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Chapter 2: Problem 80

Suppose \(A\) and \(B\) are unitary. Show that \(A^{H}, A^{-1}, A B\) are unitary.

Short Answer

Expert verified
We have given that matrices A and B are unitary. To show that \(A^H\), \(A^{-1}\), and \(AB\) are also unitary: 1. We can prove \(A^H\) is unitary by showing that \((A^H)^{H}A^H = I\). Using properties of conjugate transposes and given that \(AA^H = I\), we successfully show that \((A^H)^H A^H = I\). 2. To prove \(A^{-1}\) is unitary, we need to show that \((A^{-1})^H A^{-1} = I\). Using properties of conjugate transposes and inverses of unitary matrices, we can prove that \((A^{-1})^H (A^{-1}) = I\). 3. To show that \(AB\) is unitary, we need to prove that \((AB)^H(AB) = I\). Using properties of conjugate transposes and given that \(AA^H = A^H A = I\) and \(BB^H = B^H B = I\), we successfully show that \(B^H A^H AB = I\) and \(AB B^H A^H = I\). Thus, we conclude that \(A^H\), \(A^{-1}\), and \(AB\) are also unitary.

Step by step solution

01

Show that \(A^H\) is unitary

We need to show that \((A^H A^H) = (A^H)^2 = I\). Now we know that \(AA^H = I\), and taking the conjugate transpose of both sides, we get: \((AA^H)^H = I^H\) \((A^H)^{H}A^H = I\) Since the conjugate transpose of the identity matrix is itself, we have: \((A^H)^{H}A^H = I\) Comparing this with the definition of a unitary matrix, we can conclude that \(A^H\) is unitary.
02

Show that \(A^{-1}\) is unitary

To show that \(A^{-1}\) is unitary, we need to prove that \((A^{-1})^H A^{-1} = A^{-1}(A^{-1})^H = I\). Using the definition of the inverse matrix, we know that \(AA^{-1} = A^{-1}A = I\). Taking the conjugate transpose of both sides: \((AA^{-1})^H = I^H\) \((A^{-1})^H A^H = I\) Going back to the definition of unitary matrices, since \(AA^H = I\), multiplying both sides by \(A^{-1}\): \(A^{-1}AA^H = A^{-1}I\) \(A^{-1}A^H = A^{-1}\) Now, we have: \((A^{-1})^H A^H = I\) \((A^{-1})^H A^{-1} = I\) Comparing this with the definition of a unitary matrix, we can conclude that \(A^{-1}\) is unitary.
03

Show that \(AB\) is unitary

To show that the product of matrices A and B, \(AB\), is unitary, we need to prove that \((AB)^H(AB) = (AB)(AB)^H = I\). Using the properties of the transpose conjugate, we know: \((AB)^H = B^H A^H\) Now, we must prove that: \(B^H A^H AB = I\) \(AB B^H A^H = I\) We know that \(AA^H = A^H A = I\) and \(BB^H = B^H B = I\), which means: \(B^H A^H AB = B^H (A^H A)B = B^H I B = BB^H B = (BB^H) B = I\) Similarly, \(AB B^H A^H = A(B B^H) A^H = A I A^H = AA^H = I\) Comparing these results with the definition of a unitary matrix, we conclude that the product of unitary matrices A and B, denoted as \(AB\), is also unitary.

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Most popular questions from this chapter

Let \(e_{i}=[0, \ldots, 0,1,0, \ldots, 0],\) where 1 is the ith entry. Show (a) \(e_{P} A=A_{i},\) ith row of \(A\). (b) \(\quad B e^{T}=B^{j}, j\) th column of \(B\). (c) If \(e_{i} A=e_{i} B,\) for each \(i,\) then \(A=B\). (d) \(\operatorname{If} A e_{j}^{T}=B e_{j}^{T},\) for each \(j,\) then \(A=B\).

Let \(\mathrm{V}=\mathrm{P}_{n}(F)\), and let \(c_{0}, c_{1}, \ldots, c_{n}\) be distinct scalars in \(F\). (a) For \(0 \leq i \leq n\), define \(\mathrm{f}_{i} \in \mathrm{V}^{*}\) by \(\mathrm{f}_{i}(p(x))=p\left(c_{i}\right)\). Prove that \(\left\\{\mathrm{f}_{0}, \mathrm{f}_{1}, \ldots, \mathrm{f}_{n}\right\\}\) is a basis for \(\mathrm{V}^{*}\). Hint: Apply any linear combination of this set that equals the zero transformation to \(p(x)=\) \(\left(x-c_{1}\right)\left(x-c_{2}\right) \cdots\left(x-c_{n}\right)\), and deduce that the first coefficient is zero. (b) Use the corollary to Theorem \(2.26\) and (a) to show that there exist unique polynomials \(p_{0}(x), p_{1}(x), \ldots, p_{n}(x)\) such that \(p_{i}\left(c_{j}\right)=\delta_{i j}\) for \(0 \leq i \leq n\). These polynomials are the Lagrange polynomials defined in Section 1.6. (c) For any scalars \(a_{0}, a_{1}, \ldots, a_{n}\) (not necessarily distinct), deduce that there exists a unique polynomial \(q(x)\) of degree at most \(n\) such that \(q\left(c_{i}\right)=a_{i}\) for \(0 \leq i \leq n .\) In fact, $$ q(x)=\sum_{i=0}^{n} a_{i} p_{i}(x) $$ (d) Deduce the Lagrange interpolation formula: $$ p(x)=\sum_{i=0}^{n} p\left(c_{i}\right) p_{i}(x) $$ for any \(p(x) \in \mathrm{V}\). (e) Prove that $$ \int_{a}^{b} p(t) d t=\sum_{i=0}^{n} p\left(c_{i}\right) d_{i} $$ where $$ d_{i}=\int_{a}^{b} p_{i}(t) d t $$ Suppose now that $$ c_{i}=a+\frac{i(b-a)}{n} \text { for } i=0,1, \ldots, n \text {. } $$ For \(n=1\), the preceding result yields the trapezoidal rule for evaluating the definite integral of a polynomial. For \(n=2\), this result yields Simpson's rule for evaluating the definite integral of a polynomial.

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\\}$

Let \(A\) and \(B\) be \(n \times n\) invertible matrices. Prove that \(A B\) is invertible and \((A B)^{-1}=B^{-1} A^{-1}\).

Find a basis for the solution space of each of the following differential equations. (a) \(y^{\prime \prime}+2 y^{\prime}+y=0\) (b) \(y^{\prime \prime \prime}=y^{\prime}\) (c) \(y^{(4)}-2 y^{(2)}+y=0\) (d) \(y^{\prime \prime}+2 y^{\prime}+y=0\) (e) \(y^{(3)}-y^{(2)}+3 y^{(1)}+5 y=0\)
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