## Tag Archives: vector space

### Show that a given linear transformation has no eigenvectors

Let $V = \mathbb{R}^\mathbb{N}$ be a countable-dimensional $\mathbb{R}$ vector space, and define $T : V \rightarrow V$ by $T((a_i))_j = 0$ if $j = 0$ and $a_{j-i}$ otherwise. Show that $T$ has no eigenvectors.

Suppose there exists $a \in V$, $r \in \mathbb{R}$ nonzero, such that $T(a) = ra$. We claim that $a_i = 0$ for all $i$, and prove it by induction. For the base case $i = 0$, we have $ra_0 = 0$. Since $r \neq 0$, $a_0 = 0$. For the inductive step, if $a_i = 0$, then $ra_{i+1} = a_i = 0$. Again since $r \neq 0$, $a_{i+1} = 0$. So in fact $a = 0$.

So $T$ has no eigenvectors.

### Some properties of a QQ-vector space given the existence of a linear transformation with given properties

Let $V$ be a finite dimensional vector space over $\mathbb{Q}$ and suppose $T$ is a nonsingular linear transformation on $V$ such that $T^{-1} = T^2 + T$. Prove that the dimension of $V$ is divisible by 3. If the dimension of $V$ is precisely 3, prove that all such transformations are similar.

If $T$ is such a transformation, then we have $1 = T^3 + T^2$, and so $T^3 + T^2 - 1 = 0$. So the minimal polynomial of $T$ divides $p(x) = x^3 + x^2 - 1$. Now $p(x)$ is irreducible over $\mathbb{Q}$ by the Rational Root Test (Prop. 11 on page 308 in D&F). So the minimal polynomial of $T$ is precisely $p(x)$. Now the characteristic polynomial of $T$ divides some power of $p(x)$ (Prop. 20 in D&F) and so has degree $3k$ for some $k$. But of course the degree of the characteristic polynomial is precisely the dimension of $V$, as desired.

Now if $V$ has dimension 3, then the minimal polynomial of $V$ is the characteristic polynomial, so that the invariant factors of $T$ are simply $p(x)$. In particular, all such $T$ are similar.

### The degree of the minimal polynomial of a matrix of dimension n is at most n²

Let $A \in \mathsf{Mat}_n(F)$ be a matrix over a field $F$. Prove that the dimension of the minimal polynomial $m_A(x)$ of $A$ is at most $n^2$.

If $V = F^n$, note that the set $\mathsf{End}_F(V)$ is isomorphic to $\mathsf{Mat}_n(F)$ once we select a pair of bases. The set of all $n \times n$ matrices over $F$ is naturally an $F$-vector space of dimension $n^2$. Consider now the powers of $A$: $1 = A^0$, $A^1$, $A^2$, …, $A^{n^2}$. These matrices, as elements of $\mathsf{End}_F(V)$, are necessarily linearly dependent, so that $\sum r_i A^i = p(A) = 0$ for some $r_i \in F$. The minimal polynomial $m_A(x)$ divides this $p(x)$, and so has degree at most $n^2$.

### Over a field of characteristic not two, the tensor square of a vector space decomposes as the direct sum of the symmetric and exterior tensor squares

Let $F$ be a field of characteristic not 2, and let $V$ be an $n$-dimensional vector space over $F$. Recall that in this case, we can realize $\mathcal{S}^2(V)$ and $\bigwedge^2(V)$ as subspaces of $V \otimes_F V$. Prove that $V \otimes_F V = \mathcal{S}^2(V) \oplus \bigwedge^2(V)$.

Recall that $\mathsf{dim}_F\ V \otimes_F V = n^2$.

Suppose $z \in \mathcal{S}^2(V) \cap \bigwedge^2(V)$. Then we have $\mathsf{Alt}_2(z) = z = \mathsf{Sym}_2(z)$. Expanding these, we see that $z + (1\ 2) z = z - (1\ 2) z$, so that $2(1\ 2) z = 0$, so that $z = 0$. That is, $\mathcal{S}^2(V)$ and $\mathsf{Alt}_2(V)$ intersect trivially.

Counting dimensions, recall that $\mathsf{dim}_F\ \mathcal{S}^2(V) = \frac{n(n+1)}{2}$ and $\mathsf{dim}_F\ \bigwedge^2(V) = \frac{n(n-1)}{2}$.

Thus $V \otimes_F V = \mathcal{S}^2(V) \oplus \bigwedge^2(V)$.

### Facts about alternating bilinear maps on vector spaces

Let $F$ be a field, let $V$ be an $n$-dimensional $F$-vector space, and let $f : V \times V \rightarrow W$ be a bilinear map, where $W$ is an $F$-vector space.

1. Prove that if $f$ is an alternating $F$-bilinear map on $V$ then $f(x,y) = -f(y,x)$ for all $x,y \in V$.
2. Suppose $\mathsf{char}\ F \neq 2$. Prove that $f(x,y)$ is an alternating bilinear map on $V$ if and only if $f(x,y) = -f(y,x)$ for all $x,y \in V$.
3. Suppose $\mathsf{char}\ F = 2$. Prove that every alternating bilinear form $f(x,y)$ on $V$ is symmetric. (I.e. $f(x,y) = f(y,x)$ for all $x,y \in V$.) Prove that there exist symmetric bilinear maps which are not alternating.

Let $x,y \in V$, and suppose $f$ is an alternating bilinear map. Now $0 = f(x-y,x-y)$ $= f(x,x) - f(x,y) - f(y,x) + f(y,y)$, so that $f(x,y) = -f(x,y)$.

Suppose $\mathsf{char}\ F \neq 2$; in particular, 2 is a unit in $F$. If $f$ is bilinear such that $f(x,y) = -f(y,x)$ for all $x,y \in V$, then in particular we have $f(x,x) = -f(x,x)$, so that $2f(x,x) = 0$. Thus $f(x,x) = 0$ for all $x \in V$, and so $f$ is alternating. Conversely, if $f$ is alternating then by part (1) above we have $f(x,y) = -f(y,x)$ for all $x, y \in V$.

Now suppose $\mathsf{char}\ F = 2$. Note that $(x+y) \otimes (x+y) = x \otimes x - x \otimes y + y \otimes x - y \otimes y$. Mod $\mathcal{A}^2(V)$, we have $x \otimes y - y \otimes x = 0$. In particular, the submodule $\mathcal{C}^2(V)$ generated by all tensors of the form $x \otimes y - y \otimes x$ is contained in $\mathcal{A}^2(V)$.

We have already seen that $\mathsf{dim}_F\ \mathcal{S}^2(V) = {{n+1} \choose {n-1}} = \frac{n(n+1)}{2}$ and $\mathsf{dim}_F\ \bigwedge^2(V) = {n \choose 2} = \frac{n(n-1)}{2}$. Thus the containment $\mathcal{C}^2(V) \subsetneq \mathcal{A}^2(V)$ is proper.

### If V is an infinite dimensional vector space, then its dual space has strictly larger dimension

Let $V$ be an infinite dimensional vector space over a field $F$, say with basis $A$. Prove that the dual space $\widehat{V} = \mathsf{Hom}_F(V,F)$ has strictly larger dimension than does $V$.

We claim that $\widehat{V} \cong_F \prod_A F$. To prove this, for each $a \in A$ let $F_a$ be a copy of $F$. Now define $\varepsilon_a : \widehat{V} \rightarrow F_a$ by $\varepsilon_a(\widehat{v}) = \widehat{v}(a)$. By the universal property of direct products, there exists a unique $F$-linear transformation $\theta : \widehat{V} \rightarrow \prod_A F_a$ such that $\pi_a \circ \theta = \varepsilon_a$ for all $a \in A$. We claim that $\theta$ is an isomorphism. To see surjectivity, let $(v_a) \in \prod_A F_a$. Now define $\varphi \in \widehat{V}$ by letting $\varphi(a) = v_a$ and extending linearly; certianly $\theta(\varphi) = (v_a)$. To see injectivity, suppose $\varphi \in \mathsf{ker}\ \theta$. Then $\theta(\varphi) = 0$, so that $(\pi_a \circ \theta)(\varphi) = 0$, and thus $\varepsilon_a(\varphi) = 0$ for all $a$. Thus $\varepsilon(a) = 0$ for all $a \in A$. Since $A$ is a basis of $V$, we have $\varphi = 0$. Thus $\theta$ is an isomorphism, and we have $\widehat{V} \cong_F \prod_A F$.

By this previous exercise, $\widehat{V}$ has strictly larger dimension than does $V$.

### The dual basis of an infinite dimensional vector space does not span the dual space

Let $F$ be a field and let $V$ be an infinite dimensional vector space over $F$; say $V$ has basis $B = \{v_i\}_I$. Prove that the dual basis $\{\widehat{v}_i\}_I$ does not span the dual space $\widehat{V} = \mathsf{Hom}_F(V,F)$.

Define a linear transformation $\varphi$ on $V$ by taking $v_i$ to 1 for all $i \in I$. Note that for all $i \in I$, $\varphi(v_i) \neq 0$. Suppose now that $\varphi = \sum_{i \in K} \alpha_i\widehat{v}_i$ where $K \subseteq I$ is finite; for any $j \notin K$, we have $(\sum \alpha_i \widehat{v}_i)(v_j) = 0$, while $\varphi(v_j) = 1$. Thus $\varphi$ is not in $\mathsf{span}\ \{\widehat{v}_i\}_I$.

So the dual basis does not span $\widehat{V}$.

### The annihilator of a subset of a dual vector space

Let $V$ be a vector space over a field $F$ and let $\widehat{V} = \mathsf{Hom}_F(V,F)$ denote the dual vector space of $V$. Given $S \subseteq \widehat{V}$, define $\mathsf{Ann}(S) = \{v \in V \ |\ f(v) = 0\ \mathrm{for\ all}\ f \in S \}$. (This set is called the annihilator of $S$ in $V$.)

1. Prove that $\mathsf{Ann}(\widehat{S})$ is a subspace of $V$ for all $\widehat{S} \subseteq \widehat{V}$.
2. Suppose $\widehat{W}_1$ and $\widehat{W}_2$ are subspaces of $\widehat{V}$. Prove that $\mathsf{Ann}(\widehat{W}_1 + \widehat{W}_2) = \mathsf{Ann}(\widehat{W}_1) \cap \mathsf{Ann}(\widehat{W}_2)$ and $\mathsf{Ann}(\widehat{W}_1 \cap \widehat{W}_2) = \mathsf{Ann}(\widehat{W}_1) + \mathsf{Ann}(\widehat{W}_2)$.
3. Let $\widehat{W}_1, \widehat{W}_2 \subseteq \widehat{V}$ be subspaces. Prove that $\mathsf{Ann}(\widehat{W}_1) = \mathsf{Ann}(\widehat{W}_2)$ if and only if $\widehat{W}_1 = \widehat{W}_2$.
4. Prove that, for all $\widehat{S} \subseteq \widehat{V}$, $\mathsf{Ann}(\widehat{S}) = \mathsf{Ann}(\mathsf{span}\ \widehat{S})$.
5. Assume $V$ is finite dimensional with basis $B = \{v_i\}_{i=1}^n$, and let $\widehat{B} = \{\widehat{v}_i\}_{i=1}^n$ denote the basis dual to $B$. Prove that if $\widehat{S} = \{\widehat{v}_i\}_{i=1}^k$ for some $1 \leq k \leq n$, then $\mathsf{Ann}(\widehat{S}) = \mathsf{span} \{v_i\}_{i=k+1}^n$.
6. Assume $V$ is finite dimensional. Prove that if $\widehat{W} \subseteq \widehat{V}$ is a subspace, then $\mathsf{dim}\ \mathsf{Ann}(\widehat{W}) = \mathsf{dim}\ V - \mathsf{dim}\ \widehat{W}$.

[This needs to be cleaned up.]

Recall that a bounded lattice is a tuple $(L, \wedge, \vee, \top, \bot)$, where $\wedge$ and $\vee$ are binary operators on $L$ and $\top$ and $\bot$ are elements of $L$ satisfying the following:

1. $\wedge$ and $\vee$ are associative and commutative,
2. $\top$ and $\bot$ are identity elements with respect to $\wedge$ and $\vee$, respectively, and
3. $a \wedge (a \vee b) = a$ and $a \vee (a \wedge b) = a$ for all $a,b \in L$. (Called the “absorption laws”.)

If $L_1$ and $L_2$ are bounded lattices, a bounded lattice homomorphism is a mapping $\varphi : L_1 \rightarrow L_2$ that preserves the operators- $\varphi(a \wedge b) = \varphi(a) \wedge \varphi(b)$, $\varphi(a \vee b) = \varphi(a) \vee \varphi(b)$, $\varphi(\bot) = \bot$, and $\varphi(\top) = \top$. As usual, a lattice homomorphism which is also bijective is called a lattice isomorphism.

The interchangability of $\wedge$ and $\vee$ (and of $\bot$ and $\top$) immediately suggests the following definition. Given a bounded lattice $L$, we define a new lattice $\widehat{L}$ having the same base set as $L$ but with the roles of $\wedge$ and $\vee$ (and of $\bot$ and $\top$) interchanged. This $\widehat{L}$ is called the dual lattice of $L$.

Let $V$ be a vector space (of arbitrary dimension) over a field $F$. We let $\mathcal{S}_F(V)$ denote the set of all $F$-subspaces of $V$. We claim that $(\mathcal{S}_F(V), \cap, +, V, 0)$ is a bounded lattice. The least obvious of the axioms to check are the absorption laws. Indeed, note that for all subspaces $U,W \subseteq V$, we have $U \cap (U + W) = U$ and $U + (U \cap W) = U$.

Now let $V$ be a vector space (again of arbitrary dimension) over a field $F$, and let $\widehat{V} = \mathsf{Hom}_F(V,F)$ denote its dual space. If $S \subseteq \widehat{V}$ is an arbitrary subset and $\mathsf{Ann}(S)$ is defined as above, note that $f(0) = 0$ for all $f \in S$, and that if $x,y \in \mathsf{Ann}(S)$ and $r \in F$, we have $f(x+ry) = f(x)+rf(y) = 0$ for all $f \in S$. By the submodule criterion, $\mathsf{Ann}(S) \subseteq V$ is a subspace.

Now define $A : \mathcal{S}_F(\widehat{V}) \rightarrow \widehat{\mathcal{S}_F(V)}$ by $A(\widehat{W}) = \mathsf{Ann}(\widehat{W})$. We claim that if $V$ is finite dimensional, then $A$ is a bounded lattice homomorphism.

1. ($A(\widehat{0}) = V$) Note that for all $v \in V$, we have $\widehat{0}(v) = 0$. Thus $V = \mathsf{Ann}(\widehat{0}) = A(\widehat{0})$. ($\widehat{0}$ is the zero function $V \rightarrow F$.)
2. ($A(\widehat{V}) = 0$) Suppose there exists a nonzero element $v \in \mathsf{Ann}(\widehat{V})$. Then there exists a basis $E$ of $V$ containing $v$, and we may construct a homomorphism $\varphi : V \rightarrow F$ such that $\varphi(v) \neq 0$. In particular, $v \notin A(\widehat{V})$. On the other hand, it is certainly the case that $0 \in A(\widehat{V})$. Thus we have $A(\widehat{V}) = 0$.
3. ($A(\widehat{W}_1 + \widehat{W}_2) = A(\widehat{W}_1) \cap A(\widehat{W}_2)$) $(\subseteq)$ Let $v \in A(\widehat{W}_1 + \widehat{W}_2)$. Then for all $f + g \in \widehat{W}_1 + \widehat{W}_2$, we have $(f+g)(v) = f(v) + g(v) = 0$. In particular, if $f \in \widehat{W}_1$, then $f(v) = (f+0)(v) = 0$, so that $v \in A(\widehat{W}_1)$. Similarly, $v \in A(\widehat{W}_2)$, and thus $v \in A(\widehat{W}_1) \cap A(\widehat{W}_2)$. $(\supseteq)$ Suppose $v \in A(\widehat{W}_1) \cap A(\widehat{W}_2)$. Then for all $f+g \in \widehat{W}_1 + \widehat{W}_2$, we have $(f+g)(v) = f(v)+g(v) = 0$; thus $v \in A(\widehat{W}_1+\widehat{W}_2)$. Thus $A(\widehat{W}_1 + \widehat{W}_2) = A(\widehat{W}_1) \cap A(\widehat{W}_2)$.
4. ($A(\widehat{W}_1 \cap \widehat{W}_2) = A(\widehat{W}_1) + A(\widehat{W}_2)$) $(\supseteq)$ Suppose $v \in A(\widehat{W}_1)$. Then for all $f \in \widehat{W}_1$, $f(v) = 0$. In particular, for all $f \in \widehat{W}_1 \cap \widehat{W}_2$. Thus $v \in A(\widehat{W}_1 \cap \widehat{W}_2)$. Similarly we have $A(\widehat{W}_2) \subseteq A(\widehat{W}_1 \cap \widehat{W}_2)$; thus $A(\widehat{W}_1) + A(\widehat{W}_2) \subseteq A(\widehat{W}_1 \cap \widehat{W}_2)$. $(\subseteq)$ First, we claim that this inclusion holds for all pairs of one dimensional subspaces. If $\widehat{W}_1$ and $\widehat{W}_2$ intersect in a dimension 1 subspace (that is, $\widehat{W}_1 = \widehat{W}_2$), then certianly $A(\widehat{W}_1 \cap \widehat{W}_2) \subseteq A(\widehat{W}_1) + A(\widehat{W}_2)$. If they intersect trivially, then we have $\widehat{W}_1 = F\widehat{t}_1$ and $\widehat{W}_2 = F\widehat{t}_2$, and $A(\widehat{W}_1) = \mathsf{ker}\ \widehat{t}_1$ and $A(\widehat{W}_2) = \mathsf{ker}\ \widehat{t}_2$. Now $\widehat{t}_1$ and $\widehat{t}_2$ are nonzero linear transformations $V \rightarrow F$, and so by the first isomorphism theorem for modules their kernels have dimension $(\mathsf{dim}\ V) - 1$. Note that linear transformations $V \rightarrow F$ are realized (after fixing a basis) by matrices of dimension $1 \times \mathsf{dim}\ V$; in particular, if $\widehat{t}_1$ and $\widehat{t}_2$ have the same kernel, then they are row equivalent, and so are $F$-multiples of each other. Thus we have $A(\widehat{W}_1) + A(\widehat{W}_2) = V$. Now suppose $\widehat{W}_1 = \sum \widehat{W}_{1,i}$ and $\widehat{W}_2 = \sum \widehat{W}_{2,i}$ are sums of one dimensional subspaces. We have $A(\widehat{W}_1 \cap \widehat{W}_2) = A((\sum \widehat{W}_{1,i}) \cap (\sum \widehat{W}_{2,j}))$ $= A(\sum (\widehat{W}_{1,i} \cap \widehat{W}_{2,j}))$ $= \bigcap A(\widehat{W}_{1,i} \cap \widehat{W}_{2,j})$. From the one-dimensional case, this is equal to $\bigcap (A(\widehat{W}_{i,1} + A(\widehat{W}_{2,j}) = (\bigcap A(\widehat{W}_{1,i})) + (\bigcap A(\widehat{W}_{2,i}))$ $= A(\widehat{W}_1) + A(\widehat{W}_2)$. (Note that our proof depends on $V$ being finite dimensional.)

Thus $A$ is a bounded lattice homomorphism. We claim also that $A$ is bijective. To see surjectivity, let $W \subseteq V$ be a subspace. Define $\widehat{W} = \{ f \in \widehat{V} \ |\ \mathsf{ker}\ f \supseteq W \}$. We claim that $A(\widehat{W}) = W$. To see this, it is clear that $W \subseteq A(\widehat{A})$. Moreover, there is a mapping $f \in \widehat{W}$ whose kernel is exactly $W$, and thus $A(\widehat{W}) = W$. Before we show injectivity, we give a lemma.

Lemma: Let $\widehat{W} \subseteq \widehat{V}$ be a subspace with basis $\{\widehat{v}_i\}_{i=1}^k$, and extend to a basis $\{\widehat{v}_i\}_{i=1}^n$. Let $\{v_i\}_{i=1}^n$ be the dual basis to $\{\widehat{v}_i\}_{i=1}^n$, obtained using the natural isomorphism $V \cong \widehat{\widehat{V}}$. Then $A(\widehat{W}) = \mathsf{span}\ \{v_i\}_{i=k+1}^n$. Proof: Let $\sum \alpha_i v_i \in A(\widehat{W})$. In particular, we have $\widehat{v}_j(\sum \alpha_i v_i) = \alpha_j = 0$ for all $1 \leq j \leq k$. Thus $\sum \alpha_iv_i \in \mathsf{span}\ \{v_i\}_{i=k+1}^n$. Conversely, note that $\widehat{v}_j(v_i) = 0$ for all $k+1 \leq i \leq n$, so that $\mathsf{span}\ \{v_i\}_{i=k+1}^n \subseteq A(\widehat{W})$. $\square$

In particular, we have $\mathsf{dim}\ \widehat{W} + \mathsf{dim}\ A(\widehat{W}) = \mathsf{dim}\ V$. Now suppose $A(\widehat{W}) = \mathsf{span}\ \{v_i\}_{i=1}^k$, and extend to a basis $\{v_i\}_{i=1}^n$ of $V$. Let $\{\widehat{v}_i\}_{i=1}^n$ denote the dual basis. Note that for all $f \in \widehat{W}$, writing $f = \sum \alpha_i \widehat{v}_i$, we have $\alpha_j = f(v_j) = 0$ whenever $1 \leq j \leq k$. In particular, $\widehat{W} \subseteq \mathsf{span}\ \{\widehat{v}_i\}_{i=k+1}^n$. Condiering dimension, we have equality. Now to see injectivity for $A$, note that if $A(\widehat{W}_1) = A(\widehat{W}_2)$, then $\widehat{W}_1$ and $\widehat{W}_2$ share a basis- hence $\widehat{W}_1 = \widehat{W}_2$, and so $A$ is injective.

Thus, as lattices, we have $\mathcal{S}_F(\widehat{V}) \cong \widehat{\mathcal{S}_F(V)}$.

Finally, note that it is clear we have $\mathsf{Ann}(\mathsf{span}\ S) \subseteq \mathsf{Ann}(S)$. Conversely, if $v \in \mathsf{Ann}(S)$ and $f = \sum \alpha_i s_i \in \mathsf{span}\ S$, then $f(v) = 0$. Thus $\mathsf{Ann}(S) = \mathsf{Ann}(\mathsf{span}\ S)$.

### Express a given linear transformation in terms of a dual basis

Let $V \subseteq \mathbb{Q}[x]$ be the $\mathbb{Q}$-vector space consisting of those polynomials having degree at most 5. Recall that $B = \{1,x,x^2,x^3,x^4,x^5\}$ is a basis for this vector space over $\mathbb{Q}$. For each of the following maps $\varphi : V \rightarrow \mathbb{Q}$, verify that $\varphi$ is a linear transformation and express $\varphi$ in terms of the dual basis $B^\ast$ on $V^\ast = \mathsf{Hom}_F(V,F)$.

1. $\varphi(p) = p(\alpha)$, where $\alpha \in \mathbb{Q}$.
2. $\varphi(p) = \int_0^1 p(t)\ dt$
3. $\varphi(p) = \int_0^1 t^2p(t)\ dt$
4. $\varphi(p) = p^\prime(\alpha)$, where $\alpha \in \mathbb{Q}$. (Prime denotes the usual derivative of a polynomial.)

Let $v_i$ be the element of the dual basis $B^\ast$ such that $v_i(x^j) = 1$ if $i = j$ and 0 otherwise. I’m going to just assume that integration over an interval is linear.

1. Note that $\varphi(p+rq) = (p+rq)(\alpha) = p(\alpha) + rq(\alpha) = \varphi(p) + r \varphi(q)$; thus $\varphi$ is indeed a linear transformation. Moreover, note that $(\sum \alpha^i v_i)(\sum c_jx^j) = \sum \alpha^i v_i(\sum c_jx^j)$ $= \sum \alpha^i \sum c_j v_i(x^j)$ $= \sum \alpha^i c_i$ $= (\sum c_ix^i)(\alpha)$. Thus $\varphi = \sum \alpha^i v_i$.
2. Note that $\varphi(\sum \alpha_i x^i) = \sum \frac{\alpha_i}{i+1}$. Now $(\sum \frac{1}{i+1} v_i)(\sum \alpha_j x^j) = \sum \frac{1}{i+1} v_i(\sum \alpha_j x^j)$ $= \sum \frac{1}{i+1} \sum \alpha_j v_i(x^j)$ $= \sum \frac{\alpha_i}{i+1}$ $= \varphi(\sum \alpha_i x^i)$. So $\varphi = \sum \frac{1}{i+1} v_i$.
3. Note that $\varphi(\sum \alpha_i x^i) = \sum \frac{\alpha_i}{i+3}$. Now $(\sum \frac{1}{i+3} v_i)(\sum \alpha_j x^j) = \sum \frac{1}{i+3} v_i(\sum \alpha_j x^j)$ $= \sum \frac{1}{i+3} \sum \alpha_j v_i(x^j)$ $= \sum \frac{\alpha_i}{i+3}$ $= \varphi(\sum \alpha_i x^i)$. Thus $\varphi = \sum \frac{1}{i+3} v_i$.
4. Since differentiation (of polynomials) is linear and the evaluation map is linear, this $\varphi$ is linear. Note that $(\sum (i+1)\alpha^i v_{i+1})(\sum c_jx^j) = \sum (i+1)\alpha^i v_{i+1}(\sum c_jx^j)$ $= \sum (i+1)\alpha^i \sum c_j v_{i+1}(x^j)$ $= \sum (i+1)\alpha^i c_{i+1}$ $= \varphi(\sum c_ix^i)$. Thus $\varphi = \sum (i+1)\alpha^iv_{i+1}$.

### The endomorphism rings of a vector space and its dual space are isomorphic as algebras over the base field

Let $F$ be a field and let $V$ be a vector space over $F$ of some finite dimension $n$. Show that the mapping $\Omega : \mathsf{End}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V,F))$ given by $\Omega(\varphi)(\tau) = \tau \circ \varphi$ is an $F$-vector space isomorphism but not a ring isomorphism for $n > 1$. Exhibit an $F$-algebra isomorphism $\mathsf{End}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V,F))$.

We begin with a lemma.

Lemma: Let $R$ be a unital ring and let $M,A,B$ be left unital $R$-modules. If $\varphi : M \times A \rightarrow B$ is $R$-bilinear, then the induced map $\Phi : M \rightarrow \mathsf{Hom}_R(A,B)$ given by $\Phi(m)(a) = \phi(m,a)$ is a well-defined $R$-module homomorphism. Proof: To see well definedness, we need to verify that $\Phi(m) : A \rightarrow B$ is a module homomorphism. To that end note that $\Phi(m)(x+ry) = \varphi(m,x+ry) = \varphi(m,x) + r \varphi(m,y)$ $= \Phi(m)(x) + r\Phi(m)(y)$. Similarly, to show that $\Phi$ is a module homomorphism, note that $\Phi(x+ry)(a) = \varphi(x+ry,a) = \varphi(x,a)+ r\varphi(y,a)$ $= \Phi(x)(a) + r\Phi(y)(a)$ $= (\Phi(x)+r\Phi(y))(a)$, so that $\Phi(x+ry) = \Phi(x) + r\Phi(y)$. $\square$

[Note to self: In a similar way, if $R$ is a unital ring and $M,N,A,B$ unital modules, and $\varphi : M \times N \times A \rightarrow B$ is trilinear, then $\Phi : M \times N \rightarrow \mathsf{Hom}_R(A,B)$ is bilinear. (So that the induced map $M \rightarrow \mathsf{Hom}_R(N,\mathsf{Hom}_R(A,B))$ is a module homomorphism, or unilinear- if you will.) That is to say, in a concrete fashion we can think of multilinear maps as the uncurried versions of higher order functions on modules. (!!!) (I just had a minor epiphany and it made me happy. Okay, so the usual isomorphism $V \rightarrow \mathsf{Hom}_F(V,F)$ is just this lemma applied to the dot product $V \times V \rightarrow F$… that’s cool.) Moreover, if $A = B$ and if $M$ and $\mathsf{End}_R(A)$ are $R$-algebras, then the induced map $\Phi$ is an algebra homomorphism if and only if $\varphi(m_1m_2,a) = \varphi(m_1,\varphi(m_2,a))$ and $\varphi(1,a) = a$.]

Define $\overline{\Omega} : \mathsf{End}_F(V) \times \mathsf{Hom}_F(V,F) \rightarrow \mathsf{Hom}_F(V,F)$ by $\overline{\Omega}(\varphi,\tau) = \tau \circ \varphi$. This map is certainly bilinear, and so by the lemma induces the linear transformation $\Omega : \mathsf{Hom}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V))$. Since $V$ has finite dimension, and since its dual space $\mathsf{Hom}_F(V,F)$ has the same dimension, to see that $\Omega$ is an isomorphism of vector spaces it suffices to show that the kernel is trivial. To that end, suppose $\varphi \in \mathsf{ker}\ \Omega$. Then we have $\Omega(\varphi)(\tau) = \tau \circ \varphi = 0$ for all $\tau$. In particular, we have $\mathsf{im}\ \varphi \subseteq \mathsf{ker}\ \tau$ for all $\tau$. If there exists a nonzero element $v \in \mathsf{im}\ \varphi$, then by the Building-up lemma there is a basis $B$ of $V$ containing $v$. In particular, there is a linear transformation $\tau$ such that $\tau(v) \neq 0$. That is, we have $\mathsf{im}\ \varphi = 0$, so that $\varphi = 0$. Hence $\Omega$ is injective, and so an isomorphism of vector spaces.

Note that $\Omega(\varphi \circ \psi)(\tau) = \tau \circ \varphi \circ \psi$, while $(\Omega(\varphi) \circ \Omega(\psi))(\tau) = \Omega(\varphi)(\Omega(\psi)(\tau))$ $= \Omega(\varphi)(\tau \circ \psi)$ $= \tau \circ \psi \circ \varphi$. If $V$ has dimension greater than 1, then $\mathsf{End}_F(V)$ is not a commutative ring. Thus these expressions need not be equal in general. In fact, if we choose $\tau$, $\varphi$, and $\psi$ such that $M(\varphi) = \left[ \begin{array}{c|c} 0 & 1 \\ \hline 0 & 0 \end{array} \right]$, $M(\psi) = \left[ \begin{array}{c|c} 0 & 0 \\ \hline 1 & 0 \end{array} \right]$, and $M(\tau) = [1 | 0]$, then clearly $\tau \circ \varphi \circ \psi \neq 0$ and $\tau \circ \psi \circ \varphi = 0$. In particular, $\Omega$ is not a ring isomorphism if $n > 1$. On the other hand, if $n = 1$, then $\mathsf{End}_F(V) \cong F$ is commutative, and $\Omega$ is a ring isomorphism.

On the other hand, these rings are clearly isomorphic since $V$ and $\mathsf{Hom}_F(V,F)$ are vector spaces of the same dimension.

Note that $\mathsf{End}_F(V)$ and $\mathsf{End}_F(\mathsf{Hom}_F(V,F))$ are both $F$-algebras via the usual scalar multiplication by $F$. Fix a basis $B$ of $V$, and identify the linear transformation $\varphi \in \mathsf{End}_F(V)$ with its matrix $M^B_B(\varphi)$ with respect to this basis. (Likewise for $\mathsf{Hom}_F(V,F)$.) Now define $\Theta : \mathsf{End}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V,F))$ by $\Theta(M)(N) = NM^\mathsf{T}$. It is clear that $\Theta$ is well defined, and moreover is an $F$-vector space homomorphism. Note also that $\Theta(M_1M_2)(N) = N(M_1M_2)^\mathsf{T}$ $= NM_2^\mathsf{T}M_1^\mathsf{T}$ $= \Theta(M_1)(\Theta(M_2)(N))$, so that $\Theta(M_1M_2) = \Theta(M_1)\Theta(M_2)$. Thus $\Theta$ is a ring homomorphism; since $\Theta(I)(N) = N$, we have $\Theta(I) = 1$, and indeed $\Theta$ is an $F$-algebra homomorphism. It remains to be seen that $\Theta$ is an isomorphism; it suffices to show injectivity. To that end, suppose $\Theta(M)(N) = NM^\mathsf{T} = 0$ for all $N$. Then $MN^\mathsf{T} = 0$ for all $N$, and so $M = 0$. Thus $\Theta$ is an $F$-algebra isomorphism $\mathsf{End}_F(V) \rightarrow \mathsf{End}_F(\mathsf{Hom}_F(V,F))$. Note that $\Theta$ depends essentially on our choice of a basis $B$, and so is not “natural”.