I am struggling with a problem in Linear Algebra.
Problem#1 Here I tried presenting the product $AX$ as a multiplication of $A$ and $x$. We know that $X$ is invertible, so if we show that $\lambda X$ is the inverse of $A X$ that will mean that both $A$ and $X$ are invertible (I guess...) I tried using the following theorems:
If $A$ is invertible => $\ AA^{-1} = A^{-1}A=I $
If $A$ and $B$ are invertible => $\ (AB)^{-1} = B^{-1}A^{-1} $
P.S.: This is my first post here on the forum.
$A x_i = \lambda_i x_i\\ X = [x_1,x_2\cdots,x_n]$
$AX = X\begin{bmatrix} \lambda_1\\&\lambda_2\\&&\lambda_3\\&&&\ddots\end{bmatrix}$
Lets call this diagonal matrix $\Lambda$
Since every $\lambda_i$ is non-zero, $\det \Lambda \ne 0$
And since $X$ is invertable. The determinant of $X$ is not $0$
$\det(AX) = \det(A)\det(X) = \det(X)\det(\Lambda)\ne 0\\ \det(A)\ne 0$
And $A$ is invertable
What is $A^{-1}$
$(AX)^{-1} = X^{-1} A^{-1} = \Lambda^{-1}X^{-1} = (X\Lambda)^{-1}\\ A^{-1} = X\Lambda^{-1}X^{-1} $
And what do you suppose $\Lambda^{-1}$ is?