Question 3

Question 3.

Let

\begin{equation*} a_0 = \left( \begin{array}{r} -1 \\ 2 \\ 3 \end{array} \right) \mbox{ and } b_0 = \left( \begin{array}{r} 1 \\ 0 \\ -2 \end{array} \right). \end{equation*}

Compute \(a_0 b_0^T = \)
Let

\begin{equation*} a_1 = \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \mbox{ and } b_1 = \left( \begin{array}{r} 0 \\ 1 \\ 2 \end{array} \right). \end{equation*}

Compute \(a_1 b_1^T =\)
Compute \(\left( \begin{array}{r r} -1 \amp 1 \\ 2 \amp -1 \\ 3 \amp 2 \end{array} \right) \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \\ 0 \amp 1 \amp 2 \end{array} \right) =\)

\begin{equation*} a_0 b_0^T = \left( \begin{array}{r} -1 \\ 2 \\ 3 \end{array} \right) \left( \begin{array}{r} 1 \\ 0 \\ -2 \end{array} \right)^T = \left( \begin{array}{r} -1 \\ 2 \\ 3 \end{array} \right) \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \end{array} \right) = \left( \begin{array}{r r r} -1 \amp 0 \amp 2 \\ 2 \amp 0 \amp -4 \\ 3 \amp 0 \amp -6 \end{array} \right) \end{equation*}
\begin{equation*} a_1 b_1^T = \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \left( \begin{array}{r} 0 \\ 1 \\ 2 \end{array} \right)^T = \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \left( \begin{array}{r r r} 0 \amp 1 \amp 2 \end{array} \right) = \left( \begin{array}{r r r} 0 \amp 1 \amp 2 \\ 0 \amp -1 \amp -2 \\ 0 \amp 2 \amp 4 \end{array} \right) \end{equation*}
Compute \(\left( \begin{array}{r r} -1 \amp 1 \\ 2 \amp -1 \\ 3 \amp 2 \end{array} \right) \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \\ 0 \amp 1 \amp 2 \end{array} \right) = \left( \begin{array}{r r r} -1 \amp 1 \amp 4 \\ 2 \amp -1 \amp -6 \\ 3 \amp 2 \amp -2x \end{array} \right) \)

Solution

When computing an outer product, we like to think of it as scaling rows

\begin{equation*} x y^T = \left( \begin{array}{c} \chi_0 \\ \hline \vdots \\ \hline \chi_{m-1} \end{array} \right) y^T = \left( \begin{array}{c} \chi_0 y^T \\ \hline \vdots \\ \hline \chi_{m-1} y^T \end{array} \right) \end{equation*}

or scaling columns

\begin{equation*} \begin{array}{rcl} x y^T \amp = \amp x \left( \begin{array}{c} \psi_0 \\ \vdots \\ \psi_{n-1} \end{array} \right)^T = x \left( \begin{array}{c | c | c} \psi_0 \amp \cdots \amp \psi_{n-1} \end{array} \right) \\ \amp = \amp \left( \begin{array}{c | c | c} x \psi_0 \amp \cdots \amp x \psi_{n-1} \end{array} \right) = \left( \begin{array}{c | c | c} \psi_0 x \amp \cdots \amp \psi_{n-1} x \end{array} \right) . \end{array} \end{equation*}

This makes it easy to systematically compute the results:

\begin{equation*} \begin{array}[t]{rcl} a_0 b_0^T \amp = \amp \left( \begin{array}{r} -1 \\ 2 \\ 3 \end{array} \right) \left( \begin{array}{r} 1 \\ 0 \\ -2 \end{array} \right)^T = \left( \begin{array}{r} -1 \\ 2 \\ 3 \end{array} \right) \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \end{array} \right) \\ \amp = \amp \left( \begin{array}{r} (-1) \times \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \end{array} \right) \\ \hline (2) \times \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \end{array} \right) \\ \hline (3) \times \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \end{array} \right) \end{array} \right) = \left( \begin{array}{r r r} -1 \amp 0 \amp 2 \\ 2 \amp 0 \amp -4 \\ 3 \amp 0 \amp -6 \end{array} \right) \end{array} \end{equation*}
\begin{equation*} \begin{array}[t]{rcl} a_1 b_1^T \amp = \amp \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \left( \begin{array}{r} 0 \\ 1 \\ 2 \end{array} \right)^T = \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \left( \begin{array}{r | r | r} 0 \amp 1 \amp 2 \end{array} \right) \\ \amp = \amp \left( \begin{array}{r | r | r} (0) \times \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \amp (1) \times \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \amp (2) \times \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \end{array} \right) \\ \amp = \amp \left( \begin{array}{r r r} 0 \amp 1 \amp 2 \\ 0 \amp -1 \amp -2 \\ 0 \amp 2 \amp 4 \end{array} \right) \end{array} \end{equation*}
Here there are two insights that make computing the result easier, given what has already been computed:
- Notice that
  
  \begin{equation*} \begin{array}{rcl} A B \amp =\amp \left( \begin{array}{c | c | c | c} a_0 \amp a_1 \amp \cdots \amp a_{k-1} \end{array} \right) \left( \begin{array}{c} b_0^T \\ \hline b_1^T \\ \hline \vdots \\ \hline b_{k-1}^T \end{array} \right)\\ \amp= \amp a_0 b_0^T + a_1 b_1^T + \cdots a_{k-1} b_{k-1}^T. \end{array} \end{equation*}
  
  Thus, we can use the results from the previous parts of this problem:
  
  \begin{equation*} \begin{array}{l} \left( \begin{array}{r | r} -1 \amp 1 \\ 2 \amp -1 \\ 3 \amp 2 \end{array} \right) \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \\ \hline 0 \amp 1 \amp 2 \end{array} \right) \\ ~~~ = \left( \begin{array}{r} -1 \\ 2 \\ 3 \end{array} \right) \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \end{array} \right) + \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) \left( \begin{array}{r r r} 0 \amp 1 \amp 2 \end{array} \right) \\ ~~~ = \left( \begin{array}{r r r} -1 \amp 0 \amp 2 \\ 2 \amp 0 \amp -4 \\ 3 \amp 0 \amp -6 \end{array} \right) + \left( \begin{array}{r r r} 0 \amp 1 \amp 2 \\ 0 \amp -1 \amp -2 \\ 0 \amp 2 \amp 4 \end{array} \right) \\ ~~~ = \left( \begin{array}{r r r} -1 \amp 1 \amp 4 \\ 2 \amp -1 \amp -6 \\ 3 \amp 2 \amp -2 \end{array} \right). \end{array} \end{equation*}
- Alternatively, we can use the insight that if
  
  \begin{equation*} B = \left( \begin{array}{ c | c | c } B_0 \amp \cdots \amp B_{N-1} \end{array} \right) \end{equation*}
  
  then
  
  \begin{equation*} A B = A \left( \begin{array}{ c | c | c } B_0 \amp \cdots \amp B_{N-1} \end{array} \right) = \left( \begin{array}{ c | c | c } A B_0 \amp \cdots \amp A B_{N-1} \end{array} \right) \end{equation*}
  
  so that
  
  \begin{equation*} \begin{array}{l} \left( \begin{array}{r r} -1 \amp 1 \\ 2 \amp -1 \\ 3 \amp 2 \end{array} \right) \left( \begin{array}{r r r} 1 \amp 0 \amp -2 \\ 0 \amp 1 \amp 2 \end{array} \right) \\ ~~~ = \left( \begin{array}{r r} -1 \amp 1 \\ 2 \amp -1 \\ 3 \amp 2 \end{array} \right) \left( \begin{array}{c | c} \left( \begin{array}{r r} 1 \amp 0 \\ 0 \amp 1 \end{array} \right) \amp \left( \begin{array}{r r r} -2 \\ 2 \end{array} \right) \end{array} \right) \\ ~~~ = \left( \begin{array}{c | c} \left( \begin{array}{r r} -1 \amp 1 \\ 2 \amp -1 \\ 3 \amp 2 \end{array} \right) \left( \begin{array}{r r} 1 \amp 0 \\ 0 \amp 1 \end{array} \right) \amp \left( \begin{array}{r r} -1 \amp 1 \\ 2 \amp -1 \\ 3 \amp 2 \end{array} \right) \left( \begin{array}{r r r} -2 \\ 2 \end{array} \right) \end{array} \right) \\ ~~~ = \left( \begin{array}{r r | r} -1 \amp 1 \amp 4 \\ 2 \amp -1 \amp -6 \\ 3 \amp 2 \amp -2 \end{array} \right) = \left( \begin{array}{r r r} -1 \amp 1 \amp 4 \\ 2 \amp -1 \amp -6 \\ 3 \amp 2 \amp -2 \end{array} \right). \end{array} \end{equation*}

2 Relevance

Relevance

Again, we emphasize that understanding how to look at matrix-matrix multiplication in different ways is of utmost importance. In particular, viewing a matrix-matrix multiplication as the sum of a sequence of outer products (often called "rank-1 updates") is central to understanding important concepts like the Singular Value Decomposition (SVD) in ALAFF.

3 Resources

Resources

How to compute with matrices that have been "sliced and diced" (partitioned) as illustrated in the solution to this problem is discussed in great detail in Week 4 and Week 5 of LAFF.

Advanced Linear Algebra: Are you ready?

Subsection 2.3 Question 3

Question 3.