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Josiah's Math Notes

Linear Algebra

Topics

• Difference Matrix
\[ \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \end{bmatrix} \]

The inverse of the difference matrix is the sum matrix.

\[ x(t) = t^{2} \]

Types of differences:

Forward difference

\[ f'(x_{0}) = \frac{f(x_{1}) - f(x_{0})}{x_{1} - x_{0}} \] \[ x(t+1) - x(t) = (t+1)^{2} - t^{2} = (t^{2}+2t+1) - t^{2} = 2t + 1 \]

Backward difference

\[ f'(x_{0}) = \frac{f(x_{0}) - f(x_{-1})}{x_{0} - x_{-1}} \] \[ x(t) - x(t-1) = t^{2} - (t-1)^{2} = t^{2} - (t^{2}-2t+1) = 2t - 1 \]

Centered difference

\[ f'(x_{0}) = \frac{f(x_{1}) - f(x_{-1})}{x_{1} - x_{-1}} \] \[ \frac{x(t+1) - x(t-1)}{2} = \frac{(t+1)^{2} - (t-1)^{2}}{2} = \frac{(t^{2}+2t+1) - (t^{2}-2t+1)}{2} = \frac{4t}{2} = 2t \]
• Sum Matrix



• Miscellaneous

Skew-Hermitian Matrix

\[ \textbf{A} = -\textbf{A}^{H} \]



• Miscellaneous Theorems

Theorem 1

\[ \det(\textbf{Q}) = \pm1 \] where \(\textbf{Q}\) is an orthogonal matrix.

Proof:

Suppose \(\textbf{Q}\) is an orthogonal matrix.

\[ \begin{align} 1 &= \det(\textbf{I}) \\ &= \det(\textbf{Q}\textbf{Q}^{-1}) \\ &= \det(\textbf{Q})\det(\textbf{Q}^{-1}) \\ &= \det(\textbf{Q})\det(\textbf{Q}^{T}) \\ &= \det(\textbf{Q})\det(\textbf{Q}) \\ &= [\det(\textbf{Q})]^{2} \end{align} \]

We have

\[ [\det(\textbf{Q})]^{2} = 1 \]

Therefore

\[ \det(\textbf{Q}) = \pm \sqrt{1} = \pm 1 \]

Q.E.D.





The previous theorem leads to the next fact:

Theorem 2

Given an eigenvalue \(\lambda\) of an orthogonal matrix,

\[ |\lambda| = 1 \]



Theorem 3

A matrix of pure imaginary entries is skew symmetric if, and only if it is an Hermitian matrix.

Proof:

Suppose \(\textbf{A}\) is a skew-symmetric matrix of pure imaginary entries.

Notice that when we conjugate, we get

\[ \overline{\textbf{A}} = -\textbf{A} \]

Now

\[ \textbf{A} = -\textbf{A}^{T} \] \[ \textbf{A} = \overline{\textbf{A}^{T}} \] \[ \textbf{A} = \textbf{A}^{H} \]

Thus \(\textbf{A}\) is a Hermitian matrix.

Q.E.D.





Property

All main diagonal entries of a skew-symmetric matrix are zero. Why? Skew symmetric matrices must satisfy \(\textbf{A}^{T} = -\textbf{A}\). Suppose the main diagonal entries are \(b_{1}, b_{2}, \ldots, b_{n}\). When taking the transpose of a matrix, then entries along the main diagonal remain the same. That means if a matrix is skew-symmetric, the entries along the diagonal \(b_{1}, b_{2}, \ldots, b_{n}\) must satisfy \(b_{i} = -b_{i}\). The only solution to this equation is \(b_{i}=0\).