Let $A\in\mathbb{R}^{n\times n}$, let $I_n$ denote the identity matrix of order $n$, and let $ \mathrm{col}$ denote column space.
I'm interested in understanding for what values of $\lambda \in \mathbb{C} $ there exists a full column rank matrix $B\in\mathbb{R}^{n\times m}$, with $m<n$, such that $\mathrm{col}(\lambda I_n-A)\subseteq \mathrm{col}(B) $.
Clearly $\lambda$ must be an eigenvalue of $A$ because otherwise $\mathrm{col}(\lambda I_n-A)$ is $n$-dimensional and hence cannot be in the $m$-dimensional subspace $\mathrm{col}(B) $.
If $\lambda$ is a real eigenvalue of $A$, we can certainly find a $B$ such that $\mathrm{col}(\lambda I_n-A)\subseteq \mathrm{col}(B) $.
What if $\lambda$ is a (non-real) complex eigenvalues of $A$?
It is not possible to find this matrix $B_{n\times m}$, $m<n$, for any real matrix $A_{n\times n}$ if $\lambda$ is not real.
Let $A_1,\ldots,A_n$ be the columns of $A$. Let $\lambda=a+bi$ with $b\neq 0$.
Let $C_1,\dots,C_n$ be the columns of $\lambda Id-A$. Thus $C_j=(ae_j-A_j)+i(be_j)$, where $\{e_1,\ldots,e_n\}$ is the canonical basis of $\mathbb{R}^n$.
Now suppose the complex vector space spanned by the columns of $B_{n\times m}$, $\{B_1,\ldots,B_m\}\subset \mathbb{R}^n$, contains the complex vector space spanned by the columns $C_1,\dots,C_n$.
Thus, $C_j=(a_1^j+ib_1^j)B_1+\ldots+(a_m^j+ib_m^j)B_m=(a_1^jB_1+\ldots+a_m^jB_m)+i(b_1^jB_1+\ldots+b_m^jB_m)$, where $a_s^k,b_s^k\in\mathbb{R}$.
So $ae_j-A_j=a_1^jB_1+\ldots+a_m^jB_m$ and $be_j= b_1^jB_1+\ldots+b_m^jB_m.$
Therefore, the real vector space spanned by $\{B_1,\ldots,B_m\}$ must contain the real vector space spanned by $\{be_1,\ldots,be_n\}$, which is $\mathbb{R}^n$, since $b\neq 0$.
Thus, $m$ must be bigger or equal to n.