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Fundamental notions of vector space theory, linear independence, basis, span, scalar product, orthogonal bases. Includes a survey of matrix algebra, solution of systems linear equations, rank, kernel, eigenvalues and eigenvectors, the LU- and QR-factorizations, and least squares approximation. Selected applications in mathematics, science, engineering and business. Prereq: MATH 426. (Not offered for credit if credit is received for MATH 545 or MATH 762.)
Given columns $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$, and $\mathbf{a}_4$ in $\mathbb{R}^3$, write $2 \mathbf{a}_1-3 \mathbf{a}_2+5 \mathbf{a}_3+\mathbf{a}_4$ in the form $A \mathbf{x}$ where $A$ is a matrix and $\mathbf{x}$ is a vector.
Solution. Here the column of coefficients is $\mathbf{x}=\left[\begin{array}{r}2 \ -3 \ 5 \ 1\end{array}\right]$. Hence Definition $2.5$ gives
$$
A \mathbf{x}=2 \mathbf{a}_1-3 \mathbf{a}_2+5 \mathbf{a}_3+\mathbf{a}_4
$$
where $A=\left[\begin{array}{llll}\mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3 & \mathbf{a}_4\end{array}\right]$ is the matrix with $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$, and $\mathbf{a}_4$ as its columns.
If $A$ is any matrix, then $I A=A$ and $A I=A$, and where $I$ denotes an identity matrix of a size so that the multiplications are defined.
Solution. These both follow from the dot product rule as the reader should verify. For a more formal proof, write $A=\left[\begin{array}{llll}\mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n\end{array}\right]$ where $\mathbf{a}_j$ is column $j$ of $A$. Then Definition $2.9$ and give
$$
I A=\left[\begin{array}{llll}
I \mathbf{a}_1 & I \mathbf{a}_2 & \cdots & I \mathbf{a}_n
\end{array}\right]=\left[\begin{array}{llll}
\mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n
\end{array}\right]=A
$$
If $\mathbf{e}_j$ denotes column $j$ of $I$, then $A \mathbf{e}_j=\mathbf{a}_j$ for each $j$ by Example Hence gives:
$$
A I=A\left[\begin{array}{llll}
\mathbf{e}_1 & \mathbf{e}_2 & \cdots & \mathbf{e}_n
\end{array}\right]=\left[\begin{array}{llll}
A \mathbf{e}_1 & A \mathbf{e}_2 & \cdots & A \mathbf{e}_n
\end{array}\right]=\left[\begin{array}{llll}
\mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n
\end{array}\right]=A
$$
If $A$ is an invertible matrix, show that the transpose $A^T$ is also invertible. Show further that the inverse of $A^T$ is just the transpose of $A^{-1}$; in symbols, $\left(A^T\right)^{-1}=\left(A^{-1}\right)^T$.
Solution. $A^{-1}$ exists (by assumption). Its transpose $\left(A^{-1}\right)^T$ is the candidate proposed for the inverse of $A^T$. Using the inverse criterion, we test it as follows:
$$
\begin{aligned}
A^T\left(A^{-1}\right)^T & =\left(A^{-1} A\right)^T=I^T=I \
\left(A^{-1}\right)^T A^T & =\left(A A^{-1}\right)^T=I^T=I
\end{aligned}
$$
Hence $\left(A^{-1}\right)^T$ is indeed the inverse of $A^T$; that is, $\left(A^T\right)^{-1}=\left(A^{-1}\right)^T$.
