# Find dimension of ℒ $(V)$ and polynomial that brings every linear transformation to $0$

Here's the prompt:

Let V be a vector space of finite dimensions $n$ over the field $\mathbb{F}$, and let $\tau \in$ ℒ $(V)$. What is the dimension of ℒ $(V)$ as a vector space over $\mathbb{F}$? With $\tau^0$ denoting the identity linear transformation, and using the fact that $\tau^i \in$ ℒ $(V)$ for every $i \geq 0$, prove that there is a polynomial $f \in \mathbb{F}[x]$ such that $f(\tau) (v) = 0$ for every $v \in V$, i.e., $f(\tau) = 0 \in$ ℒ $(V)$.

I know that the dimension must be less than or equal to the dimension of the parent vector space, in this case it has to be less than n. Correct? And as for the polynomial, I'm having trouble grasping this in terms of abstract vector spaces.

Any help in the form of explanations or examples to help me grasp this will be very much appreciated.

• Could you define L(V)? – Ivan Di Liberti Apr 23 '14 at 6:29

For the first part of the question, note that there is an isomorphism between $\mathcal{L}(V,V)$ and $\mathbb{F}_{n \times n}$ (matrices of order $n \times n$ over $\mathbb{F}$), which we can denote by $\Phi$ sending each $\tau \in \mathcal{L}(V,V)$ to its matrix representation with respect to some basis $\beta$ of $V$. So dim$(\mathcal{L}(V,V))=$ dim$(\mathbb{F}_{n \times n})=n^2$.

The second part is the Caley-Hamilton theorem applied for linear operators. Essentially, for any vector $v$ you have the $\tau-$ cyclic subspace generated by $v$, and this subspace is spanned by $\{v,\tau(v),\ldots,\tau^{k-1}(v)\}$ supposing that this subspace is of dimension $k$ (always less than or equal to $n$). So then there exists scalars $a_0,\ldots, a_{k-1}$, not all zero so that $$a_0v+\cdots+a_{k-1}\tau^{k-1}(v)+\tau^k(v)=0 \quad (1).$$ Now for these cyclic subspaces the standard matrix representation with respect to the spanning vectors above is called a companion matrix, and from there you can work out (separate theorem) that the characteristic polynomial of $\tau$ restricted to this subspace is $(-1)^k(a_0+\cdots+a_{k-1}x^{k-1}+x^k)$, and if you combine it with $(1)$ above you get $$(-1)^k(a_0\tau^0+\cdots+a_{k-1}\tau^{k-1}+\tau^k)(v)=0.$$ Now again by another theorem you can prove that the characteristic polynomial of $\tau$ restricted to the subspace above divides the characteristic polynomial of $\tau$, and then you can come to the conclusion that for any $v$, $c_{\tau}(\tau)(v)=0$. For the full detail on all the theorems involved I can refer you to Linear Algebra, 4th ed, by Friedberg et al, section 5.4. or for alternative proofs just search for Caley-Hamilton theorem. Most of the time the proof is for matrices though and not directly for operators, but in this case you could make use of the isomorphism as described in the first part above.

The dimension of $\mathcal{L}(V)$ is $(\dim V)^2$ by using matrices. For the mere existence of a polynomial, consider the following powers of $\tau$: $$\tau^0,\tau^1,\tau^2,\dots,\tau^{m}.$$ where $m=(\dim V)^2$. This is a set of $m+1$ elements in a vector space of dimension $m$, so it is linearly dependent, hence there are scalars $a_0, a_1,\dots,a_{m}$ such that $$a_0\tau^0+a_1\tau^2+\dots+a_{m}\tau^m=0$$ With the Hamilton-Cayley theorem you can lower the upper bound for the degree of a polynomial to $\dim V$.

• yes! in fact for part two of the OP's question I (highly) recommend this answer...since the question merely asks for existance this answers it perfectly, and it is a LOT less work than proving the Cayley-Hamilton theorem. – Christiaan Hattingh Apr 24 '14 at 6:02

$\dim \mathscr L(V) = n^2$. This may be seen as follows: let $\{ e_i \}$, $1 \le i \le n$, be a basis for $V$ over $\Bbb F$. Define a family of $n^2$ elements $E_{ij} \in \mathscr L(V)$, $1 \le i, j \le n$, by setting $E_{ij}(e_k) = \delta_{jk}e_i$ and extending each $E_{ij}$ to all of $V$ by linearity; with $v = \sum_kv^ke_k$:

$E_{ij}(v) = E_{ij}(\sum_k v^k e_k) = \sum_k v^k E_{ij}(e_k) = \sum_k v^k \delta_{jk}(e_i) = v^je_i; \tag{1}$

it is easy to see that (1) imples $E_{ij}(\alpha v) = \alpha E_{ij}(v)$ and $E_{ij}(v + w) = E_{ij}(v) + E_{ij}(w)$. I claim that the set $\{ E_{ij} \}$ forms a basis of $\mathscr L(V)$. For let $A \in \mathscr L(V)$ and consider the action of $A$ on the basis elements $e_i$. We may write

$A(e_j) = \sum_j A^i_j e_i \tag{2}$

for any $e_j$; now consider

$\sum_{i,j} A^i_j E_{ij}(e_k) = \sum_{i, j} A^i_j \delta_{jk}e_i = \sum_i A^i_k e_i = A(e_k) \tag{3}$

by (2); this shows that

$A(v) = \sum_{i, j} A^i_j E_{ij}(v) \tag{4}$

holds for all $v \in V$ by linearity, since it holds for all vectors in the basis $\{ e_i \}$. It remains to show the $E_{ij}$ are linearly independent elements of $\mathscr L(V)$; suppose then that

$\sum_{i, j}B^i_j E_{ij} = 0 \tag{5}$

for some collection of $B^i_j \in \Bbb F$; then by the same computation used in (3) we find that

$0 = \sum_{i, j}B^i_j E_{ij}(e_k) = \sum_k B^i_k e_k; \tag{6}$

by the linear independence of the $e_k$, we have $B^i_k = 0$ for all $i, k$; thus the $E_{ij}$ are independent and form a basis for $\mathscr L(V)$. Since there are $n^2$ operators $E_{ij}$ in this basis, we see that indeed $\dim \mathscr L(V) = n^2$. It is worth observing that in a representation of $\mathscr L(V)$ as $M_{n \times n}(\Bbb F)$ in the usual manner then the entries of $E_{ij}$ are given by

$(E_{ij})_{lk} = \delta_{il} \delta_{jk}, \tag{7}$

i.e., the matrix representing $E_{ij}$ has $1$ at the intersection of row $i$ and column $j$, and zeroes everywhere else.

Having demonstrated that $\dim \mathscr L(V) = n^2$, we turn to the second of the OP's questions:

For any of the basis elements $e_i$, consider the successive powers of $\tau$ applied to $e_i$; we have $\tau^0 e_i = e_i$, $\tau^1 e_i = \tau e_i$, $\tau^2 e_i$, . . . , $\tau^{n - 1} e_i$, $\tau^n e_i$, and so forth. Since $\dim V = n$, we see that there must be a non-trivial linear dependence between the $n + 1$ vectors $\tau^j e_i$, $0 \le j \le n$; that is, there exist $\alpha_j \in \Bbb F$, not all zero, such that

$\sum_0^n \alpha_j \tau^j e_i = 0, \tag{8}$

which shows that the matrix polynomial $\sum_0^n \alpha_j \tau^j$ annihilates $e_i$; since $\tau^0 e_i = e_i \ne 0$, we see that the degree of this polynomial must be at least one, and the degree is manifestly at most $n$. We therefore define $p_i(x) \in \Bbb F[x]$ via

$p_i(x) = \sum_0^n \alpha_j \tau^j; \tag{9}$

we note in passing that if $\deg p_i = 1$, then $e_i$ is an eigenvector of $\tau$, and that we can if so desired select $p_i(x)$ ti be of minimal degree among all non-zero-polynomials $q(x) \in \Bbb F[x]$ satisfying $q(\tau)e_i = 0$; but this is not essential to completing our argument, and we need make no further assumptions concerning $p_i(x)$. Having defined non-trivial polynomials $p_j(x)$ such that $p_j(\tau) e_j = 0$ for each $j$, $1 \le j \le n$, we set

$P(x) = \prod_{j = 1}^{j = n} p_j(x) = (\prod_{j=1, j \ne k}^{j = n} p_j(x)) p_k(x) \tag{10}$

for each $k$. From what we have seen, $n \le deg P \le n^2$, and for any basis element $e_k$ we have

$P(\tau)e_k = (\prod_{j =1, j \ne k}^{j = n} p_j(\tau)) p_k(\tau) e_k = 0 \tag{11}$

by virtue of (10). Since $P(\tau) e_k$ vanishes for each basis vector $e_k$, we have

$P(\tau)v = 0 \tag{12}$

for each $v \in V$; this establishes the existence of the sought-for polynomial in $\tau$. QED.

Note: It occurred to me as I was completing the above write-up that there is a swifter path to the existence of $P(x)$, based upon the previously established fact that $\dim \mathscr L(V) = n^2$: we simply take the sequence $\tau^k \in \mathscr L(V)$, $k \ge 0$, noting that $\tau^0 = I$ and that, since $\dim \mathscr L(V) = n^2$, there must exist a non-trivial linear dependence among the $n^2 + 1$ operators $\tau^k$, $0 \le k \le n^2$; thus there exist $\beta_k \in \Bbb F$ such that

$P(\tau) = \sum_0^{n^2} \beta_k \tau^k = 0; \tag{13}$

then $P(x) \in \Bbb F[x]$ is the requisite polynomial. End of Note.

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!