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23. (Rank Factorization) Suppose an \[m \times n\] matrix A admits a factorization \[A = CD\] where C is \[m \times 4\] and D is \[4 \times n\].

  1. Show that Ais the sum of four outer products. (See section 2.4.)
  2. Let \[{\bf{m}} = {\bf{400}}\] and \[{\bf{n}} = {\bf{100}}\]. Explain why a computer programmer might prefer to store the data from A in the form of two matrices C and D.

Short Answer

Expert verified
  1. A can be written as the sum of four outer products.
  2. Storing C and D together needs fewer entries than those required for A. Hence, a computer programmer might prefer to store the data from A in the form of two matrices C and D.

Step by step solution

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01

Use the multiplication rule for partitioned matrices

(a)

Express each row of D as the transpose of a column vector.

\[D = \left[ {\begin{array}{*{20}{c}}{{d_1}^T}\\{{d_2}^T}\\{{d_3}^T}\\{{d_4}^T}\end{array}} \right]\]

Then,

\[\begin{array}{c}A = CD\\ = \left[ {\begin{array}{*{20}{c}}{{c_1}}&{{c_2}}&{{c_3}}&{{c_4}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{d_1}^T}\\{{d_2}^T}\\{{d_3}^T}\\{{d_4}^T}\end{array}} \right]\\A = {c_1}{d_1}^T + {c_2}{d_2}^T + {c_3}{d_3}^T + {c_4}{d_4}^T\end{array}\]

This implies that Ais the sum of four outer products.

02

Explain how a computer programmer works

(b)

Here, \[m = 400\] and \[n = 100\]. So, A has \[400 \times 100 = 40000\] entries. The matrices C with \[400 \times 4 = 1600\]entries and D with \[100 \times 4 = 400\] entries, hence, storing C and D together needs only 2000 entries, which is much smaller than 40000. So, the computer programmer needs only 2000 entries to store A directly.

03

Conclusion

Therefore, a computer programmer might prefer to store the data from A in the form of two matrices C and D.

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Most popular questions from this chapter

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

Suppose \(CA = {I_n}\)(the \(n \times n\) identity matrix). Show that the equation \(Ax = 0\) has only the trivial solution. Explain why Acannot have more columns than rows.

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

\(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{5}}&{\bf{4}}\\{\bf{2}}&{ - {\bf{3}}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{2}}}\\{ - {\bf{2}}}&{\bf{1}}\end{aligned}} \right)\)

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can Tmap \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

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