Let $A$ be an invertible $n \times n$ matrix over $\mathbb{R}$, prove that there exist positive constants $c_1$ and $c_2$ such that $$c_1X^tX\leq X^tA^tAX\leq c_2X^tX $$ for all $X \in \mathbb{R^{n\times 1}}$
If we consider the standard inner product on $\mathbb{R^{n\times 1}}$ : $$<X|Y>=X^tY \ \ \ \ \ \ \forall X,Y \in \mathbb{R^{n\times 1}}$$ We are required to prove that $$c_1||X||^2 \leq ||AX||^2 \leq c_2 ||X||^2$$ It is kind of asking by how much $A$ can increase or shrink the length of vectors in $\mathbb{R^{n\times 1}}$. If the field was the complex numbers, then we were sure about existence of eigenvalues and we could use eigenvalues as scaling factors of vectors. How can one use the fact that $A$ is invertible in this problem ?
I already appreciate your helpful comments/answers.
Consider $B=A^tA$ it is a symmetric definite positive matrix; by spectral theorem there is a orthonormal basis formed by the eigenvectors of $B$, denote it by $(V_1,\ldots,V_n)$ we denote $\lambda_1,\ldots,\lambda_n$ the corresponding eigenvalues ($BV_i=\lambda_iV_i$). Now let $X\in \Bbb R^{n\times 1}$, there exists $x_1,\ldots,x_n\in \Bbb R$ such that $X=x_1V_1+\ldots+x_nV_n$, then $BX=x_1\lambda_1V_1+\ldots+x_n\lambda_nV_n$, so $X^tBX=\lambda_1x_1^2+\ldots+\lambda_nx_n^2$, thus $\min(\lambda_i)\|X\|_2^2\leq X^tBX\leq \max(\lambda_n)\|X\|_2^2$.