Proof of non-singularity of triple matrix product

[IniquiTrance]

Homework Statement

Suppose $A\in \mathbb{R}^{n\times n}$ is symmetric positive definite, and therefore non-singular. Let $M\in\mathbb{R}^{m\times n}$. Show that the matrix $M^T A M$ is non-singular if and only if the columns of $M$ are linearly independent.

Homework Equations

The Attempt at a Solution

This is what I have:

If the column vectors of $M$ are linearly independent, then since $M^T AM= col(m_i^T A m_i)_{i=1:n}$, and $A$ is SPD, the columns of $M^T A M$ are all independent non-zero vectors. Thus it is non-singular.

In the other direction, if $M^T A M$ is non-singular, then it's column vectors must be linearly independent. Since its columns are all $col(m_i^T A m_i)_{i=1:n}$, the non-singularity of $M^T A M$ implies they are all linearly independent of each other, which can only be true if the $m_i$ are linearly independent of each other.

Does this seem right?

 

[mighty2000]

Your solution is correct. The key point here is that a matrix M^T A M is non-singular if and only if its columns are linearly independent. This is because a non-singular matrix has full rank, meaning that its columns span the entire vector space. Therefore, if the columns are not linearly independent, there would be redundant information and the matrix would not have full rank. On the other hand, if the columns are linearly independent, they span the entire vector space and the matrix has full rank.

Your explanation of how the columns of M^T A M are related to the columns of M is also clear and concise. Keep up the good work!