I am trying to solve the following problem:  Show that the trace functional on $n \times n$ matrices is unique in the following sense. If $W$ is the space of $n \times n$ matrices over the field $F$ and if $f$ is a linear functional on $W$ such that $f(AB)=f(BA)$ for each $A$ and $B$ in $W$, then $f$ is a scalar multiple of the trace function.  I know that one way to prove this is to show that the null space of $f$ equals the null space of the trace function, but I found this solution rather complicated; any other suggestions?
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We know that the trace of a matrix is a linear map for all square matrices and that $\operatorname{trace}(AB)=\operatorname{trace}(BA)$ when the multiplication makes sense.   On the , under properties, it says that these properties characterize the trace completely in the following sense: If $f$ is a linear function on the space of square matrices satisfying $f(xy)=f(yx)$, then $f$ and $\operatorname{tr}$ are proportional. A  of the page gives the justification, but I do not understand the logic of it. Thanks
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Let $F$ be a subfield of the complex numbers. We define $n$ linear functionals on $F^n$ ($n \geq 2$) by $f_k(x_1, \dotsc, x_n) = \sum_{j=1}^n (k-j) x_j$, $1 \leq k \leq n$. What is the dimension of the subspace annihilated by $f_1, \dotsc, f_n$?   Approaches I've tried so far:   Construct a matrix $A$ whose $k$th row's entries are the coefficients of $f_k$, i.e., $A_{ij} = i - j$, and compute the rank of the matrix. Empirically, the resulting matrix has rank 2 for $n = 2$ to $n = 6$, but I don't see a convenient pattern to follow for a row reduction type proof for the general case. Observe that $f_k$, $k \geq 2$ annihilates $x = (x_1,\dotsc, x_n)$ iff $$\sum_{j=1}^{k-1} (k-j)x_j + \sum_{j = k+1}^n (j-k)x_j = 0$$ iff $$k\left(\sum_{j=1}^{k-1} x_j - \sum_{j=k+1}^n x_j\right) = \sum_{j=1}^{k-1} jx_j - \sum_{j=k+1}^n jx_j,$$ and go from there, but I do not see how to proceed.     Note: This is Exercise 10 in Section 3.5 ("Linear Functionals") in Linear Algebra by Hoffman and Kunze; eigenvalues and determinants have not yet been introduced and so I am looking for direction towards an elementary proof.
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For invertible $A_{n\times n}$ find $\min(\operatorname{trace}(AA^T))$  (a) $0$ (b) $1$ (c) $n$ (d) $n^2$  Clearly for $A=I$, it is $n$, and I am unable to get any lower values, but how do I prove it.
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$\DeclareMathOperator{\tr}{trace}$ I am reading the  on the trace operator. The section titled Coordinate-Free Definition defines the trace as follows.  Let $V$ be a finite dimensional vector space over a field $F$ and define a bilinear map $f:V\times V^*\to F$ as $f(v, \omega)=\omega(v)$ for all $(v, \omega)\in V\times V^*$. This maps induces a unique linear map $\tr:V\otimes V^*\to F$. Since $\text{End}(V)$ has a canonical isomorphism with $V\otimes V^*$, we have now a notion of trace of a linear operator on $V$.  The Question: The second paragraph of the section in the article says that  This also clarifies why $\tr(AB)=\tr(BA)$.  I can't see how $\tr(AB)=\tr(BA)$ follows from this definition at all. Can somebody give me a hint?
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Good day, I want to prove the following theorem:  1) For any nonzero vector $ v \in V $, there exists a linear funtional $ f \in V^*$ for wich $f(v) \neq 0 $  I know that if $f$ is a lineal functional then we have 2 posibilities   1)$\dim \ker(f)=\dim V$   2)$\dim \ker(f)=\dim V-1$   I've tried to suppose that, for all $v \neq 0 $ and $f \in V^*$ We have $f(v)=0$,  Then  $Im(f)=0$ for all $f \in V^*$ and $ v \in V$ but then the only linear functional is $f=0$ ($V^*=\{0:V \to F\}$) and from this follows: $\dim \ker(f)=\dim V$.  But 2) suggest the existence of a linear functional which $\dim \ker(f)=\dim V-1$, so ¿is that the contradiction?
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I am trying to find a basis for the set of all $n \times n$ matrices with trace $0$.  I know that part of that basis will be matrices with $1$ in only one entry and $0$ for all others for entries outside the diagonal, as they are not relevant.  I don't understand though how to generalize for the entries on the diagonal.  Maybe just one matrix with $1$ in the $(1, 1)$ position and a $-1$ in all other $n - 1$ positions?
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(I'll explain what I know first and then I'll ask the questions). Given a finite dimensional vector space $V$, it is often remarked that there is no "natural" isomorphism from $V$ to $V^*$ (I guess this means a basis independent isomorphism?). I understand that one typically constructs an isomorphism $V \to V^*$ by fixing a basis for $V$, call this $B = \{v_1, v_2, \ldots, v_n\}$, and then mapping $v_i \mapsto \delta_i$ where $\delta_i$ is the linear functional given by $\delta_i(v_i) = \delta_{ij}$. Here are my questions:  (1) Why is there no natural isomorphism $V \to V^*$?  (2) If $V$ is a finite dimensional inner product space, we can map each $v \mapsto \langle v, \cdot \rangle$. Is this not "natural"? (I know we can consider sesquilinear forms, but let's keep the discussion to this inner product for simplicity).
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I need to prove that: If a nonzero linear functional $f$ on a Banach Space $X$ is discontinuous then the nullspace $N_f$ is dense in $X$.   To prove that $N_f$ is dense, it suffices to show that $\overline N_f = X$ which is equivalent to $(X \setminus N_f)^o=\emptyset$. (the interior of complement of $N_f$ is null set.) Since $f$ is a linear functional and is discontinuous, it has to be unbounded. I don't know exactly how to utilize these observations.   Also on a related topic, I'm a little confused about how to exploit the a Linear Functional $f:X \to R$ or a Linear Operator $T:X \to Y$ being unbounded. Can I say that if a linear operator is unbounded then exists a sequence $&lt;x_n&gt;$ in $X$ s.t.  $||Tx_n|| &gt; n^2||x_n||$ for each $n$ or $||Tx_n|| &gt; n||x_n||$ ?
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This is a small exercise that I just can't seem to figure out. When I see it I'll probably go 'ahhh!', but so far I haven't made any progress.  I'd like to prove that any linear functional $\phi$ on $\ell^\infty$ that is positive, i.e. for any $(x_n) \in \ell^\infty$ such that $x_n \geq 0$ for all $n \in \mathbb{N}$ we have $\phi((x_n)) \geq 0$, is a bounded linear operator and has $\|\phi \| = \phi((1,1,...))$.  Nothing I have tried worked so far, so any hints would be appreciated, thanks!
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Let $Z_n$ be the set of all $n \times n$ matrices that commute with all $n \times n $ matrices. Show that $$Z_n = \{\lambda I_n \ | \  \lambda \in \mathbb R\}$$  ($I_n$ is the $n \times n$ identity matrix)  I don't know how to use $E_{ij}$ (matrix with $1$ in $(i,j)$ and $0$ elsewhere) and the elementary matrix $P_{ij}$ to prove this question. Can anyone explain it please?
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Let $f:V\times V\rightarrow F$ be a bilinear form in a finite inner product space V.  If $F=R$, how can I prove that there exists a single linear transformation $T:V \rightarrow V$ so that for each $v,u\in V$, $f(u,v) = (Tu, v)$ ?  Would that sill be the case if $F=C$?
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I'm just getting back into having to do linear algebra and I am having some trouble with some elementary questions, any help is much appreciated.  Suppose that $A = [a_{ij}]$ is an $m\times n$ matrix and $B = [b_{ij}]$ and is an $n\times m$ matrix and $BA = I$ the identity matrix that is $n \times n$.  Show that if for some $b \in \Bbb{R}^m$ the equation $Ax = b$ has a solution, then the solution is unique.
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First of all, am I being crazy in thinking that if $\lambda$ is an eigenvalue of $AB$, where $A$ and $B$ are both $N \times N$ matrices (not necessarily invertible), then $\lambda$ is also an eigenvalue of $BA$?  If it's not true, then under what conditions is it true or not true?  If it is true, can anyone point me to a citation?  I couldn't find it in a quick perusal of Horn &amp; Johnson.  I have seen a couple proofs that the characteristic polynomial of $AB$ is equal to the characteristic polynomial of $BA$, but none with any citations.  A trivial proof would be OK, but a citation is better.
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This is a variation of the  questioned some time ago.  For a complex Hilbert space $H$ let $T: H \rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum  $$\sum_{i}{\langle |T|e_{i}, e_{i} \rangle} &lt; \infty$$  converges, where $|T| = (T T^{*})^{\frac{1}{2}}$ is the absolute value of the operator.  Assume that  $$\sum_{i}{\langle Te_{i}, e_{i} \rangle}$$  converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?  The progress on the problem is the following: Given an arbitrary bounded operator $T: H \rightarrow H$, one can use the following decomposition $$T = \big( \frac{T + T^{*}}{2} \big) ^{*} + i \big( \frac{T^{*} - T}{2i} \big) ^{*}$$  The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.  For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to  $$(UT U^{-1})(f(x)) = g(x) f(x)$$  where  $$U: H \rightarrow L^{2}(X, \mu)$$ $$g \in L^{\infty}(X, \mu)$$  Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
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The following is an interesting problem from Linear Algebra 2nd Ed - Hoffman &amp; Kunze (3.5 Q17).   Let $W$ be the subspace spanned by the commutators of $M_{n\times n}\left(F\right)$: $$C=\left[A, B\right] = AB-BA$$  Prove that $W$ is exactly the subspace of matrices with zero trace.  Assuming this is true, one can construct $n^2 - 1$ linearly independent matrices, in particular $$[e_{i,n}, e_{n,i}]\ \text{for $1\le i\le n-1$}$$ $$[e_{i,n}, e_{j,n}]\ \text{for $i\neq j$}$$ where $e_{i,j}$ are the standard basis with $0$ entry everywhere except row $i$ column $j$ which span the space of traceless matrices.  However, I have trouble showing (or rather, believing, since this fact seems to be given) that the set of commutators form a subspace. In particular, I am having difficulty showing that the set is closed under addition. Can anyone shed some light?
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Let $X$ be an infinite-dimensional vector space, and let $x_0$ be an element of $X$ such that $f(x_0)=0$ for every linear functional $f$ defined on $X$. Then can we prove that $x_0$ is the zero vector in $X$? Or, is it possible to find a non-zero $x_0$ whose image under every linear functional is zero?   In the case of $X$ being finite-dimensional, we can prove that if $f(x_0)=0$ for every linear functional $f$ on $X$, then $x_0$ must be the zero vector. What is the situation in the infinite-dimensional case?
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This is the fourth part of a four-part problem in Charles W. Curtis's book entitled Linear Algebra, An Introductory Approach (p. 216).  I've succeeded in proving the first three parts, but the most interesting part of the problem eludes me.  Part (a) requires the reader to prove that $\operatorname{Tr}{(AB)} = \operatorname{Tr}{(BA)}$, which I was able to show by writing out each side of the equation using sigma notation.  Part (b) asks the reader to use part (a) to show that similar matrices have the same trace.  If $A$ and $B$ are similar, then  $\operatorname{Tr}{(A)} = \operatorname{Tr}{(S^{-1}BS)}$  $= \operatorname{Tr}(BSS^{-1})$  $= \operatorname{Tr}(B)$,  which completes part (b).  Part (c) asks the reader to show that the vector subspace of matrices with trace equal to zero have dimension $n^2 - 1$.  Curtis provides the hint that the map from $M_n(F)$ to $F$ is a linear transformation.  From this, I used the theorem that $\dim T(V) + \dim n(T) = \dim V$ to obtain the dimension of the null space.  Part (d), however, I'm stuck on.  It asks the reader to show that subspace described in part (c) is generated by matrices of the form $AB - BA$, where $A$ and $B$ are arbitrary $n \times n$ matrices.  I tried to form a basis for the subspace, but wasn't really sure what it would look like since an $n \times n$ matrix has $n^2$ entries in it, but the basis would need $n^2 - 1$ matrixes.  I also tried to think of a linear transformation whose image would have the form of $AB - BA$, but this also didn't help me.  I'm kind of stuck...  Many thanks in advance!