# How do I prove dim($A^t A$)=dim($A$) on proving rank($A^TA$)=rank($A$)?

While I'm studying on Linear Algbera, I was stucked in:

Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$

I understood what above means, and I am now curious about

Yes, I understood : N($$A^TA$$)=N($$A$$)

Then, how can I convince that rank($$A^TA$$)=rank($$A$$)?

In the above question, they say it is by rank-nullity theorem

But in my book, it seems like there is a no rank-nullity theroem, but dimension theorem:

Let $$V,W$$ be vector spaces, and let $$T:V->W$$ be linear. If $$V$$ is finite-dimensional, then

nullity($$T$$) +rank($$T$$) = dim($$V$$)

I thought like:

nullity($$A^TA$$) + rank($$A^TA$$) =dim($$A^TA$$) and

nullity($$A$$) +rank($$A$$) = dim($$A$$) So both of the rank would be the same.

But I think the dimensions of $$A^TA$$ and $$A$$ are different.

$$A^TA$$ becomes $${n \times n}$$ matrix and $$A$$ is $${m \times n}$$ matrix so the dimension is different.

Where I was wrong? Or I am misunderstood?

• What's $\operatorname{dim}A$?
– user239203
Mar 4 at 6:34
• @Gae.S. I meant dimension Mar 4 at 6:35
• The rank-nullity theorem states that the rank plus nullity equals the number of columns. Mar 4 at 6:35
• If the "dimension" of an m×n matrix is defined to be n, then indeed m×n and n×n have same dimension and everything works Mar 4 at 6:42

Rank is the maximum number of linearly independent columns of $$A$$, and dim($$A$$) is the number of columns of $$A$$. Same goes for $$A^TA$$. Since they both have $$n$$ columns, the proof you linked is correct.
If you look at $$A$$ and $$A^TA$$ as linear transformations, then we have $$A: \mathbb R^n \rightarrow \mathbb R^m$$, and $$A^TA: \mathbb R^n \rightarrow \mathbb R^n$$. So the right hand side of the dimension theorem is equal to $$n$$ for both $$A$$ and $$A^TA$$.
• So what you mean is $R^n$ means the dimensions when matrix as a linear transformation above? Mar 4 at 7:14
• Yes, because $A$ takes vectors in $\mathbb R^n$ as input. It acts on an $n$-dimensional vector ($A^TA$ acts on the same vector, but the image of $A$ and $A^TA$ are in different spaces).