Let $A$ be an $m\times n$ matrix. Suppose that the columns of $A$ are linearly independent.
(a) Is $A^TA$ invertible? Explain.
(b) Is $AA^T$ invertible? Explain
I have a solution for A but not sure about B.
a) $A$ is a $m\times n$ matrix, so $A^T$ is a $n\times m$ matrix. So $A^TA$ is a $n\times n$ square matrix. So a square matrix with linearly independent columns -> RREF will have $n$ pivot columns be $n\times n$. $I_K$ is invertible.
Vector $v$ in $N(A^TA) \Longrightarrow v^Ta^TAv = 0v^T = 0v = 0$
$v^TA^T=(Av)^T$
$(Av)^TAv=0 \Longrightarrow (Av)(Av) = 0 \Longrightarrow ||Av||^2 = 0 $
$Av= 0 $
If vector $v$ is in $N(A^TA)$ then $v$ is in $N(A)$. $N(A^TA)=N(A)=0$
$(A^TA)x=0$ only solution is $x=0$.
Columns of $A^TA$ are linearly independent and a square matrix. So the RREF of $(A^TA) = I_K$. Therefor $A^TA$ is invertible.
Can someone help me with question b?
If $m \ne n$ and $A$ has independent columns, then the columns of $A^T$ must be dependent. Hence there exists $x\ne 0$ such that $A^Tx=0$, and hence $AA^Tx=0$, and hence $AA^T$ is not invertible.