Prove that $\operatorname{null} T_1 = \operatorname{null} T_2$ iff there exists an invertible operator $S$ - a linear transformation from $W$ to $W$ such that $T_1=ST_2$.

Here we're assuming $L$ is a vector space and $W$ is finite-dimensional vector space. $T_1$ and $T_2$ are linear transformations from $L$ to $W$.

Proving one direction is easy. Suppose $T_1=ST_2$. $S$ is invertible, so we know its nullspace contains only the zero element. So take any non-zero $v$ in the nullspace of $T_1$. So $T_1(v)=0=ST_2(v)$. $S$ only maps the zero element to zero, so $T_2(v)$ must be zero. This proves $\operatorname{null} T_1$ is contained in $\operatorname{null} T_2$. $S$ is invertible, so we can just as easily write $S^{-1}T_1=T_2$, and by the same logic prove $\operatorname{null} T_2$ is contained in $\operatorname{null} T_1$, so $\operatorname{null} T_1 = \operatorname{null} T_2$.

I haven't made any progress on proving the other direction though. Any suggestions?


1 Answer 1


Hint: Here's a proof for finite dimensional spaces: let $\{v_1,\dots,v_k\}$ be a basis for the shared null-space of $T_1$ and $T_2$. Extend this to a basis $\{v_1,\dots,v_n\}$ of $L$.

We want $S:W \to W$ to satisfy $T_1 v_j = ST_2 v_j$ for $j = (k+1),\dots,(n-1),n$. Why can we guarantee that such a map exists?

Try to recast this as a proof via quotient spaces. In particular, we have $\ker(T_1) = \ker(T_2) = K$, and $T_1$ and $T_2$ both induce maps $\tilde T_i: L/K \to W$.


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