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Suppose $f:\mathbb{R}^n\setminus\{0\}\to \mathbb{R}^m\setminus\{0\}$ is a linear map inducing a map $g:\mathbb{R}P^{n-1}\to \mathbb{R}P^{m-1}$ and hence a homomorphism $$ g^*:\mathbb{Z}/2\mathbb{Z}[x]/(x^m)\to \mathbb{Z}/2\mathbb{Z}[y]/(y^n) $$ on cohomology rings with $\mathbb{Z}/2\mathbb{Z}$-coefficients. Since this is a graded ring map, we have $x\mapsto \lambda y$.

Why does linearity imply that $\lambda=1$? Is there an argument without going into the calculation of the cohomology of the projective space?

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    Notice that taking lambda equal to 1 does not necessarily give a ring morphism. – 2017-01-29
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    I don't really see why... – 2017-01-29
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    @MarianoSuárez-Álvarez Could you please say a little bit more on your comment? Do you mean if $m – 2017-01-29

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First of all notice that $g: \mathbb{R}P^n \to \mathbb{R}P^m$ is well defined provided that $f$ is injective. Consequently $m=n+k$ for some $k \geq 0$.

Since the cohomology ring $H^*(\mathbb{R}P^n)=\mathbb{Z}/2\mathbb{Z}[x]/(x^{n+1})$ is generated by the Poincaré dual of (the homology class of) an hyperplane, we have that $$g^*(x)= g^*PD[\text{hyperplane}]=PD[g^{-1}(\text{generic hyperplane})]=y,$$ where the last identity is because the image of $g$ intersects the generic hyperplane in a $n-1$ dimensional subspace.

Another possibility is to prove that $g$ is isotopic to the inclusion $\mathbb{R}P^n \subset \mathbb{R}P^m$.

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    ...and... ${}{}$ – 2017-01-29
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    I edited, hope it is more clear now – 2017-01-29
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    Sorry I changed notation here $n=n-1$ and $m=m-1$ – 2017-01-29
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    Thank you for the answer. Do you know what the concern ''does not necessarily give a ring morphism'' from the comment above is referring to? I am worried that I understand something incorrectly. – 2017-01-31
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    Well, I don't know. I think he didn't notice that $m\geq n$. – 2017-01-31