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Chapter 2: Q6SE (page 93)

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{ - {\bf{1}}}\end{aligned}} \right)\),\(B = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}\end{aligned}} \right)\).These are Pauli spin matrices used in the study of electron spin in quantum mechanics. Show that \({A^{\bf{2}}} = I\), \({B^{\bf{2}}} = I\), and \(AB = - BA\). Matrices such that \(AB = - BA\) are said to anticommute.

Short Answer

Expert verified

Hence, \({A^2} = I,{B^2} = I,\) and \(AB = - BA\) are proved.

Step by step solution

01

Compute \({A^{\bf{2}}}\)

\(\begin{aligned}{c}{A^2} = AA\\ = \left( {\begin{aligned}{*{20}{c}}1&0\\0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\0&{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right)\\{A^2} = I\end{aligned}\)

02

Compute \({B^{\bf{2}}}\)

\(\begin{aligned}{c}{B^2} = BB\\ = \left( {\begin{aligned}{*{20}{c}}0&1\\1&0\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0&1\\1&0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right)\\{B^2} = I\end{aligned}\)

03

Check \(AB =  - BA\)

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}1&0\\0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0&1\\1&0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&1\\{ - 1}&0\end{aligned}} \right)\end{aligned}\)

\(\begin{aligned}{c} - BA = - \left( {\begin{aligned}{*{20}{c}}0&1\\1&0\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\0&{ - 1}\end{aligned}} \right)\\ = - \left( {\begin{aligned}{*{20}{c}}0&{ - 1}\\1&0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&1\\{ - 1}&0\end{aligned}} \right)\end{aligned}\)

Thus \( - BA = AB\).

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Most popular questions from this chapter

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 \(AB\).

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

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

36. Write the command(s) that will create a \(6 \times 4\) matrix with random entries. In what range of numbers do the entries lie? Tell how to create a \(3 \times 3\) matrix with random integer entries between \( - {\bf{9}}\) and 9. (Hint:If xis a random number such that 0 < x < 1, then \( - 9.5 < 19\left( {x - .5} \right) < 9.5\).

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.

Explain why the columns of an \(n \times n\) matrix Aare linearly independent when Ais invertible.

Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.
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