I have to show that the matrix $M^TM$ is SPD if and only if the columns of the matrix M are linearly independent I have been given M be an $m\times n$ matrix. I have to show that the matrix $M^TM$ is symmetric positive definite if and only if the columns of the matrix $M$ are linearly independent
My thoughts are as below $M^TM$ looks like a cholesky property and we know it is applicable as the matrix is SPD, still not able to connect how and which specific property to use to show if and only if columns of M are linearly independent... 
Any hint as to how I should approach this...
 A: It is obvious that $M^T M$ is symmetric.
Suppose the matrix $M^T M$ is SPD but $M$ does not have linearly independent columns. Then $M$ is not injective, and consequently $M^T M$ is not definite.
Suppose $M$ has linearly indepedent columns, then $M^T$ has linearly indepedent rows. For any $x$ let $y = Mx$, then $x^T M^T M x = y^T y$. This product is zero if and only if $y$ is zero. Then $x$ must have been zero. Whence $x^T M^T M x = 0$ if and only if $x$ is zero. Whence the matrix is symmetric positive-definite.
A: If and only if proofs go 2 ways:
Since this is homework, I'll throw in some hints:
Given that columns are independent, prove $M^TM$ is symmetric positive definite.
For any matrix (Singular or not), $M^TM$ is always symmetric. (Try transposing it to see what you get).
Now, if columns are independent, what can be said about its eigenvalues? 
So, $Ax=\lambda x$, premultiply both sides with $A^T$,
$A^TAx=A^T\lambda x = \lambda (A^Tx) = ?? $.

Given that $M^TM$ is SPD, prove columns are independent. 
If $M^TM$ is SPD, then again the symmetric part is fairly useless; the useful part is the PD. Let $x$ be a eigenvector of M.
$M^TMx = M^T\lambda x = \lambda M^Tx = \lambda^2x$
So now, if M was singular. There would be a $\lambda=0 \text{ but } x \neq 0 \text { s.t. } M^TMx=0$
How does this violate the our assumption of SPD ?
