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Chapter 2: Problem 32

Show that \(A=\left[\begin{array}{cc}\frac{1}{3}-\frac{2}{3} i & \frac{2}{3} i \\\ -\frac{2}{3} i & -\frac{1}{3}-\frac{2}{3} i\end{array}\right]\) is unitary.

Short Answer

Expert verified
Finding the conjugate transpose of matrix A, we get \[A* = \left[\begin{array}{cc} \frac{1}{3}+\frac{2}{3}i & -\frac{2}{3}i \\ \frac{2}{3}i & -\frac{1}{3}+\frac{2}{3}i \end{array}\right]\]. Computing the product A * A* gives \[A * A* = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\]. Since the result equals the identity matrix, matrix A is unitary.

Step by step solution

01

Find the conjugate transpose of matrix A

To find the conjugate transpose of matrix A, we need to first transpose the matrix and then find the complex conjugate of each element. The transpose of matrix A is given by: $A^T = \left[\begin{array}{cc} \frac{1}{3}-\frac{2}{3}i & -\frac{2}{3}i \\ \frac{2}{3}i & -\frac{1}{3}-\frac{2}{3}i \end{array}\right]^T = \left[\begin{array}{cc} \frac{1}{3}-\frac{2}{3}i & \frac{2}{3}i \\ -\frac{2}{3}i & -\frac{1}{3}-\frac{2}{3}i \end{array}\right]$ Now we can find the complex conjugate of each element: $A* = \left[\begin{array}{cc} \frac{1}{3}+\frac{2}{3}i & -\frac{2}{3}i \\ \frac{2}{3}i & -\frac{1}{3}+\frac{2}{3}i \end{array}\right]$
02

Compute the product A * A*

Now we need to find the product of A and A*: $A * A* = \left[\begin{array}{cc} \frac{1}{3}-\frac{2}{3}i & \frac{2}{3}i \\ -\frac{2}{3}i & -\frac{1}{3}-\frac{2}{3}i \end{array}\right] * \left[\begin{array}{cc} \frac{1}{3}+\frac{2}{3}i & -\frac{2}{3}i \\ \frac{2}{3}i & -\frac{1}{3}+\frac{2}{3}i \end{array}\right]$ $= \left[\begin{array}{cc} \left((\frac{1}{3}-\frac{2}{3}i)\left(\frac{1}{3}+\frac{2}{3}i\right) + (\frac{2}{3}i)(\frac{2}{3}i)\right) & \left((\frac{1}{3}-\frac{2}{3}i)(-\frac{2}{3}i) + (\frac{2}{3}i)(-\frac{1}{3}+\frac{2}{3}i)\right) \\ \left((-\frac{2}{3}i)(\frac{1}{3}+\frac{2}{3}i) + (-\frac{1}{3}-\frac{2}{3}i)(\frac{2}{3}i)\right) & \left((-\frac{2}{3}i)(-\frac{2}{3}i) + (-\frac{1}{3}-\frac{2}{3}i)(-\frac{1}{3}+\frac{2}{3}i)\right) \end{array}\right]$ $= \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]$
03

Verify the result

Since A * A* is equal to the identity matrix, we can conclude that the given matrix A is unitary.

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Most popular questions from this chapter

Let \(V\) and \(W\) be vector spaces with subspaces \(V_{1}\) and \(W_{1}\), respectively. If \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) is linear, prove that \(\mathrm{T}\left(\mathrm{V}_{1}\right)\) is a subspace of \(\mathrm{W}\) and that $\left\\{x \in \mathrm{V}: \mathrm{T}(x) \in \mathrm{W}_{1}\right\\}\( is a subspace of \)\mathrm{V}$.

Compute \(A B\) using block multiplication, where $$A=\left[\begin{array}{ccc} 1 & 2 & 1 \\ 3 & 4 & 0 \\ 0 & 0 & 2 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cccc} 1 & 2 & 3 & 1 \\ 4 & 5 & 6 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right].$$

Let \(V\) and \(W\) be vector spaces, let \(T: V \rightarrow W\) be linear, and let \(\left\\{w_{1}, w_{2}, \ldots, w_{k}\right\\}\) be a linearly independent set of \(k\) vectors from \(\mathbf{R}(\mathbf{T})\). Prove that if $S=\left\\{v_{1}, v_{2}, \ldots, v_{k}\right\\}$ is chosen so that \(\mathrm{T}\left(v_{i}\right)=w_{i}\) for \(i=1,2, \ldots, k\), then \(S\) is linearly independent. Visit goo.g1/kmaQS2 for a solution.

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 \(B=\left[\begin{array}{cc}1 & 0 \\ 1-i & i\end{array}\right]\)

For each matrix \(A\) and ordered basis \(\beta\), find \(\left[\mathrm{L}_{A}\right]_{\beta}\). Also, find an invertible matrix \(Q\) such that \(\left[\mathrm{L}_{A}\right]_{\beta}=Q^{-1} A Q\). (a) \(A=\left(\begin{array}{ll}1 & 3 \\ 1 & 1\end{array}\right)\) and $\beta=\left\\{\left(\begin{array}{l}1 \\\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 2\end{array}\right)\right\\}$ (b) \(A=\left(\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right)\) and $\beta=\left\\{\left(\begin{array}{l}1 \\\ 1\end{array}\right),\left(\begin{array}{r}1 \\ -1\end{array}\right)\right\\}$ (c) $A=\left(\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 1 \\ 1 & 1 & 0\end{array}\right) \quad\( and \)\quad \beta=\left\\{\left(\begin{array}{l}1 \\\ 1 \\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 1 \\\ 2\end{array}\right)\right\\}$ (d) $A=\left(\begin{array}{rrr}13 & 1 & 4 \\ 1 & 13 & 4 \\ 4 & 4 & 10\end{array}\right)\( and \)\beta=\left\\{\left(\begin{array}{r}1 \\ 1 \\\ -2\end{array}\right),\left(\begin{array}{r}1 \\ -1 \\\ 0\end{array}\right),\left(\begin{array}{l}1 \\ 1 \\\ 1\end{array}\right)\right\\}$
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