Skip to main content

Math 211 Activities Workbook: Adapted from: Understanding Linear Algebra by David Austin

David Austin, Erica Bernstein

Section 8.2 The matrix transpose

Definition 8.2.1.

The transpose of the \(m\times n\) matrix \(A\) is the \(n\times m\) matrix \(A^T\) whose rows are the columns of \(A\text{.}\)

Example 8.2.2.

If \(A=\begin{bmatrix} 4 \amp -3 \amp 0 \amp 5 \\ -1 \amp 2 \amp 1 \amp 3 \\ \end{bmatrix} \text{,}\) then \(A^T=\begin{bmatrix} 4 \amp -1 \\ -3 \amp 2 \\ 0 \amp 1 \\ 5 \amp 3 \\ \end{bmatrix}\)

Activity 8.2.1.

This activity illustrates how multiplying a vector by \(A^T\) is related to computing dot products with the columns of \(A\text{.}\) You’ll develop a better understanding of this relationship if you compute the dot products and matrix products in this activity without using technology.
  1. If \(B = \begin{bmatrix} 3 \amp 4 \\ -1 \amp 2 \\ 0 \amp -2 \\ \end{bmatrix} \text{,}\) write the matrix \(B^T\text{.}\)
  2. Suppose that
    \begin{equation*} \vvec_1=\threevec20{-2},\hspace{24pt} \vvec_2=\threevec112,\hspace{24pt} \wvec=\threevec{-2}23\text{.} \end{equation*}
    Find the dot products \(\vvec_1\cdot\wvec\) and \(\vvec_2\cdot\wvec\text{.}\)
  3. Now write the matrix \(A = \begin{bmatrix} \vvec_1 \amp \vvec_2 \end{bmatrix}\) and its transpose \(A^T\text{.}\) Find the product \(A^T\wvec\) and describe how this product computes both dot products \(\vvec_1\cdot\wvec\) and \(\vvec_2\cdot\wvec\text{.}\)
  4. Suppose that \(\xvec\) is a vector that is orthogonal to both \(\vvec_1\) and \(\vvec_2\text{.}\) What does this say about the dot products \(\vvec_1\cdot\xvec\) and \(\vvec_2\cdot\xvec\text{?}\) What does this say about the product \(A^T\xvec\text{?}\)
  5. Use the matrix \(A^T\) to give a parametric description of all the vectors \(\xvec\) that are orthogonal to \(\vvec_1\) and \(\vvec_2\text{.}\)
  6. Remember that \(\nul(A^T)\text{,}\) the null space of \(A^T\text{,}\) is the solution set of the equation \(A^T\xvec=\zerovec\text{.}\) If \(\xvec\) is a vector in \(\nul(A^T)\text{,}\) explain why \(\xvec\) must be orthogonal to both \(\vvec_1\) and \(\vvec_2\text{.}\)
  7. Remember that \(\col(A)\text{,}\) the column space of \(A\text{,}\) is the set of linear combinations of the columns of \(A\text{.}\) Therefore, any vector in \(\col(A)\) can be written as \(c_1\vvec_1+c_2\vvec_2\text{.}\) If \(\xvec\) is a vector in \(\nul(A^T)\text{,}\) explain why \(\xvec\) is orthogonal to every vector in \(\col(A)\text{.}\)
Solution.
  1. \(\displaystyle A^T = \begin{bmatrix} 3 \amp -1 \amp 0 \\ 4 \amp 2 \amp -2 \end{bmatrix}\)
  2. \begin{align*} \vvec_1\cdot\wvec \amp{}={} 2(-2) + 0(2) + (-2)3 \amp{}={} -10\\ \vvec_2\cdot\wvec \amp{}={} 1(-2) + 1(2) + 2(3) \amp{}={} 6 \end{align*}
  3. \begin{equation*} A^T\wvec = \twovec{2(-2)+0(2)+(-2)3}{1(-2)+1(2)+2(3)} = \twovec{-10}{6} =\twovec{\vvec_1\cdot\wvec}{\vvec_2\cdot\wvec} \end{equation*}
  4. Both dot products are 0 so we have \(A^T\xvec = \zerovec\text{.}\)
  5. We need to solve the equation \(A^T\xvec = \zerovec\) so we find the reduced row echelon form
    \begin{equation*} A^T \sim \begin{bmatrix} 1 \amp 0 \amp -1 \\ 0 \amp 1 \amp 3\\ \end{bmatrix}\text{.} \end{equation*}
    The vectors orthogonal to both \(\vvec_1\) and \(\vvec_2\) have the form \(\xvec=x_3\threevec1{-3}1\text{.}\)
  6. \(A^T\xvec = \twovec{\vvec_1\cdot\xvec}{\vvec_2\cdot\xvec} = \twovec00\) tells us that \(\vvec_1\cdot\xvec=0\) and \(\vvec_2\cdot\xvec=0\text{.}\)
  7. Since \(\xvec\) is orthogonal to both \(\vvec_1\) and \(\vvec_2\text{,}\) we have
    \begin{equation*} (c_1\vvec_1+c_2\vvec_2)\cdot\xvec = c_1\vvec_1\cdot\xvec + c_2\vvec_2\cdot\xvec = 0 \end{equation*}
The previous activity demonstrates an important connection between the matrix transpose and dot products. More specifically, the components of the product \(A^T\xvec\) are simply the dot products of the columns of \(A\) with \(\xvec\text{.}\) We will make frequent use of this observation so let’s record it as a proposition.
PrevTopNext