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If $ m > n$, where $m$ and $n$ are columns and rows of A respectively, then $A^T*A$ is always a symmetric matrix which is not invertible. I am sure this must be an outcome of a simple theorem which I am unable to figure out. Can you please help.

honeybadger
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  • What theorems have you learned thus far? Have you tried searching via Google? What are your thoughts on it, other than guessing there must be a theorem or proof? It looks to me as though **you are being asked to prove this, given the material already covered in class, and/or in a text... – amWhy Aug 10 '17 at 21:14

4 Answers4

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Interpreting in terms of linear maps:

If $A^tA$ were invertible, I would correspond to a bijective linear map, hence $A$ would correspond to an injective linear map from $K^m$ to $K^n$ ($K$ is the base field) and $A^T$ to a surjective linear map from $K^n$ to $K^m$. Both are impossible if $m>n$.

Bernard
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The rank of any $m \times n$ matrix is at most $\min\{m,n\}$. Additionally, the rank of the product of two matrices $AB$ is bounded above by the ranks of $A$ and $B$. Putting these together, we have that $A^TA$ has rank at most $n$, which is strictly less than $m$; since it is an $m \times m$ matrix, it does not have full rank and cannot be inverted.

platty
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$A$ is an $n \times m$ matrix.

$A^TA$ is an $m \times m$ matrix.

However, $\operatorname{rank}(A^TA)=\operatorname{rank} (A) \leq n < m$, hence it is not invertible.

Siong Thye Goh
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$\ker A\subseteq \ker(A^TA)$ and $\dim\ker A\ge m-n$ by rank-nullity.