# Is $\mathrm{col}(\lambda I_n-A)\subseteq \mathrm{col}(B)$ for a complex $\lambda$?

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 $A$?

• You should be clear about over which field you are working when you talk about column space. The column space over the complex numbers (a complex vector space) is not the same as the column space over the real numbers (not a complex vector space). Working over $\Bbb R$, there is no $\mathrm{col}(\lambda I_n-A)$ unless $\lambda\in\Bbb R$. But if working over $\Bbb C$, restricting $B$ to real matrices is a strange requirement. – Marc van Leeuwen Nov 29 '14 at 12:24

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.