Combining $4$ outer products into $3$ Let $\mathbf{a}$ and $\mathbf{b}$ be $n\times 1$ vectors, and, for some $u<n$,
$$\mathbf{\Lambda} = \begin{bmatrix}\mathbf{I}_{u} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{bmatrix}_{n\times n}.$$
Further, suppose that the last $n-u$ elements of $\mathbf{a}$ are equal to zero, so that $\mathbf{\Lambda}\mathbf{a}=\mathbf{a}$.
I want to show that, for arbitrary scalars $\chi_1$, $\chi_2$, $\chi_3$, and $\chi_4$, there exists $n\times 1$ vectors $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$ such that
\begin{align}
\mathbf{x}\mathbf{x}'+\mathbf{y}\mathbf{y}'+\mathbf{z}\mathbf{z}'=&\;
\chi_1^2(\mathbf{b}-\mathbf{a})(\mathbf{b}'-\mathbf{a}')+\chi_2^2\mathbf{\Lambda}(\mathbf{b}-\mathbf{a})(\mathbf{b}'-\mathbf{a}')\mathbf{\Lambda}+\\[1.5ex]
&\;\chi_3^2(\mathbf{I}_n-\mathbf{\Lambda})(\mathbf{b}-\mathbf{a})(\mathbf{b}'-\mathbf{a}')(\mathbf{I}_n-\mathbf{\Lambda})+\chi_4^2(\mathbf{a}\mathbf{b'}\mathbf{\Lambda}+\mathbf{\Lambda}\mathbf{b}\mathbf{a}').\tag1
\end{align}
If it wasn't for the last term on the right-hand side, $\chi_4^2(\mathbf{a}\mathbf{b'}\mathbf{\Lambda}+\mathbf{\Lambda}\mathbf{b}\mathbf{a}')$, this could be trivially achieved by setting
$$
\mathbf{x}=\chi_1(\mathbf{b}-\mathbf{a}),\quad
\mathbf{y}=\chi_2\mathbf{\Lambda}(\mathbf{b}-\mathbf{a})\quad\text{and}\quad
\mathbf{z}=\chi_3(\mathbf{I}_n-\mathbf{\Lambda})(\mathbf{b}-\mathbf{a}).
$$
So, the question boils down to how to incorporate the last term. Any help would be much appreciated.

Context
I am trying to prove the following conjecture which I've checked is true numerically. If I can prove the result above I think I have a proof for the conjecture.
In addition to the definitions above, let $\mathbf{B}$ be an $m\times n$ matrix, with $m<n$, and $\mathbf{h}$ be a $m\times 1$ vector that satisfies the following fixed point problem
\begin{align}
\mathbf{h}'=&\;\mathbf{a}'\mathbf{\Omega}\mathbf{B}'\left(\mathbf{B}\mathbf{\Omega}\mathbf{B}'\right)^{-1}\\[1.5ex]
\mathbf{\Omega}=&\;\mathbf{I}_n+\left(\mathbf{\Delta}^{-1}+ \mathbf{G} \right)^{-1},
\end{align}
where $\mathbf{\Delta}$ is a block diagonal matrix with each block
being positive semidefinite
$$
\mathbf{\Delta}=\begin{bmatrix}
\mathbf{\Delta}_{1} & \mathbf{0}_{u,n-u}\\
\mathbf{0}_{n-u,u} & \mathbf{\Delta}_{2}%
\end{bmatrix},
$$
and the matrix $\mathbf{G}$ is given by the right-hand side of equation $(1)$ with $\mathbf{b}=\mathbf{B}'\mathbf{h}$.
The conjecture is that the vector $\mathbf{h}$ satisfies
$$
\mathbf{h}'=\mathbf{a}'\mathbf{\Gamma}\mathbf{B}'\left(\mathbf{B}\mathbf{\Gamma}\mathbf{B}'\right)^{-1},
$$
where the matrix $\mathbf{\Gamma}$ is given by
$$
\mathbf{\Gamma}=\mathbf{I}_n+w_{1}
\begin{bmatrix}
\mathbf{\Delta}_{1} & \mathbf{0}_{u,n-u}\\
\mathbf{0}_{n-u,u} & \mathbf{0}_{n-u,n-u}
\end{bmatrix}
+w_{2}
\begin{bmatrix}
\mathbf{0}_{u,u} & \mathbf{0}_{u,n-u}\\
\mathbf{0}_{n-u,u} & \mathbf{\Delta}_{2}
\end{bmatrix}
+w_{3}\mathbf{a}\mathbf{a}'\mathbf{\Delta}
$$
for some constants $w_1$, $w_2$ and $w_3$.
 A: The statement is true if $χ_2^2+\frac{(χ_1χ_3)^2}{χ_1^2+χ_3^2} ≥ χ_4^2/2$, otherwise it might be false.
First, let $c = b-a$ so that the equation to solve becomes
$$\begin{align*}
xx' + yy' + zz' &= χ_1^2cc' + χ_2^2Λcc'Λ + χ_3^2(I_n-Λ)cc'(I_n-Λ)+χ_4^2(a(c+a)'Λ+Λ(c+a)a')\\
&= (χ_1^2+χ_3^2)cc' + (χ_2^2+χ_3^2)Λcc'Λ - χ_3^2(Λcc' + cc'Λ) + 2χ_4^2aa' + χ_4^2 (ac'Λ+Λca') \text{.}
\end{align*}$$
We will guess that $x,y,z$ are of the form $α_ic + β_iΛc + γ_ia$ for $i=1,2,3$, since this is the simplest way to build vectors out of the data we are given. Then
$$\begin{align*}
xx' + yy' + zz' &= \sum_{i=1}^3 α_i^2 cc' + β_i^2 Λcc'Λ + γ_i^2aa' + α_iβ_i(cc'Λ+Λcc') + α_iγ_i(ca'+ac') + β_iγ_i(Λca'+ac'Λ)\\
&= α^2 cc' + β^2 Λcc'Λ + γ^2aa' + αβ(cc'Λ + Λcc') + αγ(ca'+ac')+βγ(Λca'+ac'Λ) \text{.}
\end{align*}$$
(I write $αβ$ for the scalar product of $α$ and $β$.) Matching this expression against the target expression gives the following system.
$$\begin{align*}α^2 &= χ_1^2+χ_3^2\\
β^2 &= χ_2^2+χ_3^2\\
γ^2 &= 2χ_4^2\\
αβ &= -χ_3^2\\
αγ &= 0\\
βγ &= χ_4^2
\end{align*}$$
Let's suppose we are not in an edge case with $χ_i = 0$. We can solve this system geometrically. Choose two orthogonal vectors $α$ and $γ$ with the desired lengths. Then $αβ = -χ_3^2$ and $βγ = χ_4^2$ tell us the projection of $β$ on the plane generated by $α$ and $γ$. The minimal square-norm of $β$ is (I skip the computation) $χ_4^2/2 + \frac{χ_3^4}{χ_1^2+χ_3^2}$. Since $β$ should have a square-norm of $χ_2^2+χ_3^2$, in order to have a solution, we must have $χ_2^2+χ_3^2 ≥ χ_4^2/2 + \frac{χ_3^4}{χ_1^2+χ_3^2}$, or in other words $χ_2^2+\frac{(χ_1χ_3)^2}{χ_1^2+χ_3^2} ≥ χ_4^2/2$. If this is the case, there is a solution.
To find a counter-example, we need to take $χ_2^2+\frac{(χ_1χ_3)^2}{χ_1^2+χ_3^2} < χ_4^2/2$. For instance, with $n=3$ and $u=1$, take $c = (1,1,1)$, $a=(1,2,0)$, $χ_1 = 1$, $χ_2 = 0$, $χ_3 = 1$, $χ_4 = 3$. Then if my computations are correct, we find
$$\begin{pmatrix}
37&64&1\\
64&109&1\\
1&1&2
\end{pmatrix}$$
which is not positive semidefinite.
The condition $χ_2^2+\frac{(χ_1χ_3)^2}{χ_1^2+χ_3^2} ≥ χ_4^2/2$ might even be necessary. To show that, one could try to use the fact that even if this condition fails, there is a solution $β$ but with complex coefficients. This can be used to build a solution $x,y,z$, but there are complex coefficients: can this be used in some way to deduce that $xx'+yy'+zz'$ is not positive definite?
PS: Thank you for the context.
A: This is not true in general because the $\mathbf {a}\mathbf b'\Lambda + \Lambda\mathbf b\mathbf {a}'$ is not semidefinite positive. For example $$\mathbf a = \begin{bmatrix}1\\ 0\end{bmatrix}, \mathbf b = \begin{bmatrix}-1\\ 0\end{bmatrix}$$
