# How to show that the linear mappings $\phi_\mu^\nu$ generates the set $L(E;V)$

In the book of Linear Algebra By Werner Greub at p 76, to show that $$\dim L(E;F) = \dim E \times \dim F$$, it constructs a map $\phi_\mu^\nu : E \to F$ by

$$\phi_\sigma^\lambda (x_\nu) = \delta_\nu^\lambda y_\sigma$$, and claims that these maps forms a basis for $L(E;F)$, but it doesn't give and proof to show that these are linearly independent maps, I'm trying to prove this.

How to show that the linear mappings $\phi_\mu^\nu$ generates the set $L(E;V)$

To show that $$\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu = 0_{L(E;V)}$$, we need to show that $$(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu)(x) = 0\quad \forall x\in E$$

I actually made a proof about it, and I was writing here, so that you can check for any flow, but I have found a mistake during the writing, so how can we prove this ?

To pick it up from where you left off: let's prove that the mappings $\phi_\mu^\nu$ are linearly independent.

We want to show that if $\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu = 0_{L(E;V)}$, then all coefficients must be zero. That is, we need to show that $$\left(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu\right)(x) = 0\quad \forall x\in E \implies \alpha_{\mu}^\nu = 0 \quad \forall \mu, \nu$$ With that, we can plug in $x = x_\mu$ and note that $$\left(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu\right)(x_\sigma) = \sum_{\mu}\alpha^\sigma_\mu x_\mu = 0$$ However, the vectors $x_\mu$ form a basis, so we conclude that $\alpha^\sigma_\mu = 0$ for all $\mu$. Noting that $\sigma$ was also arbitrary, we conclude that every $\alpha_\mu^\sigma = 0$.

Now, there's the separate questions of showing that these maps generate the set $L(E;F)$. That is: given an arbitrary map $\phi:E \to F$, we want to show that there exist coefficients such that $$\phi = \sum_{\mu,\nu} \alpha_\mu^\nu \phi_\mu^\nu$$ The proof is as follows: again, write out $\left(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu\right)(x_\sigma)$. By the definition of a basis, there is a unique set of coefficients $\alpha_\mu^\nu$ that guarantee these equations hold.

With that, we want to show that under these constraints, we have $\phi = \sum \alpha_\mu^\nu \phi_\mu^\nu$. To that end, it suffices to note that these are two linear maps which act the same way on a basis.

• but you need to show the result holds for all x, not just for the basis.Of course you can write any x as a linear combination of the basis, but when the coefficient in that linear combination comes into play in $\left(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu\right)(x) = 0\quad \forall x\in E \implies \alpha_{\mu}^\nu = 0 \quad \forall \mu, \nu$, it just doesn't directly imply that $\alpha_\mu^\sigma = 0.$
– Our
Jun 16, 2017 at 14:58
• @Leth you have the direction of the proof wrong. You know that the map is zero for all $x$, and you want to conclude that the coefficients are zero. Jun 16, 2017 at 18:09
• exactly which point is wrong, I didn't get it, we are saying the same thing ?
– Our
Jun 17, 2017 at 5:02
• You're saying that our goal is to conclude that the map is zero for all $x$. I'm saying we assume that the map is zero for all $x$, and our goal is to conclude that the coefficients are zero Jun 17, 2017 at 5:24
• What I was missing was that we assumed some linear combination of those maps is equal to zero map, so that equality holds for all x, this was our assumption, so since it holds for all x, I we plug in $x = x_\nu$, it should also hold.In your last sentence, saying that we didn't show that equality holds allow me to realised our assumption, so if you have said that as I have said above "so that equality holds for all x, this was our assumption, so since it holds for all x, I we plug in $x = x_\nu$, it should also hold", it could have been clearer.
– Our
Jun 17, 2017 at 17:18