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Chapter 2: Problem 81

Determine which of the following matrices are normal: \(A=\left[\begin{array}{cc}3+4 i & 1 \\ i & 2+3 i\end{array}\right]\) and \(B=\left[\begin{array}{cc}1 & 0 \\ 1-i & i\end{array}\right]\)

Short Answer

Expert verified
In conclusion, neither matrix \(A=\left[\begin{array}{cc}3+4 i & 1 \\\ i & 2+3 i\end{array}\right]\) nor matrix \(B=\left[\begin{array}{cc}1 & 0 \\\ 1-i & i\end{array}\right]\) is normal, as their respective products \(AA^*\) and \(A^*A\), and \(BB^*\) and \(B^*B\) are not equal.

Step by step solution

01

Compute A* (Conjugate transpose of A)

To find the conjugate transpose of A, transpose A and take the complex conjugate of each element: \(A^* = \left[\begin{array}{cc}3-4i & -i \\ 1 & 2-3i \end{array}\right]\)
02

Multiply A by A* (AA*)

Multiply the original matrix A with its conjugate transpose A*: \(AA^*=\left[\begin{array}{cc}3+4i & 1 \\ i & 2+3i \end{array}\right]\left[\begin{array}{cc}3-4i & -i \\ 1 & 2-3i \end{array}\right]=\left[\begin{array}{cc}25 & 5-12i \\ 5+12i & 14 \end{array}\right]\)
03

Multiply A* by A (A*A)

Multiply the conjugate transpose A* by the original matrix A: \(A^*A=\left[\begin{array}{cc}3-4i & -i \\ 1 & 2-3i \end{array}\right]\left[\begin{array}{cc}3+4i & 1 \\ i & 2+3i \end{array}\right]=\left[\begin{array}{cc}25 & 5+12i \\ 5-12i & 14 \end{array}\right]\)
04

Compare AA* and A*A for matrix A

Since \(AA^*\neq A^*A\), matrix A is not a normal matrix. For matrix B:
05

Compute B* (Conjugate transpose of B)

To find the conjugate transpose of B, transpose B and take the complex conjugate of each element: \(B^*=\left[\begin{array}{cc}1 & 1+i \\ 0 & -i \end{array}\right]\)
06

Multiply B by B* (BB*)

Multiply the original matrix B with its conjugate transpose B*: \(BB^* =\left[\begin{array}{cc}1 & 0 \\ 1-i & i \end{array}\right]\left[\begin{array}{cc}1 & 1+i \\ 0 & -i \end{array}\right]=\left[\begin{array}{cc}1 & 1+i \\ 1+i & 2 \end{array}\right]\)
07

Multiply B* by B (B*B)

Multiply the conjugate transpose B* by the original matrix B: \(B^*B=\left[\begin{array}{cc}1 & 1+i \\ 0 & -i \end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 1-i & i \end{array}\right]=\left[\begin{array}{cc}1 & 1+i \\ 1-i & -1 \end{array}\right]\)
08

Compare BB* and B*B for matrix B

Since \(BB^*\neq B^*B\), matrix B is not a normal matrix. In conclusion, neither matrix A nor matrix B is normal.

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Most popular questions from this chapter

Prove Theorem \(2.2(\) iii) and (iv): (iii) \((B+C) A=B A+C A\) (iv) \(k(A B)=(k A) B=A(k B)\)

Suppose that $\left\\{\mathrm{U}_{1}, \mathrm{U}_{2}, \ldots, \mathrm{U}_{n}\right\\}$ is a collection of pairwise commutative linear operators on a vector space \(\mathrm{V}\) (i.e., operators such that \(\mathrm{U}_{i} \mathrm{U}_{j}=\) \(\mathrm{U}_{j} \mathrm{U}_{i}\) for all $i, j\( ). Prove that, for any \)i(1 \leq i \leq n)$, $\mathrm{N}\left(\mathrm{U}_{i}\right) \subseteq \mathrm{N}\left(\mathrm{U}_{1} \mathrm{U}_{2} \cdots \mathrm{U}_{n}\right) .$

Prove that there exists a linear transformation \(T: R^{2} \rightarrow R^{3}\) such that \(\mathrm{T}(1,1)=(1,0,2)\) and \(\mathrm{T}(2,3)=(1,-1,4)\). What is \(\mathrm{T}(8,11)\) ?

Assume the notation in Theorem \(2.13 .\) (a) Suppose that \(z\) is a (column) vector in \(\mathrm{F}^{p}\). Use Theorem 2.13(b) to prove that \(B z\) is a linear combination of the columns of \(B\). In particular, if \(z=\left(a_{1}, a_{2}, \ldots, a_{p}\right)^{t}\), then show that $$ B z=\sum_{j=1}^{p} a_{j} v_{j} . $$ (b) Extend (a) to prove that column \(j\) of \(A B\) is a linear combination of the columns of \(A\) with the coefficients in the linear combination being the entries of column \(j\) of \(B\). (c) For any row vector \(w \in \mathrm{F}^{m}\), prove that \(w A\) is a linear combination of the rows of \(A\) with the coefficients in the linear combination being the coordinates of \(w\). Hint: Use properties of the transpose operation applied to (a). (d) Prove the analogous result to (b) about rows: Row \(i\) of \(A B\) is a linear combination of the rows of \(B\) with the coefficients in the linear combination being the entries of row \(i\) of \(A\).

Suppose \(A\) is a complex matrix. Show that \(A A^{H}\) and \(A^{H} A\) are Hermitian.
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