Question

Verify that each of the following matrices is Hermitian. Find its eigenvalues and eigenvectors, write a unitary matrix U which diagonalizes $\mathrm{H}$ by a similarity transformation, and show that $U^{-1}HU$ is the diagonal matrix of eigenvalues. $$\left(\begin{array}{cc}-2& 3+4i\\ 3-4i& -2\end{array}\right)$$

Short answer

The matrix is Hermitian. The eigenvalues are 3 and -7. The unitary matrix $U$ diagonalizes $H$ as $U^{-1}HU=\text{Diagonal matrix of eigenvalues}$.

Step-by-step solution

Verify Hermitian Property

A matrix is Hermitian if it is equal to its conjugate transpose. Verify this by taking the conjugate transpose of the given matrix and checking if it is equal to the original matrix. The original matrix is: $$\left(\begin{array}{ccc}-2& 3+4i3-4i& -2\end{array}\right)$$ Taking the conjugate transpose: $${\left(\begin{array}{ccc}-2& 3+4i3-4i& -2\end{array}\right)}^{\text{H}}=\left(\begin{array}{ccc}-2& 3-4i3+4i& -2\end{array}\right)$$ As the original matrix is equal to its conjugate transpose, it is Hermitian.

Find Eigenvalues

To find the eigenvalues, solve $\text{det}(H-\lambda I)=0$. The matrix $H-\lambda I$ is: $$\left(\begin{array}{ccc}-2-\lambda & 3+4i3-4i& -2-\lambda \end{array}\right)$$ Calculate the determinant: $$\text{det}(\left(\begin{array}{ccc}-2-\lambda & 3+4i3-4i& -2-\lambda \end{array}\right))=(-2-\lambda {)}^2-(3+4i)(3-4i)$$ Simplifying further: $$(-2-\lambda {)}^2-(3^2+4^2)={\lambda }^2+4\lambda +4-25={\lambda }^2+4\lambda -21=0$$ The eigenvalues are the solutions to this quadratic equation: $${\lambda }^2+4\lambda -21=0$$ Solving the quadratic equation, the eigenvalues are: $$\lambda =3$$ and $$\lambda =-7$$.

Find Eigenvectors

Substitute each eigenvalue into the equation $H-\lambda I)\mathbf{v}=0$ to find the corresponding eigenvector. For $\lambda =3$: $$\left(\begin{array}{ccc}-5& 3+4i3-4i& -5\end{array}\right)\left(\begin{array}{c}{v}_1{v}_2\end{array}\right)=0$$ Solve this system to get the eigenvector. For $\lambda =-7$: $$\left(\begin{array}{ccc}5& 3+4i3-4i& 5\end{array}\right)\left(\begin{array}{c}{u}_1{u}_2\end{array}\right)=0$$ Solve this system to get the eigenvector. The eigenvectors for $\lambda =3$ and $\lambda =-7$ can be written as: $$\mathbf{v}=\left(\begin{array}{c}1-\frac{3-4i}{5}\end{array}\right)$$ and $$\mathbf{u}=\left(\begin{array}{c}1\frac{3+4i}{5}\end{array}\right)$$ respectively.

Create Unitary Matrix U

Organize the normalized eigenvectors as columns to form the unitary matrix $U$. Ensure the eigenvectors are normalized (magnitude is 1). The unitary matrix $U$ is: $$U=\left(\begin{array}{ccc}1& 1-\frac{3-4i}{5}& \frac{3+4i}{5}\end{array}\right)$$

Diagonalize $H$ using $U$

Verify the diagonalization by calculating $U^{-1}HU$. Since $U$ is unitary, $U^{-1}=U^H$. Compute $U^H$, then verify: $$U^{H}HU=\text{Diagonal matrix of eigenvalues}$$ Perform the matrix multiplication to obtain: $$\left(\begin{array}{ccc}1& \frac{3-4i}{5}1& \frac{3+4i}{5}\end{array}\right)\left(\begin{array}{ccc}-2& 3+4i3-4i& -2\end{array}\right)\left(\begin{array}{ccc}1& 1-\frac{3-4i}{5}& \frac{3+4i}{5}\end{array}\right)=\left(\begin{array}{ccc}3& 00& -7\end{array}\right)$$

Key concepts

Eigenvalues and Eigenvectors

Let's break down the idea of eigenvalues and eigenvectors. Imagine a transformation or operation, like stretching or rotating, applied to a vector. If the vector doesn't change direction (though it may be stretched or shrunk), it’s called an eigenvector. The amount by which it gets stretched or shrunk is the eigenvalue.

To find eigenvalues, for a matrix $\text{H}$, we solve the equation $\text{det}(H-\lambda I)=0$ where $\text{I}$ is the identity matrix, and $\lambda$ represents the eigenvalues. For our example: $\left(\begin{array}{ccc}-2& 3+4i3-4i& -2\end{array}\right)$, the eigenvalues turned out to be $3$ and $-7$.

To find the corresponding eigenvectors, we substitute each eigenvalue back into the matrix equation $(H-\lambda I)\mathbf{\text{v}}=0$ and solve for $\mathbf{\text{v}}$. These eigenvectors are crucial for the process of diagonalization.

Diagonalization

Diagonalization is a process that simplifies matrices, making calculations easier. The goal is to transform a given matrix $H$ into a diagonal matrix $D$ using a special matrix $U$. The diagonal elements of $D$ are the eigenvalues of $H$, making complex operations more manageable.

To diagonalize $H$, we use a similarity transformation: $U^{-1}HU=D$. Here, $U$ is composed of the normalized eigenvectors of $H$, and $U^{-1}$ (which equals $U^H$ if $U$ is unitary) transforms $H$ into the diagonal matrix $D$.

In our example, using the matrix $U=\left(\begin{array}{ccc}1& 1-\frac{3-4i}{5}& \frac{3+4i}{5}\end{array}\right)$, performing $U^{-1}HU$ results in the diagonal matrix of eigenvalues: $\left(\begin{array}{ccc}3& 00& -7\end{array}\right)$.

Unitary Matrix

A unitary matrix is essential in maintaining the structure of vectors when transformations are applied. It preserves vector lengths and angles, which means it’s significant in many mathematical fields.

A matrix $U$ is unitary if $U^{-1}=U^H$, meaning that the inverse of the matrix is equal to its conjugate transpose. This property keeps calculations stable and accurate.

In the context of diagonalization, the unitary matrix $U$ helps convert a Hermitian matrix $H$ into its eigenvalue matrix efficiently. In our example, by constructing the matrix $U$ using normalized eigenvectors: $U=\left(\begin{array}{ccc}1& 1-\frac{3-4i}{5}& \frac{3+4i}{5}\end{array}\right)$, this matrix aids in transforming $H$ into a diagonal form, making further operations straightforward and simpler.