# Abstract Nonsense

## Interesting Linear Algebra and Matrix Analysis Problems

Point of post: In this post I will just do some neat problems from basic linear algebra and matrix analysis. Feel free to comment/critique any.

NOTE: It would be very appreciated if anyone could give me some more linear algebra problems to do.

1.

Problem: Let $\mathscr{V}$ be a $n$-dimensional $F$-space and denote it’s scalar multiplication by $\ast_1$ and $K$ a subfield of $F$. Then:

a) $F$ is a vector space over $K$ with usual addition and multiplication.

b) If $\dim_K F=m$ then $\mathscr{V}$ is a vector space over $K$ with multiplication $\ast_2=\left(\ast_1\right)_{\mid K\times\mathscr{V}}$ and $\dim_K\mathscr{V}=mn$

Proof:

a) This is immediate from the field axioms.

b) The fact that $\mathscr{V}$ over $K$ forms a vector space is clear. Now, to prove the claim about the dimensionality let $\{\beta_1,\cdots,\beta_m\}$ be a basis for $F$ over $K$ and $\{x_1,\cdots,x_n\}$ a basis for $\mathscr{V}$ over $F$. We claim that $\{\beta_1x_1,\cdots,\beta_m x_1,\cdots,\beta_1x_n,\cdots,\beta_mx_n\}$ is a basis for $\mathscr{V}$ over $K$. To see this, first suppose that $\alpha_{1,1},\cdots,\alpha_{m,n}\in K$ were such that

$\displaystyle \displaystyle \sum_{j=1}^{n}\sum_{i=1}^{m}\alpha_{i,j}\beta_ix_j=\sum_{j=1}^{n}\left(\sum_{i=1}^{m}\beta_i\alpha_{i,j}\right)=\bold{0}$

Then, by the linear independence of $\{x_1,\cdots,x_n\}$ we see that $\displaystyle \sum_{i=1}^{m}\alpha_{i,1}\beta_i=\cdots=\sum_{i=1}^{m}\alpha_{i,n}\beta_i=0$. But, since $\{\beta_1,\cdots,\beta_m\}$ are are linearly independent over $K$ we see that each of the $m$ sums above implies that $\alpha_{1,j}=\cdots=\alpha_{m,j}=0\text{ }j=1,\cdots,n$ from where the linear independence of $\{\alpha_{1,1},\cdots,\alpha_{m,n}\}$ follows.

To see that $\text{span}_K\text{ }\{\beta_1x_1,\cdots,\beta_mx_1,\cdots,\beta_1x_n,\cdots,\beta_mx_n\}=\mathscr{V}$ we may merely notice that since $\text{span}_{F}\text{ }\{x_1,\cdots,x_n\}=\mathscr{V}$ for any $x\in\mathscr{V}$ there exists $\gamma_1,\cdots,\gamma_n\in F$ such that $\displaystyle x=\sum_{j=1}^{n}\gamma_j x_j$. But, since $\text{span}_K\text{ }\{\beta_1,\cdots,\beta_m\}=F$ we know that $\displaystyle \gamma_j=\sum_{i=1}^{m}\alpha_{i,j}\beta_i,\text{ }j=1,\cdots,n$ where $\alpha_{i,j}\in K,\text{ }i=1,\cdots,m\text{ }j=1,\cdots,n$. So, then upon insertion and expansion one arrives at

$\displaystyle x=\sum_{i\in[m]\text{ and }j\in[n]}\alpha_{i,j}\beta_ix_j$

where $[m]=\{1,\cdots,m\}$ and $[n]=\{1,\cdots,n\}$

2.

Problem: Let $S\subseteq\mathbb{R}[x]$. Show that $e^x\notin\text{span }S$

Proof: This is equivalent to asking why $e^x$ is not equal to a polynomial. But, this is clear since if $p(x)$ is any polynomial then $D^{\text{deg }p+1}p=\bold{0}$ but $D^{\text{deg }p+1}e^x=e^x$ (where $D$ is the differential operator).

3.

Problem: Let $F$ be a field. Show then that if one considers $F$ as a vector space over itself that there does not exist proper subspaces $W_1$ and $W_2$ such that $F=\mathscr{W}_1\oplus \mathscr{W}_2$.

Proof: The obvious way is to note that if $F=\mathscr{W}_1\oplus \mathscr{W}_2$ that $1=\dim_F F=\dim_F \mathscr{W}_1+\dim_F \mathscr{W}_2$, and clearly this implies that, up to relabeling, $\dim_F \mathscr{W}_1=1$ and $\dim_F \mathscr{W}_2=0$. Thus, since $\mathscr{W}_1$ is a $1$-dimensional subspace of $F$ (which is itself $1$-dimensional) we may conclude that $\mathscr{W}_1=F$. Similarly, since $\dim_F \mathscr{W}_2=0$ we may conclude that $\mathscr{W}_2=\{\bold{0}\}$.

4.

Problem: Show that if $\mathbb{R}^n$ is endowed with the usual topology and $\mathscr{S}$ is a subspace of $\mathbb{R}^n$ (with the usual operations) then $\mathscr{S}$ is closed.

5.

Problem: Let $f:\mathbb{R}^n\to\mathbb{R}$ be additive ($f(x+y)=f(x)+f(y)$) and continuous.. Prove that $f$ is a linear functional.

Proof: First notice that it follows by induction that

$f\left((mx_1,\cdots,mx_n)\right)=m f\left((x_1,\cdots,x_n)\right)$

for all $m\in\mathbb{N}$. Furthermore, if $m\in\mathbb{N}$ we see that

$f\left((x_1,\cdots,x_n)\right)=f\left(\left(\frac{1}{m}mx_1,\cdots,\frac{1}{m}mx_n\right)\right)=m f\left(\left(\frac{1}{m}x_1,\cdots,\frac{1}{m}x_n\right)\right)$

Thus,

$\frac{1}{m} f\left((x_1,\cdots,x_n)\right)=f\left(\left(\frac{1}{m}x_1,\cdots,\frac{1}{m}x_n\right)\right)$

So, by combining these two we see that if $\frac{p}{q}\in\mathbb{Q}$ that

$f\left(\left(\frac{p}{q}x_1,\cdots,\frac{p}{q}x_n\right)\right)=\frac{p}{q}f\left((x_1,\cdots,x_n)\right)$

Thus, if  $\left(q_1,\cdots,q_n\right)\in\mathbb{Q}^n$  we see that

\begin{aligned}f\left((q_1,\cdots,q_n)\right) &=f\left((q_1,\cdots,0)+\cdots+(0,\cdots,q_n)\right)\\ &=f((q_1,\cdots,0))+\cdots+f((0,\cdots,q_n))\\ &=q_1f((1,\cdots,0))+\cdots+q_nf((0,\cdots,1))\end{aligned}

Thus, if

$g((x_1,\cdots,x_n)=x_1f(1,\cdots,0)+\cdots+x_nf(0,\cdots,1)$

we see that $f_{\mid\mathbb{Q}^n}=g_{\mid\mathbb{Q}^n}$. But, this means that $f$ and $g$ are continuous maps which agree on a dense subset of the domain, which means that they must agree on all of the domain. Namely, $f=g$. But, $g$ is evidently a linear functional and so the conclusion follows.

6.

Problem: Let $A\in\text{Mat}_{n\times n}$ be both normal and nilpotent. Then $A=[0]$

Proof: By the spectral theorem we have that

$\displaystyle A=U\text{ diag}(\lambda_1,\cdots,\lambda_n)U^{*}$

for some unitary matrix $U$. Thus,

$A^*=U\text{ diag}(\overline{\lambda_1},\cdots,\overline{\lambda_n})U^{*}\text{ }(1)$

Thus,

$AA^{*}=U\text{ diag }\left(|\lambda_1|^2,\cdots,|\lambda_n|^2\right)U^{*}$

and so

$\left(AA^{*}\right)^k=U\text{diag }(|\lambda_1|^{2k},\cdots,|\lambda_n|^{2k})U^{*}$

But, using normality and nilpotence we see that

$\left(AA^{*}\right)^k=\left(A^k\right)\left(A^*\right)^k=[0]$

so, upon comparision we find that

$\text{diag}\left(|\lambda_1|^{2k},\cdots,|\lambda_n|^{2k}|\right)=[0]$

and so $\lambda_1=\cdots=\lambda_n=0$. Thus,  appealing to $(1)$ we find that $A=[0]$

7.

Problem: Let $A\in\text{Mat}_{n\times n}$ be such that $A^k=[0]$ for some $k>n$. Show that $A^r=[0]$ for some $r\leqslant n$.

Proof: Note that if $\lambda\in\sigma\left(A\right)$ (where $\sigma$ is the spectrum of a matrix) with associated eigenvector $x$, then

$0=A^kx=A^{k-1}\lambda x=\cdots=\lambda^kx$

and since $x\ne0$ we may conclude that $\lambda^k=0\implies \lambda=0$. Thus, all the eigenvalues of $A$ are $0$. But, then we know that

$A=S\begin{bmatrix} J_{n_1}(0) & \cdots & 0\\ \vdots & \ddots & \vdots \\ 0 & \cdots & J_{n_k}(0)\end{bmatrix}S^{-1}$

where

$J_{n_\ell}(0)=\underbrace{\begin{bmatrix}0 & 1 &\cdots & 0\\ \vdots & \ddots & \ddots &\vdots\\ 0 & \cdots &\ddots&1\\ 0 &\cdots &0 &0\end{bmatrix}}_{n_{\ell}\times n_{\ell}}$

is the Jordan block associated with the particular eigenvalue. But, since $n_1+\cdots+n_k=n$ we evidently have that $\displaystyle \max_{1\leqslant j\leqslant k}n_j\leqslant n$. But, evidently if $\displaystyle N=\max_{1\leqslant j\leqslant k}n_j$ we have that

$J_{n_k}(0)^N=[0]$

and so

$A^N=S\begin{bmatrix} J_{n_1}(0)^N & \cdots & 0\\ \vdots & \ddots & \vdots \\ 0 & \cdots & J_{n_k}(0)^N\end{bmatrix}S^{-1}=S[0]S^{-1}=[0]$

which is what was to be proven.

8.

Problem: What are the possible Jordan canonical forms for $A\in\text{Mat}_{n\times n}$ given that $A^m=I,\text{ }m\geqslant 2$?

Proof: We know that

$A=SJS^{-1}$

for some Jordan matrix $J$. Thus,

$I=A^{m}=SJ^{m}S^{-1}\implies J^m=I$

But, notice that if

$\displaystyle J=\begin{bmatrix} J_{n_1}(\lambda_1) & \cdots &0\\ \vdots & \ddots &\vdots\\ 0 &\cdots &J_{n_k}\left(\lambda_k\right)\end{bmatrix}$

that

$J^m=\begin{bmatrix} \left[J_{n_1}(\lambda_1)\right]^m & \cdots & 0\\ \vdots & \ddots &\vdots\\ 0 & \cdots & \left[J_{n_k}(\lambda_k)\right]^m\end{bmatrix}$

And thus, for $J^m=I$ it’s evident that $\left[J_{n_\ell}\left(\lambda_\ell\right)\right]^m=I,\text{ }\ell=1,\cdots,k$. But, if $n_{\ell}>0$ it easily follows that $\left[J_{n_{\ell}}(\lambda_{\ell})\right]^m\ne I$ (just consider induction on $n_{\ell}$ and partition the matrix into smaller ones). Thus, it follows that $n_{\ell}=1,\text{ }\ell=1,\cdots k$ so that $J_{n_{\ell}}(\lambda_\ell)=\lambda_\ell$. Thus, we may conclude that

$A=S\begin{bmatrix}\ell_1 & \cdots &0\\ \vdots &\ddots & \vdots\\ 0 &\cdots &\ell_{k}\end{bmatrix}S^{-1}$

where $\lambda_1,\cdots\lambda_k$ are eigenvalues of $A$, and thus $m$-th roots of unity.

References:

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

2. Brown, William C. A Second Course in Linear Algebra. New York: Wiley, 1988. Print.

3.  Horn, Roger A., and Charles R. Johnson. Matrix Analysis. Cambridge [u.a.: Cambridge Univ., 2009. Print.

4.  Roman, Steven. Advanced Linear Algebra. New York: Springer, 2005. Print.