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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 $k = \operatorname{dim} \operatorname{span} \{a_1, ..., a_n\} \in \{0, 1, ..., n\}$.

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

1 Answer 1

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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 $S$.

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  • $\begingroup$ 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 $b_i \in \ker A$. $\endgroup$
    – Andreo
    Jul 14, 2019 at 4:36
  • $\begingroup$ I've added a solution with annihilators in the main question. It is too long for the comments. Could you please take a look? $\endgroup$
    – Andreo
    Jul 15, 2019 at 2:40

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