Prove that $im(T^{k+1}) ⊆ im(T^k)$ for every non-negative integer $k$ for any linear transformation $T: R^n → R^n$

Question

Prove that $im(T^{k+1}) ⊆ im(T^k)$ for every non-negative integer $k$ for any linear transformation $T: R^n → R^n.$

I planned to try and prove this by induction, but I'm very unsure where to even begin with actually proving this. I know that $T^{k+1} = T^k*T$, and that the image is the set of linearly independent column vectors of $A$ for $T(x) = Ax.$ Doing the base case would show that $im(T) ⊆ R^n$, since $T^0 = I_n$, and $im(I_n) = R^n$. Any suggestions?

Answer 1

The image is not the set of linearly independent column vectors of $A$, but the span of its column vectors, or what amounts to the same, the span of a maximal set of linearly independent column vectors.

It's easier to adopt the point of view of linear maps: as $\; T^{k+1}= T^k\circ T$, it is obvious that $\;\DeclareMathOperator\Im{Im}\Im(T^{k+1})=T^{k+1}(\mathbf R^n)=T^k\bigl(T(\mathbf R^n)\bigr)$.

Now, as $T(\mathbf R^n)\subset \mathbf R^n$, transforming by $T$ we get $$T^{k+1}(\mathbf R^n)=T^k\bigl(T(\mathbf R^n)\bigr)\subset T^k(\mathbf R^n).$$

Answer 2

What you said the image is is actually the basis of the image. The image of $T$ is just a set of all possible values of $T$. Once you internalize this properly, the problem should look almost entirely trivial. The base case is trivial because it just says that the image of $T$ is a subset of $R^n$, which is true because it's a transformation from $R^n$ to $R^n$. For the inductive step, you need to show that any value of $T^{k+1}$ can be represented as a value, under some argument, of $T^k$.