Am I showing the positive-definiteness of a symmetric matrix A correctly? A is a 4x4 square, symmetric matrix.  First I computed its Cholesky Decomposition $LL^t$ since the first part of the question asked for it.  Then it asked to show A was positive-definite.  I am a little shaky about some of the last parts of my proof:
For $\vec z$ $\ne$ $\vec 0$, we have that 
$$z^tAz$$
$$=z^tLL^tz$$
$$=(z^tL)(L^tz)$$
$$=<(z^tL),(L^tz)>$$
$$=<(z^tL),(z^tL)^t>$$
$$=||z^tL||^2$$
$$=||L^tz||^2$$
$$>0$$
since both $L$ and $L^t$ have trivial null space (both are uni-triangular and invertible), and z was assumed to be a non-zero vector.  Thus the image $L^t(z)$ is not the zero vector, and so the above vector-norm squared is positive, which shows that A is positive-definite.
Is this ok? The last few equalities feel a little awkward.
Thanks,
 A: Your arguments are fine.
The only thing that might feel a little awkward is your notation of the inner product $\langle \cdot,\cdot \rangle$:
Obviously it is not the usual dot product of two column vectors since you have the row vector $z^tL$ and the column vector $L^t z$ in there together. What you mean in the 3$^\text{rd}$ line is the usual matrix product of an $1\times n$-matrix $a$ and an $n\times 1$-matrix $b$ (usual notation: $ab$) which is the same as taking the dot product of the column vectors $a^t$ and $b$ (usual notation: $a^t\cdot b$ or $\langle a^t, b\rangle$). That's probably what you wanted to express in the 4$^\text{th}$ line.
[Note that the matrix product is not an inner product in general.]
Suggestion for a fix: After your 3$^\text{rd}$ line continue with $$\ldots\\=(z^tL)(L^tz)\\=\langle (z^tL)^t, L^tz\rangle\\=\langle L^tz,L^t z\rangle\\=\lVert L^t z\rVert^2\\>0$$
with the same arguments you gave.
(If you have a different notation for the inner product that allows what you wrote there you probably may discard my concerns.)
