{lll} 5 / 6 & 1 / 6 & - 1 / 3\\ 1 / 6 & 5 / 6 & 1 / 3\\ - 1 / 3 & 1 / 3 & 1 / 3 \end{array} \right],\)

   a projection matrix onto \(C (\hat{\boldsymbol{Q}})\)

   \(\displaystyle \widehat{\boldsymbol{Q} }^T \hat{\boldsymbol{Q}} = \left[ \begin{array}{lll} 1 / \sqrt{2} & 1 / \sqrt{2} & 0\\ - 1 / \sqrt{3} & 1 / \sqrt{3} & 1 / \sqrt{3} \end{array} \right] \left[ \begin{array}{ll} 1 / \sqrt{2} & - 1 / \sqrt{3}\\ 1 / \sqrt{2} & 1 / \sqrt{3}\\ 0 & 1 / \sqrt{3} \end{array} \right] = \left[ \begin{array}{ll} 1 & 0\\ 0 & 1 \end{array} \right],\)

   the identity matrix.

6. Determine vector \(\boldsymbol{q}_3\) orthonormal to vectors \(\boldsymbol{q}_1, \boldsymbol{q}_2\) from Ex. 5.

   Solution. By observation, choose

   \(\displaystyle \boldsymbol{q}_3 = \frac{1}{\sqrt{6}} \left[ \begin{array}{l} - 1\\ 1\\ - 2 \end{array} \right] .\)

   Verify

   \(\displaystyle \boldsymbol{q}_1^T \boldsymbol{q}_3 = \frac{1}{\sqrt{12}} \left[ \begin{array}{lll} 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{l} - 1\\ 1\\ - 2 \end{array} \right] = 0, \boldsymbol{q}_2^T \boldsymbol{q}_3 = \frac{1}{\sqrt{18}} \left[ \begin{array}{lll} - 1 & 1 & 1 \end{array} \right] \left[ \begin{array}{l} - 1\\ 1\\ - 2 \end{array} \right] = 0, \boldsymbol{q}_3^T \boldsymbol{q}_3 = \frac{1}{6} \left[ \begin{array}{lll} - 1 & 1 & - 2 \end{array} \right] \left[ \begin{array}{l} - 1\\ 1\\ - 2 \end{array} \right] = 1\)

7. Establish whether vectors \(\boldsymbol{u}= \left[ \begin{array}{lll} 1 & 2 & 3 \end{array} \right]^T\), \(\boldsymbol{v}= \left[ \begin{array}{lll} - 3 & 1 & - 2 \end{array} \right]^T\), \(\boldsymbol{w}= \left[ \begin{array}{lll} 2 & - 3 & 1 \end{array} \right]^T\) all lie in the same plane within \(\mathbb{R}^3\).

   Solution. The vectors would have to be linearly dependent. Form the matrix

   \(\displaystyle \boldsymbol{A}= \left[ \begin{array}{lll} \boldsymbol{u} & \boldsymbol{v} & \boldsymbol{w} \end{array} \right] = \left[ \begin{array}{lll} 1 & - 3 & 2\\ 2 & 1 & - 3\\ 3 & - 2 & 1 \end{array} \right] .\)

   Carry out reduction to rref

   \(\displaystyle \boldsymbol{A} \sim \left[ \begin{array}{lll} 1 & - 3 & 2\\ 0 & 7 & - 7\\ 0 & 7 & - 5 \end{array} \right] \sim \left[ \begin{array}{lll} 1 & - 3 & 2\\ 0 & 7 & - 7\\ 0 & 0 & 2 \end{array} \right],\)

   a matrix of full rank, hence with linearly independent columns, implying that \(\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}\) do not all lie in the same plane.

8. Determine \(\boldsymbol{v}\) the reflection of vector \(\boldsymbol{u}= \left[ \begin{array}{ll} 1 & \sqrt{3} \end{array} \right]^T\) across vector \(\boldsymbol{w}= \left[ \begin{array}{ll} 1 & 1 \end{array} \right]^T\).

   Solution. Form unit vector

   \(\displaystyle \boldsymbol{q}= \frac{\boldsymbol{w}}{\| \boldsymbol{w} \|} = \frac{1}{\sqrt{2}} \left[ \begin{array}{l} 1\\ 1 \end{array} \right] .\)

   The reflection matrix is

   \(\displaystyle \boldsymbol{R}= 2\boldsymbol{q}\boldsymbol{q}^T -\boldsymbol{I}= \left[ \begin{array}{l} 1\\ 1 \end{array} \right] \left[ \begin{array}{ll} 1 & 1 \end{array} \right] - \left[ \begin{array}{ll} 1 & 0\\ 0 & 1 \end{array} \right] = \left[ \begin{array}{ll} 1 & 1\\ 1 & 1 \end{array} \right] - \left[ \begin{array}{ll} 1 & 0\\ 0 & 1 \end{array} \right] = \left[ \begin{array}{ll} 0 & 1\\ 1 & 0 \end{array} \right] .\)

   Compute reflection

   \(\displaystyle \boldsymbol{R}\boldsymbol{u}= \left[ \begin{array}{l} \sqrt{3}\\ 1 \end{array} \right] .\)

9. Determine \(\boldsymbol{w}\) the rotation of vector \(\boldsymbol{u}= \left[ \begin{array}{ll} 1 & \sqrt{3} \end{array} \right]^T\) by angle \(\theta = - \pi / 6\).

   Solution. The rotation matrix is

   \(\displaystyle \boldsymbol{R}= \left[ \begin{array}{ll} \cos \theta & - \sin \theta\\ \sin \theta & \cos \theta \end{array} \right] = \left[ \begin{array}{ll} \sqrt{3} / 2 & 1 / 2\\ - 1 / 2 & \sqrt{3} / 2 \end{array} \right],\)

   and the rotated vector is

   \(\displaystyle \boldsymbol{w}=\boldsymbol{R}\boldsymbol{u}= \left[ \begin{array}{ll} \sqrt{3} / 2 & 1 / 2\\ - 1 / 2 & \sqrt{3} / 2 \end{array} \right] \left[ \begin{array}{l} 1\\ \sqrt{3} \end{array} \right] = \left[ \begin{array}{l} \sqrt{3}\\ 1 \end{array} \right] .\)

10. Compute \(\boldsymbol{z}=\boldsymbol{v}-\boldsymbol{w}\) with \(\boldsymbol{v}, \boldsymbol{w}\) from Ex. 8,9.

    Solution. The difference is \(\boldsymbol{z}=\boldsymbol{v}-\boldsymbol{w}\) highlighting that rotation and reflection of a vector can produce the same result.

2Matrix operations

1. Find two linear combinations of vectors \(\boldsymbol{u}= \left[ \begin{array}{lll} 1 & 1 & 1 \end{array} \right]\), \(\boldsymbol{v}= \left[ \begin{array}{lll} 1 & 2 & 3 \end{array} \right]\) first with scaling coefficients \(\alpha = 2\), \(\beta = 1\), and then with scaling coefficients \(\alpha = 1\), \(\beta = 2\).

   Solution. Organize the multiple linear combinations as a matrix-matrix product

   \(\displaystyle \boldsymbol{A}\boldsymbol{B}= \left[ \begin{array}{ll} \boldsymbol{u} & \boldsymbol{v} \end{array} \right] \left[ \begin{array}{ll} 2 & 1\\ 1 & 2 \end{array} \right] = \left[ \begin{array}{ll} 1 & 1\\ 1 & 2\\ 1 & 3 \end{array} \right] \left[ \begin{array}{ll} 2 & 1\\ 1 & 2 \end{array} \right] = \left[ \begin{array}{ll} 3 & 3\\ 4 & 4\\ 5 & 7 \end{array} \right] .\)

2. Express the above linear combinations \(\boldsymbol{B}\) as a matrix-matrix product \(\boldsymbol{B}=\boldsymbol{A}\boldsymbol{X}\). Define the column vectors of \(\boldsymbol{A}, \boldsymbol{X}\).

   Solution. As above.

3. Consider \(\boldsymbol{A}, \boldsymbol{B} \in \mathbb{R}^{m \times m}\). Which of the following matrices are always equal to \(\boldsymbol{C}= (\boldsymbol{A}-\boldsymbol{B})^2\)?

   1. \(\boldsymbol{A}^2 -\boldsymbol{B}^2\)

   2. \((\boldsymbol{B}-\boldsymbol{A})^2\)

   3. \(\boldsymbol{A}^2 - 2\boldsymbol{A}\boldsymbol{B}+\boldsymbol{B}^2\)

   4. \(\boldsymbol{A} (\boldsymbol{A}-\boldsymbol{B}) -\boldsymbol{B} (\boldsymbol{B}-\boldsymbol{A})\)

   5. \(\boldsymbol{A}^2 -\boldsymbol{A}\boldsymbol{B}-\boldsymbol{B}\boldsymbol{A}+\boldsymbol{B}^2\)

   Solution. Expand the product taking into account that matrix multiplication is not commutative

   \(\displaystyle \boldsymbol{C}= (\boldsymbol{A}-\boldsymbol{B}) (\boldsymbol{A}-\boldsymbol{B}) =\boldsymbol{A}^2 -\boldsymbol{B}\boldsymbol{A}-\boldsymbol{A}\boldsymbol{B}+\boldsymbol{B}^2 .\)

   Of the possible choices only (b,e) are always equal to \(\boldsymbol{C}\).

4. Find the inverse of

   \(\displaystyle \boldsymbol{A}= \left[ \begin{array}{lll} 2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{array} \right] .\)

   Solution. Apply Gauss-Jordan algorithm

   \(\displaystyle \left[ \begin{array}{ll} \boldsymbol{A} & \boldsymbol{I} \end{array} \right] = \left[ \begin{array}{llllll} 2 & 1 & 1 & 1 & 0 & 0\\ 1 & 2 & 1 & 0 & 1 & 0\\ 1 & 1 & 2 & 0 & 0 & 1 \end{array} \right] \sim \left[ \begin{array}{llllll} 1 & 1 / 2 & 1 / 2 & 1 / 2 & 0 & 0\\ 1 & 2 & 1 & 0 & 1 & 0\\ 1 & 1 & 2 & 0 & 0 & 1 \end{array} \right] \sim\) \(\displaystyle \left[ \begin{array}{llllll} 1 & 1 / 2 & 1 / 2 & 1 / 2 & 0 & 0\\ 0 & 3 / 2 & 1 / 2 & - 1 / 2 & 1 & 0\\ 0 & 1 / 2 & 3 / 2 & - 1 / 2 & 0 & 1 \end{array} \right] \sim \left[ \begin{array}{llllll} 1 & 1 / 2 & 1 / 2 & 1 / 2 & 0 & 0\\ 0 & 1 & 1 / 3 & - 1 / 3 & 2 / 3 & 0\\ 0 & 1 / 2 & 3 / 2 & - 1 / 2 & 0 & 1 \end{array} \right] \sim\) \(\displaystyle \left[ \begin{array}{llllll} 1 & 1 / 2 & 1 / 2 & 1 / 2 & 0 & 0\\ 0 & 1 & 1 / 3 & - 1 / 3 & 2 / 3 & 0\\ 0 & 0 & 4 / 3 & - 1 / 3 & - 1 / 3 & 1 \end{array} \right] \sim \left[ \begin{array}{llllll} 1 & 1 / 2 & 1 / 2 & 1 / 2 & 0 & 0\\ 0 & 1 & 1 / 3 & - 1 / 3 & 2 / 3 & 0\\ 0 & 0 & 1 & - 1 / 4 & - 1 / 4 & 3 / 4 \end{array} \right] \sim\) \(\displaystyle \left[ \begin{array}{llllll} 1 & 0 & 0 & 3 / 4 & - 1 / 4 & - 1 / 4\\ 0 & 1 & 0 & - 1 / 4 & 3 / 4 & - 1 / 4\\ 0 & 0 & 1 & - 1 / 4 & - 1 / 4 & 3 / 4 \end{array} \right] \Rightarrow \boldsymbol{A}^{- 1} = \left[ \begin{array}{lll} 3 / 4 & - 1 / 4 & - 1 / 4\\ - 1 / 4 & 3 / 4 & - 1 / 4\\ - 1 / 4 & - 1 / 4 & 3 / 4 \end{array} \right] .\)

   Alternatively, notice that

   \(\displaystyle \boldsymbol{A}=\boldsymbol{I}- \left[ \begin{array}{l} - 1\\ - 1\\ - 1 \end{array} \right] \left[ \begin{array}{lll} 1 & 1 & 1 \end{array} \right],\)

   and the formula in Ex. 5 below gives

   \(\displaystyle \boldsymbol{A}^{- 1} =\boldsymbol{I}+ \frac{\boldsymbol{u}\boldsymbol{v}^T}{1 -\boldsymbol{v}^T \boldsymbol{u}} =\boldsymbol{I}+ \frac{1}{4} \left[ \begin{array}{lll} - 1 & - 1 & - 1 \end{array} \right] \left[ \begin{array}{l} 1\\ 1\\ 1 \end{array} \right] = \left[ \begin{array}{lll} 3 / 4 & - 1 / 4 & - 1 / 4\\ - 1 / 4 & 3 / 4 & - 1 / 4\\ - 1 / 4 & - 1 / 4 & 3 / 4 \end{array} \right] .\)

5. Verify that the inverse of \(\boldsymbol{A}=\boldsymbol{I}-\boldsymbol{u}\boldsymbol{v}^T\) is

   \(\displaystyle \boldsymbol{A}^{- 1} =\boldsymbol{I}+ \frac{\boldsymbol{u}\boldsymbol{v}^T}{1 -\boldsymbol{v}^T \boldsymbol{u}}\)

   when \(\boldsymbol{v}^T \boldsymbol{u} \neq 1\).

   Solution. Verify that \(\boldsymbol{A}\boldsymbol{A}^{- 1} =\boldsymbol{A}^{- 1} \boldsymbol{A}=\boldsymbol{I}.\)

   \(\displaystyle \boldsymbol{A}\boldsymbol{A}^{- 1} = (\boldsymbol{I}-\boldsymbol{u}\boldsymbol{v}^T) \left( \boldsymbol{I}+ \frac{\boldsymbol{u}\boldsymbol{v}^T}{1 -\boldsymbol{v}^T \boldsymbol{u}} \right) =\boldsymbol{I}-\boldsymbol{u}\boldsymbol{v}^T + \frac{\boldsymbol{u}\boldsymbol{v}^T}{1 -\boldsymbol{v}^T \boldsymbol{u}} - \frac{\boldsymbol{u} (\boldsymbol{v}^T \boldsymbol{u}) \boldsymbol{v}^T}{1 -\boldsymbol{v}^T \boldsymbol{u}} =\) \(\displaystyle =\boldsymbol{I}-\boldsymbol{u}\boldsymbol{v}^T + \frac{1 -\boldsymbol{v}^T \boldsymbol{u}}{1 -\boldsymbol{v}^T \boldsymbol{u}} \boldsymbol{u}\boldsymbol{v}^T =\boldsymbol{I}. \checked\) \(\displaystyle \boldsymbol{A}^{- 1} \boldsymbol{A}= \left( \boldsymbol{I}+ \frac{\boldsymbol{u}\boldsymbol{v}^T}{1 -\boldsymbol{v}^T \boldsymbol{u}} \right) (\boldsymbol{I}-\boldsymbol{u}\boldsymbol{v}^T) =\boldsymbol{I}-\boldsymbol{u}\boldsymbol{v}^T + \frac{\boldsymbol{u}\boldsymbol{v}^T}{1 -\boldsymbol{v}^T \boldsymbol{u}} - \frac{\boldsymbol{u} (\boldsymbol{v}^T \boldsymbol{u}) \boldsymbol{v}^T}{1 -\boldsymbol{v}^T \boldsymbol{u}} =\boldsymbol{I}\)

6. Find \(\boldsymbol{A}^T, \boldsymbol{A}^{- 1}, (\boldsymbol{A}^{- 1})^T, (\boldsymbol{A}^T)^{- 1}\) for

   \(\displaystyle \boldsymbol{A}= \left[ \begin{array}{ll} 1 & 0\\ 9 & 3 \end{array} \right] .\)

   Solution. By definition of transpose

   \(\displaystyle \boldsymbol{A}^T = \left[ \begin{array}{ll} 1 & 9\\ 0 & 3 \end{array} \right] .\)

   Apply Gauss-Jordan

   \(\displaystyle \left[ \begin{array}{ll} \boldsymbol{A} & \boldsymbol{I} \end{array} \right] = \left[ \begin{array}{llll} 1 & 0 & 1 & 0\\ 9 & 3 & 0 & 1 \end{array} \right] \sim \left[ \begin{array}{llll} 1 & 0 & 1 & 0\\ 0 & 3 & - 9 & 1 \end{array} \right] \sim \left[ \begin{array}{llll} 1 & 0 & 1 & 0\\ 0 & 1 & - 3 & 1 / 3 \end{array} \right] \Rightarrow \boldsymbol{A}^{- 1} = \left[ \begin{array}{ll} 1 & 0\\ - 3 & 1 / 3 \end{array} \right] .\) \(\displaystyle (\boldsymbol{A}^{- 1})^T = \left[ \begin{array}{ll} 1 & - 3\\ 0 & 1 / 3 \end{array} \right] = (\boldsymbol{A}^T)^{- 1} .\)

7. Describe within \(\mathbb{R}^3\) the geometry of the column spaces of matrices

   \(\displaystyle \boldsymbol{A}= \left[ \begin{array}{ll} 1 & 2\\ 0 & 0\\ 0 & 0 \end{array} \right], \boldsymbol{B}= \left[ \begin{array}{ll} 1 & 0\\ 0 & 2\\ 0 & 0 \end{array} \right], \boldsymbol{C}= \left[ \begin{array}{ll} 1 & 0\\ 2 & 0\\ 0 & 0 \end{array} \right] .\)

   Solution. \(C (\boldsymbol{A})\) is a line since \(\operatorname{rank} (\boldsymbol{A}) = 1\), \(C (\boldsymbol{B})\) is a plane, \(\operatorname{rank} (\boldsymbol{B}) = 2\), \(C (\boldsymbol{A})\) is a line.

8. The vector subspaces of \(\mathbb{R}^2\) are lines, \(\mathbb{R}^2\) itself and \(Z = \left\{ \left[ \begin{array}{ll} 0 & 0 \end{array} \right]^T \right\}\). What are the vector subspaces of \(\mathbb{R}^3\)?

   Solution. Lines, planes, \(\mathbb{R}^3\) itself and \(Z = \left\{ \left[ \begin{array}{lll} 0 & 0 & 0 \end{array} \right]^T \right\}\)

9. Reduce the following matrices to row echelon form

   \(\displaystyle \boldsymbol{A}= \left[ \begin{array}{lllll} 1 & 2 & 2 & 4 & 6\\ 1 & 2 & 3 & 6 & 9\\ 0 & 0 & 1 & 2 & 3 \end{array} \right], \boldsymbol{B}= \left[ \begin{array}{lll} 2 & 4 & 2\\ 0 & 4 & 4\\ 0 & 8 & 8 \end{array} \right] .\)

   Solution. Obtain

   \(\displaystyle \boldsymbol{A} \sim \left[ \begin{array}{lllll} 1 & 2 & 2 & 4 & 6\\ 0 & 0 & 1 & 2 & 3\\ 0 & 0 & 1 & 2 & 3 \end{array} \right] \sim \left[ \begin{array}{lllll} 1 & 2 & 2 & 4 & 6\\ 0 & 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 0 & 0 \end{array} \right]\)
   \(\displaystyle \boldsymbol{B} \sim \left[ \begin{array}{lll} 2 & 4 & 2\\ 0 & 4 & 4\\ 0 & 8 & 8 \end{array} \right] \sim \left[ \begin{array}{lll} 1 & 2 & 1\\ 0 & 1 & 1\\ 0 & 1 & 1 \end{array} \right] \sim \left[ \begin{array}{lll} 1 & 2 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{array} \right] \sim \left[ \begin{array}{lll} 1 & 0 & - 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{array} \right]\)

10. Determine the null space of

    \(\displaystyle \boldsymbol{A}= \left[ \begin{array}{lll} - 1 & 3 & 5\\ - 2 & 6 & 10 \end{array} \right] .\)

    Solution. By rref

    \(\displaystyle \boldsymbol{A} \sim \left[ \begin{array}{lll} - 1 & 3 & 5\\ 0 & 0 & 0 \end{array} \right],\)

    hence \(r =\operatorname{rank} (\boldsymbol{A}) = 1\), with \(\boldsymbol{A} \in \mathbb{R}^{2 \times 3} =\mathbb{R}^{m \times n}\). From FTLA \(r + z = n\) with \(z = \dim N (\boldsymbol{A}) = 3\). Two basis vectors for the null space can be chosen as

    \(\displaystyle \boldsymbol{u}= \left[ \begin{array}{l} 3\\ 1\\ 0 \end{array} \right], \boldsymbol{v}= \left[ \begin{array}{l} 8\\ 1\\ 1 \end{array} \right] .\)