Q47P Find a unitary matrix U which di... [FREE SOLUTION]

Chapter 3: Q47P (page 161)

Find a unitary matrix U which diagonalizes A in (11.31) and verify that $U^{- 1} A U$ is diagonal with the eigenvalues down the main diagonal.

Short Answer

Expert verified

Unitary matrix

U= $\frac{1}{2} (\begin{matrix} \sqrt{2} & 1 & 1 \\ 0 & i \sqrt{2} & - i \sqrt{2} \\ \sqrt{2} & - 1 & - 1 \end{matrix})$

Step by step solution

Given Information

$A = \frac{1}{2} (\begin{matrix} 1 & \sqrt{2} & 1 \\ - \sqrt{2} & 0 & \sqrt{2} \\ 1 & - \sqrt{2} & 1 \end{matrix})$

Unitary Matrix

In linear algebra, a complex square matrix $U$ is unitary if its conjugate transpose $U^{*}$ is also its inverse, that is, if

$U^{*} U = U U^{*} = U U^{- 1} = I,$

Where, $l$ is the identity matrix.

Eigenvalue equation

$A r = 位 r,$

Thus,

$|\begin{matrix} \frac{1}{2} - 位 & \frac{1}{\sqrt{2}} & \frac{1}{2} \\ - \frac{1}{\sqrt{2}} & - 位 & \frac{1}{\sqrt{2}} \\ \frac{1}{2} & - \frac{1}{\sqrt{2}} & \frac{1}{2} - x \end{matrix}| = 0$

$- 位 {(\frac{1}{2} - 位)}^{2} + \frac{1}{4} + \frac{1}{4} + \frac{位}{4} + \frac{1}{2} (\frac{1}{2} - 位) + \frac{1}{2} (\frac{1}{2} - 位) = - 位 {(\frac{1}{2} - 位)}^{2} + \frac{1}{2} + \frac{位}{4} + (\frac{1}{2} - 位) = - 位 {(1 - 2 位)}^{2} + 2 + 位 + 2 (1 - 2 位)$

$= 4 位^{2} - 4 位^{3} - 4 位 + 4 = 位^{3} - 位^{2} + 位 - 1 = 位^{2} (位 - 1) + 位 - 1 = (位 - 1) (位^{2} + 1) = 0$

The eigenvalues

$位_{1} = 1,$

And

$位_{2} = i,$

And

$位_{3} = - i,$

The eigen values

$v_{- i} = \frac{1}{2} (\begin{matrix} 1 \\ - i \sqrt{2} \\ - 1 \end{matrix})$

Unitary matrix U which diagonalizes A ,

$U = \frac{1}{2} (\begin{matrix} \sqrt{2} & 1 & 1 \\ 0 & i \sqrt{2} & - i \sqrt{2} \\ \sqrt{2} & - 1 & - 1 \end{matrix})$

The inverse of the equation

$U^{- 1} = U^{^{+}} = \frac{1}{2} (\begin{matrix} \sqrt{2} & 0 & \sqrt{2} \\ 0 & - i \sqrt{2} & - 1 \\ \sqrt{2} & - i \sqrt{2} & - 1 \end{matrix})$

$U^{- 1} A D = D = \frac{1}{8} (\begin{matrix} \sqrt{2} & 0 & \sqrt{2} \\ 1 & - i \sqrt{2} & - 1 \\ 1 & i \sqrt{2} & - 1 \end{matrix}) (\begin{matrix} 1 & \sqrt{2} & 1 \\ - \sqrt{2} & 0 & \sqrt{2} \\ 1 & - \sqrt{2} & - 1 \end{matrix}) (\begin{matrix} \sqrt{2} & 1 & 1 \\ 0 & i \sqrt{2} & - i \sqrt{2} \\ \sqrt{2} & - 1 & - 1 \end{matrix}) = \frac{1}{8} (\begin{matrix} 2 \sqrt{2} & 0 & 2 \sqrt{2} \\ 2 i & 2 \sqrt{2} & - 2 i \\ - 2 i & 2 \sqrt{2} & 2 i \end{matrix}) (\begin{matrix} \sqrt{2} & 1 & 1 \\ 0 & i \sqrt{2} & - i \sqrt{2} \\ \sqrt{2} & - 1 & - 1 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - 1 \end{matrix})$

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Find a unitary matrix U which diagonalizes A in (11.31) and verify that $U^{- 1} A U$ is diagonal with the eigenvalues down the main diagonal.

Short Answer

Step by step solution

Given Information

Unitary Matrix

Eigenvalue equation

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91影视

Find a unitary matrix U which diagonalizes A in (11.31) and verify that U-1AU is diagonal with the eigenvalues down the main diagonal.

Short Answer

Step by step solution

Given Information

Unitary Matrix

Eigenvalue equation

One App. One Place for Learning.

Most popular questions from this chapter

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Find a unitary matrix U which diagonalizes A in (11.31) and verify that $U^{- 1} A U$ is diagonal with the eigenvalues down the main diagonal.