Positive Definite Matrix A=YY'

Question

Show $A$ is positive definite, and find a matrix $Y$ such that $A=YY'$.

$$A = \begin{bmatrix} 2 & -1 & 1\\ -1 & 1 & 0\\ 1 & 0 & 2 \end{bmatrix}$$

I am able to show that $$xAx^T = (x_1 - x_2)^2 + \left( x_1 + x_3 \right)^2 + x_3^2$$ which is a sum of squares, therefore non-negative; and zero if $x_1 = x_2 = x_3 = 0.$ Thus $A>0$. But I am unable to find $Y$ such that $$A =YY'$$ I need help on how to compute $Y$ explicitly.

Answer 1

You're almost there. You got \begin{align} \langle Ax,x\rangle &= (x_1 - x_2)^2 + \left( x_1 + x_3 \right)^2 + x_3^2 \\ &= \langle (x_1-x_2,x_1+x_3,x_3),(x_1-x_2,x_1+x_3,x_3)\rangle \\ &= \langle Yx,Yx\rangle \\ &= \langle Y^TYx,x\rangle \end{align} where $Y$ is the matrix which acts as $T(x_1,x_2,x_3) := (x_1-x_2,x_1+x_3,x_3)$, i.e. $$Y = \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 1\end{bmatrix}.$$ Direct computation indeed gives $Y^TY = A$.

Answer 2

You should first diagonalize $ A$ i.e. write $A$ as $QDQ^T$ where $Q$ is orthogonal and $D$ is a diagonal matrix consisting of the positive eigenvalues of $A$. Then put $Y=Q\sqrt{D}Q^T$ where $\sqrt{D}$ is the diagonal matrix formed by square rooting the diagonal entries of $D$.