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I need to prove that $A^T$$A$ is an invertible matrix.

$$ A= \begin{bmatrix} \vec{a_1} & \vec{a_2} & \ldots & \vec{a_n} \\ \end{bmatrix} $$

Can I prove this using a vector $\vec{y}$ and numbers $k_1$, $k_2$, $\ldots$, $k_n$

$A^T$$A$$\vec{y}$ = $k_1$$\vec{a_1}$ + $k_2$$\vec{a_2}$ + $\ldots$ + $k_n$$\vec{a_n}$

i.e. can I get a linear combination from $A$'s columns?

$A$ $=$ $M$x$N$ , $M$ $>$ $N$, and $A$ is not a $zero$ matrix.

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    $\begingroup$ This need not be true. For example take $A$ to be the zero matrix. You need $A$ to be full rank. $\endgroup$ – Anurag A Jan 29 '16 at 11:04
  • $\begingroup$ Moreover if $A$ is an $m \times n$ matrix one needs $m \geq n$, as $A^T A$ is an $n \times n$ matrix and $\operatorname{rank}(A^T A) \leq \min(\operatorname{rank} A^T, \operatorname{rank} A) = \operatorname{rank} A \leq \min(m, n)$. $\endgroup$ – Travis Willse Jan 29 '16 at 11:12
  • $\begingroup$ A = MxN , M > N, and A is not a zero matrix. $\endgroup$ – Hydroxis Jan 29 '16 at 12:16
  • $\begingroup$ If $A$ is not quadratic, you can't write $A^TAy$ as $k_1\vec{a}_1+...+k_n\vec{a_n}$ because of a dimension mismatch. $\endgroup$ – Gregor de Cillia Feb 1 '16 at 2:10
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Your conditions with respect to the matrix $A$ are a little bit too vague to post a proper answer, but that might help. I will show, that a matrix $A^TA$ is invertible if and only if it $A$ has a trivial kernel, i.e if

$$ Ax = \vec{0}\Rightarrow x = \vec{0},\ x\in R^n. $$

Classical proof

Suppose $A^TAx$ is $\vec{0}$, then $$\vec{0} = A^TAx \Rightarrow 0 = x^TA^TAx = (Ax)^T(Ax) = \|Ax\|^2 \Rightarrow Ax = \vec{0}$$

On the contrary, if $Ax = \vec{0}$, it holds that $A^TAx = A^T\vec{0} = 0$.


Proof using columns

To show the same result using columns, denote the columns as you did in the question with $a_1,...,a_n$. The following orthogonality can be shown.

$$ A^TAx = [a_1|a_2|...|a_n]^TAx = 0\Rightarrow Ax\perp a_i,\ i= 1,...,n $$ Now, since the orthogonal space of $Ax$ is a linear space, we have that $$ Ax \perp \text{span}\{a_i,i=1,...,n\}. $$ Since $\text{span}\{a_i,i=1,...,n\}$ is exactly the image set of $A$ (if we regard it as a map). $$ Ax\in \text{span}\{a_i,i=1,...,n\}. $$ We can conclude $Ax=\vec{0}$. The contrary can be shown as above.

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  • $\begingroup$ Your comment answered my question. Thanks. $\endgroup$ – Hydroxis Feb 1 '16 at 15:12

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