Abstract Nonsense

Crushing one theorem at a time

Halmos Sections 34 and 35:Products and Polynomials (Pt. II)

Point of post: This post is a continuation of


Problem: Let \mathscr{V} be an n-dimensional F-space and A\in\text{End}\left(\mathscr{V}\right).

a) Prove that the set of all B\in\text{End}\left(\mathscr{V}\right) for which AB=\bold{0} is a subspace of \text{End}\left(\mathscr{V}\right)

b) Show that by a suitable choice of A the dimension of the subspace described in a) can be 0,n or n^2.

c) Can every subspace of \text{End}\left(\mathscr{V}\right) be gotten this way?


a) We must merely note that if B,B' both annihilate A, then for any \alpha,\beta\in F we have that A\left(\alpha B+\beta B'\right)=\alpha AB+\beta AB'=\bold{0}. Thus, it follows that the set of all such B is a subspace of \text{End}\left(\mathscr{V}\right).

b) Let T_0\in\text{End}\left(\mathscr{V}\right). It’s fairly easy to show that \left\{v\in\mathscr{V}:T(v)=\bold{0}\right\} is a subspace of \mathscr{V}. Thus, we may find a basis \{x_1,\cdots,x_m\} for it and extend this to a basis \{x_1,\cdots,x_m,x_{m+1},\cdots,x_n\} for \mathscr{V}. We claim that

\mathscr{B}=\left\{T_{i,j}:i\in[n]\text{ and }j\in[m]\right\}

is a basis for \mathscr{G}=\left\{T\in\text{End}\left(\mathscr{V}\right):T_0T=\bold{0}\right\}. To see this we note that evidently \mathscr{B} is linearly independent and \text{span }\mathscr{B}\subseteq\mathscr{G}, so it remains to show the reverse inclusion. To do this, suppose that T\notin\text{span }\mathscr{B} but T\in\mathscr{G}. We note that since T\notin\text{span }\mathscr{B} that

\displaystyle T=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha_{i,j}T_{i,j}

where there is some i_0\in[n]\text{ and }j_0\notin[m] such that \alpha_{i,j}\ne 0.  Note then that

\displaystyle T(x_{i_0})=\sum_{j=1}^{n}\alpha_{i_0,j}T_{i_0,j}(x_{i_0})=\sum_{j=1}^{n}\alpha_{i_0,j}x_j

and thus by assumption that T\in\mathscr{G} we see that

\displaystyle 0=T_0\left(T(x_{i_0})\right)=\sum_{j=1}^{n}\alpha_{i_0,j}T_0(x_j)=\sum_{j=m+1}^{n}\alpha_{i_0,j}T_0(x_j)=T\left(\sum_{j=m+1}^n x_j\right)

and thus

\displaystyle \sum_{j=m+1}^{n}\alpha_{i_0,j}x_j\in\ker T

but by construction this is only possible if

\displaystyle \sum_{j=m+1}^{n}\alpha_{i_0,j}x_j=\bold{0}

and since \{x_{m+1},\cdots,x_n\} is a linearly independent set this implies that \alpha_{i_0,m+1}=\cdots=\alpha_{i_0,n}=0 and so in particular \alpha_{i_0,j_0}=0 which is contradictory to our assumption. Thus, we see that if T\notin\text{span }\mathscr{B} then T\notin\mathscr{G} so that

\text{End}\left(\mathscr{V}\right)-\text{span }\mathscr{B}\subseteq\text{End}\left(\mathscr{V}\right)-\mathscr{G}

which is equivalent to

\text{span }\mathscr{B}\supseteq\mathscr{G}

from where it follows that \text{span }\mathscr{B}=\mathscr{G}. Therefore


From this it’s fairly easy to find examples of transformations for which \dim_F\mathscr{G}=0,n,n^2 and furthermore it’s clear that the only possible dimensions of \mathscr{G} are 0,n,2n,3n,\cdots,(n-1)n,n^2.

c) No. Consider \mathbb{R}^4. Then, there does not exist a T_0\in\text{End}\left(\mathbb{R}^4\right) such that \left\{T\in\text{End}\left(\mathbb{R}^4\right):T_0T=\bold{0}\right\} has dimension 5 since 5 is not an integral multiple of 4.


Problem: Let A\in\text{End}\left(\mathscr{V}\right) where \mathscr{V} is an n-dimensional F-space and consider the map

T_A:\text{End}\left(\mathscr{V}\right)\to\text{End}\left(\mathscr{V}\right):X\mapsto AX

Prove that T_A is a linear transformation. Can every linear transformation on that space be obtained in this manner?

Proof: Almost by definition we see that

T_A\left(\alpha X+\beta Y\right)=A\left(\alpha X+\beta Y\right)=\alpha A X+\beta A Y=\alpha T_A\left(X\right)+\beta T_A\left(Y\right)

To prove the second part we let

\Phi:\text{End}\left(\mathscr{V}\right)\to\text{End}\left(\text{End}\left(\mathscr{V}\right)\right):A\mapsto T_A

This is evidently linear since

\begin{aligned}\Phi\left(\alpha A+\beta B\right)(X) &=T_{\alpha A+\beta B}\left(X\right)\\ &=\left(\alpha A+\beta B\right)X\\ &=\alpha A X+\beta B X\\ &=\alpha\left(\Phi\left(A\right)\right)(X)+\beta \left(\Phi\left(B\right)\right)(X)\end{aligned}


And since X\in\text{End}\left(\mathscr{V}\right) was arbitrary it follows that \Phi\left(\alpha A+\beta B\right)=\alpha \Phi\left(A\right)+\beta\Phi\left(B\right). Note then that \Phi is also injective since T_A=T_B implies that


and thus if \Phi were surjective we’d have that


but note that \dim_F\text{End}\left(\mathscr{V}\right)=n^2 and \dim_F\text{End}\left(\text{End}\left(\mathscr{V}\right)\right)=n^4. Thus, if n\ne 1 it follows that \text{End}\left(\text{End}\left(\mathscr{V}\right)\right)\not\cong\text{End}\left(\mathscr{V}\right) and so \Phi is not surjective, if n=1 then the above reasoning leads to a dimension argument that \Phi must be surjective. Thus, it follows that


if and only if n=1.


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


November 29, 2010 - Posted by | Fun Problems, Halmos, Linear Algebra, Uncategorized | , , , , ,

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