Let $A$ be any matrix. How do I prove that $\mathrm{Im}(A)=\mathrm{Im}(AA^T)$?

Question

How do I prove that $\mathrm{Im}(A)=\mathrm{Im}(AA^T)$?

Is there any property that makes this the case, because I don't know how to prove this.

Answer 1

(1) $\text{Im}(AA^{T}) \subset \text{Im}(A)$: Suppose $ y \in \text{Im}(AA^{T})$, then $y = AA^{T}x = Au \implies y \in \text{Im}(A)$.

(2) $x \in \ker (A^{T}) \implies A^{T}x = 0 \implies AA^{T}x = 0 \implies \ker (A^{T}) \subset \ker (AA^{T}).$ On the other hand, if $x \in \ker (AA^{T})$, then $$||A^{T}x||^{2} = \langle A^{T}x, A^{T}x \rangle = \langle x, AA^{T}x \rangle = \langle x, 0 \rangle = 0.$$ It follows that $A^{T}x = 0$. Therefore we have shown $\ker (A^{T}) = \ker (AA^{T})$. By the rank nullity theorem, it follows that $\text{rank}(A A^{T}) = \text{rank}(A^{T}) = \text {rank} A$. But combining this with (1) gives the result as it was promised.

It is important the matrix $A$ has real entries:

$A = \begin{bmatrix}1 & i \\ \end{bmatrix}$. Then $A^{T} = \begin{bmatrix}1 \\ i \\ \end{bmatrix}$. So that $AA^{T} = 0$ and the result obviously fails.

Answer 2

Note: this is only true for matrices over the real numbers. When matrices are taken over the complex numbers, this can fail to be true.

Hint:

The fact that $im(AA^T)\subseteq im(A)$ is trivial which I leave to you to justify.

The big part of the proof is trying to show the reverse inclusion.

This can be done by showing that $rank(A)=rank(AA^T)$ since both $im(AA^T)$ and $im(A)$ are vector spaces. If one vectorspace is a subspace of another and they are of the same finite dimension, they must be equal.

Have you yet seen the proof for why $rank(A)=rank(A^T)=rank(AA^T)=rank(A^TA)$?

Prove rank $A^TA$ = rank $A$ for any $A_{m \times n}$