# Dimension of a subspace of linear transformations. Exercise from FDVS by Paul R. Halmos

This is an exercise from Finite Dimensional Vector Spaces by Paul R. Halmos, page 61, Ex. 6:

Let $$V$$ be a $$n$$-dimensional vector space, $$\mathcal{L}(V, V)$$ set of all linear transformations on $$V$$ (also a vector space) and $$S = \{B \in \mathcal{L}(V, V) : AB = 0\}$$ for some $$A \in \mathcal{L}(V, V)$$. What values can dimension of $$S$$ have?

My solution looks very robotic. I represent linear transformations $$A$$ and $$B$$ as an $$n^2$$ dimensional vectors with coordinates $$a_{ij}$$ and $$b_{ij}$$, then I get a system of $$n^2$$ linear equations $$\sum_{j=1}^n a_{ij} \cdot b_{jk} = 0$$, for $$i,k \in \{1, ... , n\}$$. And then by choosing appropriate scalars $$a_{ij}$$ we can set dimension of $$S$$ to any value from $$0$$ to $$n^2$$ - I'm not sure, It looks like not any value, but only of the form $$l*n$$, for $$l \in \{0,1, ... , n\}$$.

Is there a better solution?

Update:

@Omnomnomnom, since in this book kernels were not yet introduced, I think the annihilators can be used to solve this problem. Could you please take a look at my solution:

Let $$\{x_1, ..., x_n\}$$ be a basis of $$V$$, linear transformation $$Ax = \sum_{i=1}^n a_i(x) \cdot x_i$$, where $$\{a_1, ..., a_n\}$$ are linear functionals on $$V$$ and $$Bx = B(\sum_{i=1}^n \xi_i \cdot x_i ) = \sum_{i=1}^n \xi_i \cdot b_i$$, where $$\{b_1, ..., b_n\}$$ are some vectors in $$V$$ (column of the matrix of $$B$$). Then $$AB=0$$ means that $$a_i(b_j)=0$$, for all $$i,j = 1..n$$ or $$b_j \in \{a_i\}^o$$ annihilator of $$a_i$$. Since $$\operatorname{span}(a_1)$$ is one dimensional (assuming $$a_1 \neq 0$$), the annihilator subspace $$\{a_1\}^o$$ is $$n-1$$ dimensional. And we can choose $$n-1$$ linearly independent vectors $$b_i$$ in that annihilator subspace. Then dimension of $$B$$ is $$n(n-1)$$. By adding more linearly independent $$a_i$$ to $$A$$, the dimension of $$B = n(n-k)$$, where $$k = \operatorname{dim} \operatorname{span} \{a_1, ..., a_n\} \in \{0, 1, ..., n\}$$.

• It seems Paul Halmos has lost his first name, and his middle initial has been promoted to a first initial. Jul 14, 2019 at 22:33
• @andreas-blass fixed :) Jul 15, 2019 at 2:37

Hint: Note that $$AB = 0$$ if and only if the image of $$B$$ is in the kernel of $$A$$. So, we can identify $$S$$ with $$\mathcal L(V,\ker A)$$.
For a more matrix-oriented approach: note that $$AB = 0$$ if and only if every column of $$B$$ lies in the null space of $$A$$. Now, using a basis for the nullspace of $$A$$, construct a basis for $$S$$.
• Kernel was not introduced at this point in the book, but I like this solution. I think that a linear transformation $B$ can be represented algebraically as a set of columns in the following way: $Bx = B(\sum_{i=1}^n \xi_i \cdot x_i ) = \sum_{i=1}^n \xi_i \cdot b_i$, where $\{x_1, ..., x_n\}$ is a basis of $V$ and $\{b_1, ..., b_n\}$ some vectors in $V$ (columns of the matrix of $B$). Then every $b_i \in \ker A$. Jul 14, 2019 at 4:36