# Abstract Nonsense

## Halmos Sections 37 and 38: Matrices and Matrices of Linear Transformations(Pt. V)

Point of post: This is a continuation of this post.

18.

Problem: For which values of $\alpha$ are the following matrices invertible? Find the inverses whenever possible.

a) $\begin{pmatrix}1 & \alpha & 0\\ \alpha & 1 & \alpha\\ 0 & \alpha & 1\end{pmatrix}$

b) $\begin{pmatrix}\alpha & 1 & 0\\ 1 & \alpha & 1\\ 0 & 1 & \alpha\end{pmatrix}$

c) $\begin{pmatrix}0 & 1 & \alpha\\ 1 & \alpha & 0\\ \alpha & 0 & 1\end{pmatrix}$

d) $\begin{pmatrix}1 & 1 & 1\\ 1 & 1 & \alpha\\ 1 & \alpha & 1\end{pmatrix}$

Proof:

a) This is invertible precisely when $\displaystyle \alpha\ne\pm\frac{\sqrt{2}}{2}$. Indeed, if $\displaystyle \alpha=\pm\frac{\sqrt{2}}{2}$ then $\begin{pmatrix}1 & \alpha & 0\\ \alpha & 1 & \alpha\\ 0 & \alpha & 1\end{pmatrix}\begin{pmatrix}1\\ \mp\sqrt{2}\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$. And, if $\alpha\ne\frac{\pm\sqrt{2}}{2}$ then $\displaystyle \frac{1}{1-2\alpha^2}\begin{pmatrix}\alpha^2-1 & \alpha & -\alpha^2\\ \alpha & -1 & \alpha\\ -\alpha^2 & \alpha & \alpha^2-1\end{pmatrix}$ serves as an inverse.

b) This is invertible precisely when $x\ne\pm\sqrt{2},0$.  Indeed, if $\alpha=\pm\sqrt{2}$ then $\begin{pmatrix}\alpha & 1 & 0\\ 1 & \alpha & 1\\ 0 & 1 & \alpha\end{pmatrix}\begin{pmatrix}1\\ \mp\sqrt{2}\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$. And, if $\alpha=0$ then $\begin{pmatrix}\alpha & 1 & 0\\ 1 & \alpha & 1\\ 0 & 1 & \alpha\end{pmatrix}\begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$. If $\alpha\ne\pm\sqrt{2},0$ then $\displaystyle \frac{1}{\alpha^2-2}\begin{pmatrix}-\frac{1-\alpha^2}{\alpha} & -1 & \frac{1}{\alpha}\\ -1 & \alpha & -1\\ \frac{1}{\alpha} & -1 & -\frac{1-\alpha^2}{\alpha}\end{pmatrix}$ serves as an inverse.

c) Assuming we’re discussing real matrices this is true precisely when $\alpha \ne -1$. Indeed, if $\alpha=-1$ then $\begin{pmatrix}0 & 1 & \alpha\\ 1 & \alpha & 0\\ \alpha & 0 & 1\end{pmatrix}\begin{pmatrix}1\\ 1\\ 1\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$. If $\alpha\ne -1$ then $\displaystyle \frac{1}{\alpha^3+1}\begin{pmatrix}-\alpha & 1 & \alpha^2\\ 1 & \alpha^2 & \alpha\\ \alpha^2 & -\alpha & 1\end{pmatrix}$ serves as an inverse.

d) This is invertible precisely when $\alpha\ne 1$. Indeed, if $\alpha=1$ then $\begin{pmatrix}1 & 1 & 1\\ 1 & 1 & \alpha\\ 1 & \alpha & 1\end{pmatrix}\begin{pmatrix}-1\\ 1\\ 0\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$. Otherwise $\begin{pmatrix}\frac{\alpha+1}{\alpha-1} & \frac{1}{1-\alpha} & \frac{1}{1-\alpha}\\ \frac{1}{1-\alpha} & 0 & \frac{1}{\alpha-1}\\ \frac{1}{1-\alpha} & \frac{1}{\alpha-1} & 0\end{pmatrix}$ serves as an inverse.

19.

Problem:

a) It is easy to extend matrix theory to linear transformations between different vector spaces. Suppose that $\mathscr{U}$ and $\mathscr{V}$ are vector spaces over the same field $F$. Let $\{x_1,\cdots,x_n\}$ and $\{y_1,\cdots,y_m\}$ be bases for $\mathscr{U}$ and $\mathscr{V}$ respectively, and $A\in\text{Hom}\left(\mathscr{U},\mathscr{V}\right)$. The matrix of $A$ is, by definition, the rectangular $m\times n$ array of scalars defined by

$\displaystyle A(x_j)=\sum_{i=1}^{m}\alpha_{i,j}y_i$

Define addition and multiplication of rectangular matrices so as to generalize as many possible of the results of section 38.

b) Suppose that $A$ and $B$ are multipliable matrices. Partition $A$ into four rectangular blocks (top left, top right, bottom left, bottom right) and then partition $B$ similarly so that the number of columns in the top left part of $A$ is the same as the number of rows in the top left part of $B$. If, in an obvious shorthand, these partitioned matrices are indicated by

$A=\begin{pmatrix}A_{1,1} & A_{1,2}\\ A_{2,1} & A_{2,2}\end{pmatrix}$

and

$B=\begin{pmatrix}B_{1,1} & B_{1,2}\\ B_{2,1} & B_{2,2}\end{pmatrix}$

Then,

$AB=\begin{pmatrix}A_{1,1}B_{1,1}+A_{1,2}B_{2,1} & A_{1,1}B_{1,2}+A_{1,2}B_{2,2}\\ A_{2,1}B_{1,1}+A_{2,2}B_{2,1} & A_{2,1}B_{1,2}+A_{2,2}B_{2,2}\end{pmatrix}$

c) Use subspaces and complements to express the result of b) in terms of linear transformations (instead of matrices).

Proof:

a) We may define addition of rectangular matrices exactly the same. For multiplication, we may only (in any normal sense) define matrix multiplication of rectangular matrices $A=[\alpha_{i,j}]$ and $B=[\beta_{i,j}]$ when the size of them are $m\times p$ and $p\times n$ respectively. From there we may define the multiplication $AB$ to be the $m\times n$ matrix whose general term is, unsurprisingly

$\displaystyle \sum_{r=1}^{p}\alpha_{i,p}\beta_{p,j}$

One can check that with this definition all the axioms of an associative non-unital algebra are held.

b) This is just tedious computation, if someone has a dying desire for me to upload this, let me know.

c) Suppose that $\mathscr{V}=\text{span}\{x_1,\cdots,x_k\}\oplus\text{span}\{x_{k+1},\cdots,x_n\}$.  Then,  for some $T\in\text{End}\left(\mathscr{V}\right)$  we can compute $[T]_{\mathcal{B}}$, where $\mathcal{B}=(x_1,\cdots,x_k,x_{k+1},\cdots,x_{n})$ by thinking of $\displaystyle T(x_j)=\sum_{i=1}^{k}\alpha_{i,j}x_i+\sum_{i=k+1}^{n}\alpha_{i,j}x_i$ and $\displaystyle T(x_i)=\sum_{r=1}^{k}\beta_{r,i}x_r+\sum_{r=k+1}^{n}\beta_{r,i}x_r$ we see then that

\displaystyle \begin{aligned}T'\left(T(x_j)\right) &= T'\left(\sum_{i=1}^{k}\alpha_{i,j}x_i+\sum_{i=k+1}^{n}\alpha_{i,j}x_i\right)\\ &= \sum_{i=1}^{n}\alpha_{i,j}\left(\sum_{r=1}^{k}\beta_{r,i}x_r+\sum_{r=k+1}^{n}\beta_{r,i}x_r\right)+\sum_{i=k+1}^{n}\alpha_{i,j}\left(\sum_{r=1}^{k}\beta_{i,r}x_r+\sum_{i=k+1}^{n}\beta_{i,r}x_r\right)\\ &=\sum_{i=1}^{k}\sum_{r=1}^{k}\alpha_{i,j}\beta_{i,r}x_r+\sum_{i=1}^{k}\sum_{r=k+1}^{n}\alpha_{i,j}\beta_{r,i}x_r+\sum_{i=k+1}^{n}\sum_{r=1}^{k}\alpha_{i,j}\beta_{r,i}x_r+\sum_{i=k+1}^{n}\sum_{r=k+1}^{n}\alpha_{i,j}\beta_{r,i}x_r\end{aligned}

from where, with the proper definitions of $A_{i,j},B_{i,j},\text{ }i,j\in[2]$ the result follows.

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print