I don't think they're bad. I think you can make the problem much simpler by doing column operations to write A = A'C where C is the matrix representing the column operations, and A' is of a special form. Of course, it means you have to pick a good special form, but hey! Math is an art.
Proving invertibility of A^TA means showing that the matrix A^TA has an inverse, meaning that it can be multiplied by another matrix to give the identity matrix. This is important in linear algebra because an invertible matrix has many useful properties and allows for the solution of equations involving that matrix.
Having linearly independent columns in A^TA is important because it ensures that the matrix is full rank, meaning that its columns span the entire space. This is necessary for the matrix to have an inverse, as a matrix with linearly dependent columns cannot have an inverse.
To prove that A^TA has linearly independent columns, you can use the determinant test. If the determinant of A^TA is non-zero, then the columns are linearly independent. Another method is to use the rank-nullity theorem, which states that the rank of a matrix is equal to the number of non-zero eigenvalues.
Proving invertibility of A^TA has many implications in linear algebra. It allows for the solution of equations involving that matrix, and also shows that the matrix has many useful properties such as being orthogonal and having a unique solution to the equation Ax = b. It is also used in various applications, such as data compression and image processing.
No, A^TA cannot be invertible if it has linearly dependent columns. This is because a matrix with linearly dependent columns will have a determinant of 0, meaning it does not have an inverse. However, it is possible for A^TA to be invertible even if A does not have linearly independent columns, as long as the columns of A^TA are linearly independent.