Can functionals be independent?

Question

I was asked this question:

Let $A$ be a 2x2 matrix with real entries. Let $\xi_1, \xi_2, \xi_3$ be functionals from $mat_{2x2}(\mathbb R)\to\mathbb R$ such that:

$\xi_1(A)= tr(A)$, $\xi_2(A)=tr(A\begin{pmatrix} 0 & 1\\2 & 0 \end{pmatrix})$, $\xi_3=a_{11}+a_{12}+a_{21}+a_{22}$.

We are asked to show if the functionals are linearly independent.

My question is...How exactly can we say if functionals are linearly independent? What does that even mean that functions are linearly independent? are we just asked to show that $\alpha \xi_1(A)+\beta \xi_2(A) +\gamma \xi_3(A)=0$ implies $\alpha = \beta = \gamma =0$?

If that's the case, can't we just say $A=\begin{pmatrix} 1 & 2 \\ -3 & -1\end{pmatrix}$, and see that $\xi_1(A)+\xi_2(A)+\xi_3(A) =0+1-1=0$ and so they are not linearly independent? (if that is indeed the definition of linearly independent for functions...)

Answer 1

Remember, if $f, g : S \to T$ are two functions between sets, then $f = g$ as functions $S \to T$ if and only if $f(s) = g(s)$ for all $s \in S$.

Hence, linear independence of $\xi_1$, $\xi_2$ and $\xi_3$ would mean that if $\alpha \xi_1 + \beta \xi_2 + \gamma \xi_3 = 0$ as a functional, i.e., if $$ (\alpha \xi_1 + \beta \xi_2 + \gamma \xi_3)(A) := \alpha \xi_1(A) + \beta \xi_2(A) + \gamma \xi_3(A) = 0 $$ for every $A$, then $\alpha = \beta = \gamma$. All that the $A$ you've found demonstrates is that $\xi_1 + \xi_2 + \xi_3$ has a non-trivial kernel.

---

To really check whether or not $\xi_1$, $\xi_2$, and $\xi_3$ are linearly independent, it's worth writing out $(\alpha \xi_1 + \beta \xi_2 + \gamma \xi_3)(A)$ for general $\alpha$, $\beta$, $\gamma$, and $A$: $$ (\alpha \xi_1 + \beta \xi_2 + \gamma \xi_3)(A) := \alpha \xi_1(A) + \beta \xi_2(A) + \gamma \xi_3(A) \\ = \alpha\operatorname{Tr}(A) + \beta\operatorname{Tr}\left(A\begin{pmatrix} 0 & 1\\ 2 & 0\end{pmatrix}\right) + \gamma(a_{11}+a_{12}+a_{21}+a_{22})\\ = \alpha(a_{11}+a_{22}) + \beta\operatorname{Tr}\begin{pmatrix}2a_{12} & a_{11} \\ 2a_{22} & a_{21} \end{pmatrix}+\gamma(a_{11}+a_{12}+a_{21}+a_{22})\\ = \alpha(a_{11}+a_{22}) + \beta(2a_{12} + a_{21}) + \gamma(a_{11}+a_{12}+a_{21}+a_{22})\\ = (\alpha+\gamma)a_{11}+(2\beta+\gamma)a_{12}+(\beta+\gamma)a_{21}+(\alpha+\gamma)a_{22}. $$ Now, the functionals $\phi_{ij} : A \to a_{ij}$ form the basis for $(M_2(\mathbb{R})^\ast$ dual to the usual basis $\{E_{ij}\}$ of $M_2(\mathbb{R})$ (i.e., $E_{ij}$ is the matrix with $1$ as its $ij$ entry, and all other entries $0$). Hence, by this computation, $$ \alpha\xi_1 + \beta\xi_2+\gamma\xi_3 = (\alpha+\gamma)\phi_{11} + (2\beta+\gamma)\phi_{12}+(\beta+\gamma)\phi_{21}+(\alpha+\gamma)\phi_{22}. $$ What can you therefore conclude?

Answer 2

Linearly dependent means that one functional can be represented as a linear combination of the others. Thus, if there is one pair $\alpha, \beta \in \mathbb{R}$ such that $$ \xi_1(A) = \alpha\xi_2(A) + \beta\xi_3(A) $$ holds for every $A$ then $\xi_1$,$\xi_2$,$\xi_3$ are linearly dependent.

To show linear independence, look for $A_1$, $A_2$, $A_3$ such that $\xi_i(A_j)$ is zero for $i\neq j$ and non-zero for $i=j$. If you find such $A_1$,$A_2$,$A_3$ then the $\xi_i$ cannot be linearly dependent.

Answer 3

I see from another question that you know the concept of a dual vector space. The functionals here are elements of the dual space of the space of $2\times2$-matrices. The question is if they are independent as elements of (“vectors in”) this space. The others have spelt that out for you.

Let me add that probably you know what the dimension of that space is and you can write down a basis for it. You can then always write down the functional in that basis (actually there is not much to do there, but that is something that you should understand) and in this way reduce it to testing the linear dependence of three vectors in some $\mathbb R^n$.