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A Omitted details from Section 3.1
A.1 Implementation of Figure 1
Figure 1 is generated from a simple three-task linear MTL problem that we constructed, using Equation (4). Specifically, we set  $\hat{y}_1 \approx (0.98, 0, 0.2)$ ,  $\hat{y}_2 \approx (-0.49, -0.85, 0.2)$ ,  $\hat{y}_3 \approx (-0.49, 0.85, 0.2)$  (the number of data points is  $n = 3$ ; this is a rotated version of the equiangular tight frame), set  $q = 1$  (the width of the network is one, i.e., under-parameterized), and plotted the achievable points of Equation (4) by sweeping  $P_Z$  (the set of rank-1 projection matrices). The software we used is Mathematica.
A.2 Over-parametrization reduces the Pareto front to a singleton
For general non-linear multi-task neural networks, we will show that increasing the width reduces the Pareto front to a singleton, where all tasks simultaneously achieve zero training loss. As a consequence, linear scalarization with any convex coefficients will be able to achieve this point.
To illustrate this point, we follow the same setting as in Section 2, except changing the model to be a two-layer ReLU multi-task network with bias terms. Concretely, for task  $i$ , the prediction on input  $x$  is given by  $f_{i}(x,W,b,a_{i}) = a_{i}^{\top}\max (Wx + b,0)$ . We have the following result.
Theorem A.1. There exists a two-layer ReLU multi-task network with hidden width  $q = nk$  and parameters  $(W, b, a_1, \dots, a_k)$ , such that
$$
f _ {i} \left(x _ {i} ^ {j}, W, b, a _ {i}\right) = y _ {i} ^ {j}, \quad \forall j \in [ n ], \forall i \in [ k ]. \tag {14}
$$
This implies that the network achieves zero training loss on each task.
Proof. For every  $i \in [k]$ , we can apply Theorem 1 in (Zhang et al., 2021) and find  $(W_i, b_i, \tilde{a}_i)$  such that
$$
\tilde {a} _ {i} ^ {\top} \max  \left(W _ {i} x _ {i} ^ {j} + b _ {i}, 0\right) = y _ {i} ^ {j}, \quad \forall j \in [ n ]. \tag {15}
$$
Here,  $W_{i}\in \mathbb{R}^{n\times p}$  and  $b_{i},\tilde{a}_{i}\in \mathbb{R}^{n}$ . Now consider
$$
W = \left( \begin{array}{c} W _ {1} \\ W _ {2} \\ \vdots \\ W _ {n} \end{array} \right) \in \mathbb {R} ^ {n k \times p}, \quad b = \left( \begin{array}{c} b _ {1} \\ b _ {2} \\ \vdots \\ b _ {n} \end{array} \right) \in \mathbb {R} ^ {n k}, \quad a _ {i} = e _ {i} \otimes \tilde {a} _ {i} \in \mathbb {R} ^ {n k} \quad \forall i \in [ n ], \tag {16}
$$
where  $e_i$  denotes the  $i$ -th unit vector in  $\mathbb{R}^n$  and  $\otimes$  stands for the Kronecker product. It is straightforward to see that
$$
a _ {i} ^ {\top} \max  \left(W _ {i} x _ {i} ^ {j} + b _ {i}, 0\right) = y _ {i} ^ {j}, \quad \forall j \in [ n ], \forall i \in [ k ]. \tag {17}
$$
This finishes the proof as desired.
Proof of Theorem 3.1. Define  $v_{i} = \langle \hat{y}_{i}, s \rangle$  and  $v = (v_{1}, \dots, v_{k})^{\top}$ . We are interested in the set
$$
S := \left\{v \mid \| s \| = 1, s \in \operatorname {s p a n} \left(\left\{\hat {y} _ {i} \right\} _ {i \in [ k ]}\right) \right\}. \tag {18}
$$
We will show  $S$  is equivalent to the following set
$$
B := \left\{v \mid v ^ {\top} Q v = 1 \right\}, \quad Q = \left(\hat {Y} ^ {\top} \hat {Y}\right) ^ {- 1}, \tag {19}
$$
which is essentially the boundary of an ellipsoid.
Step 1:  $S\subset B$  Note  $v = \hat{Y}^{\top}s$  , so it suffices to show
$$
s ^ {\top} \hat {Y} (\hat {Y} ^ {\top} \hat {Y}) ^ {- 1} \hat {Y} ^ {\top} s = 1. \tag {20}
$$
Denote  $P_{\hat{Y}} = \hat{Y} (\hat{Y}^{\top}\hat{Y})^{-1}\hat{Y}^{\top}$ , which is a projection matrix that maps to the column space of  $\hat{Y}$ . Since  $s\in \mathrm{span}(\{\hat{y}_i\}_{i\in [k]})$ , we have
$$
s ^ {\top} P _ {\hat {Y}} s = s ^ {\top} s = \| s \| ^ {2} = 1. \tag {21}
$$
Step 2:  $B \subset S$ . Since  $\hat{Y}^{\top}$  has full row rank, for every  $v \in \mathbb{R}^k$ , we can always find  $s \in \mathbb{R}^n$  such that  $\hat{Y}^{\top}s = v$ . We can further assume  $s \in \operatorname{span}(\{\hat{y}_i\}_{i \in [k]})$  by removing the component in the null space of  $\hat{Y}^{\top}$ . Assuming  $v \in B$ , we have
$$
s ^ {\top} P _ {\hat {Y}} s = 1, \quad s \in \operatorname {s p a n} \left(\left\{\hat {y} _ {i} \right\} _ {i \in [ k ]}\right). \tag {22}
$$
Since  $P_{\hat{Y}}$  is a projection matrix that maps to the column space of  $\hat{Y}$ , we have  $s^{\top}s = 1$ , which implies  $\| s \| = 1$ .
Now we have  $S = B$ . The set
$$
S ^ {\prime} := \left\{\left| v \right| \mid \| s \| = 1, s \in \operatorname {s p a n} \left(\left\{\hat {y} _ {i} \right\} _ {i \in [ k ]}\right) \right\} \tag {23}
$$
can be obtained by reflecting  $E$  to the non-negative orthant. Formally, a reflection is determined by a collection of axes  $\{i_1, \dots, i_l\} (0 \leq l \leq k)$ . The image of such reflection in the non-negative orthant is given by
$$
B _ {i _ {1} \dots i _ {l}} = \{v \mid v ^ {\top} Q _ {i _ {1} \dots i _ {l}} v = 1, v \geq 0 \}, \tag {24}
$$
where  $Q_{i_1,\dots i_l} = D_{i_1\dots i_l}QD_{i_1\dots i_l}$ . As a consequence, we have
$$
S ^ {\prime} = \bigcup_ {0 \leq l \leq k} \bigcup_ {1 \leq i _ {1} <   \dots <   i _ {l} \leq k} B _ {i _ {1} \dots i _ {l}}. \tag {25}
$$
Finally, the feasible region can be equivalently characterized as
$$
\mathcal {F} _ {1} = \left\{v ^ {2} \mid \| s \| = 1, s \in \operatorname {s p a n} \left(\left\{\hat {y} _ {i} \right\} _ {i \in [ k ]}\right) \right\}, \tag {26}
$$
which can be obtained by squaring the coordinates of  $S'$ . Therefore, if we denote
$$
E _ {i _ {1} \dots i _ {l}} = \left\{v \mid \sqrt {v} ^ {\top} Q _ {i _ {1} \dots i _ {l}} \sqrt {v} = 1 \right\}, \tag {27}
$$
then (note the square root naturally implies  $v \geq 0$ )
$$
\mathcal {F} _ {1} = \bigcup_ {0 \leq l \leq k} \bigcup_ {1 \leq i _ {1} <   \dots <   i _ {l} \leq k} E _ {i _ {1} \dots i _ {l}}. \tag {28}
$$
A.4 Proof of Theorem 3.2
Proof of Theorem 3.2. Note the orthogonal complement of a 1-dimensional subspace of  $\operatorname{span}(\{\hat{y}_i\}_{i \in [k]})$  is a  $(k - 1)$ -dimensional subspace, and that the objective function for each task is the square of the projection of  $\hat{y}_i$ . Denote  $t \coloneqq (\| \hat{y}_1 \|^2, \dots, \| \hat{y}_k \|^2)^\top$ . By the Pythagorean theorem (Bhatia, 2013, Section I.6), if  $v \in \mathcal{F}_1$ , then we must have  $t - v \in \mathcal{F}_{q - 1}$ , and vice versa. As a consequence, denote
$$
I _ {i _ {1} \dots i _ {l}} = \left\{v \mid \sqrt {t - v} ^ {\top} Q _ {i _ {1} \dots i _ {l}} \sqrt {t - v} = 1 \right\}, \tag {29}
$$
$$
\mathcal {F} _ {q - 1} = \bigcup_ {0 \leq l \leq k} \bigcup_ {1 \leq i _ {1} <   \dots <   i _ {l} \leq k} I _ {i _ {1} \dots i _ {l}} \tag {30}
$$
by Theorem 3.1.
A.5 Why scalarization fails in the presence of gradient disagreement
We explain why linear scalarization is incapable of exploring an intersection point with gradient disagreement.
We first recall the geometric interpretation (see Figure 4.9 in (Boyd et al., 2004)) of linear scalarization: a point  $P$  lying at the boundary of the feasible region can be achieved by scalarization, if and only if there exists a hyperplane at  $P$  and the feasible region lies above the hyperplane. The normal vector of the hyperplane is proportional to the scalarization coefficients. By definition, if a hyperplane at  $P$  lies below the feasible region, its normal vector must be a subgradient of the surface. When the boundary of the feasible region is differentiable and the subdifferential set is non-empty, the normal vector must be the gradient of the surface, and the hyperplane becomes the tangent plane at  $P$ .
Now suppose  $P$  lies at the intersection of two differentiable surfaces  $S_{1}$  and  $S_{2}$ , and that  $P$  can be achieved by scalarization. Applying the above argument to  $S_{1}$  and  $P$ , we know that the scalarization coefficients are proportional to the gradient w.r.t.  $S_{1}$ . Similarly, applying the above argument to  $S_{2}$  and  $P$  yields that the scalarization coefficients are proportional to the gradient w.r.t.  $S_{2}$ . This will result in a contradiction if the two gradients w.r.t.  $S_{1}$  and  $S_{2}$  at  $P$  lie in different directions, a phenomenon we refer to as gradient disagreement. In this case, scalarization cannot reach  $P$ .
B Omitted details from Section 3.2
B.1 Proof of Lemma 3.4
We begin by reviewing some basics of convex analysis. We refer the readers to Rockafellar (1997) for more background knowledge.
Definition B.1 (Convex hull). Let  $X = \{x_{i}\}_{i\in [m]}$  be a collection of points in  $\mathbb{R}^n$ . The convex hull of  $X$ , denoted as  $\operatorname{conv}(X)$ , is the set of all convex combinations of points in  $X$ , i.e.,
$$
\operatorname {c o n v} (X) := \left\{\sum_ {i \in [ m ]} \alpha_ {i} x _ {i} \mid \alpha_ {i} \geq 0, \sum_ {i \in [ m ]} \alpha_ {i} = 1 \right\}. \tag {31}
$$
Definition B.2 (Cone and convex cone). A set  $K \subset \mathbb{R}^n$  is called a cone if  $x \in K$  implies  $\alpha x \in K$  for all  $\alpha > 0$ . A convex cone is a cone that is closed under addition. The convex cone generated by  $X = \{x_i\}_{i \in [m]} \subset \mathbb{R}^n$ , denoted as  $\operatorname{cone}(X)$ , is given by
$$
\operatorname {c o n e} (X) := \operatorname {c o n v} (\operatorname {r a y} (X)), \tag {32}
$$
where  $\operatorname{ray}(X) = \{\lambda x_i \mid \lambda \geq 0, i \in [m]\}$ .
Definition B.3 (Dual cone). The dual cone of  $C \subset \mathbb{R}^n$ , denoted as  $C^*$ , is given by
$$
C ^ {*} := \{y \mid \langle y, x \rangle \geq 0, \forall x \in C \} \tag {33}
$$
Definition B.4 (Projection onto a convex set). Let  $C$  be a closed convex set in  $\mathbb{R}^n$ . The orthogonal projection of  $y \in \mathbb{R}^n$  onto the set  $C$ , denoted as  $P_C y$ , is the solution to the optimization problem
$$
P _ {C} y := \inf  _ {x \in C} \left\{\| x - y \| \right\}. \tag {34}
$$
Theorem B.5 (Bourbaki-Cheney-Goldstein inequality (Ciarlet, 2013)). Let  $C$  be a closed convex set in  $\mathbb{R}^n$ . For any  $x \in \mathbb{R}^n$ , we have
$$
\langle x - P _ {C} x, y - P _ {C} x \rangle \leq 0, \quad \forall y \in C. \tag {35}
$$
Theorem Theorem B.5 is also known as the variational characterization of convex projection. We will use it to prove the following lemma, which serves as a crucial ingredient in the proof of Lemma 3.4.
Lemma B.6. Let  $C$  be a closed convex set in  $\mathbb{R}^n$ . For any  $z \in \mathbb{R}^n$ , we have
$$
\langle z ^ {\prime}, x \rangle \geq \langle z, x \rangle , \quad \forall x \in C, \tag {36}
$$
where  $z' = 2P_{C}z - z$  is the reflection of  $z$  w.r.t.  $C$ .
Proof. Plugging in  $z' = 2P_{C}z - z$ , it suffices to show
$$
\langle P _ {C} z - z, x \rangle \geq 0, \quad \forall x \in C. \tag {37}
$$
This is true because
$$
\left\langle P _ {C} z - z, x - P _ {C} z \right\rangle \geq 0, \quad \forall x \in C \tag {38}
$$
by Theorem B.5, and that
$$
\langle z - P _ {C} z, P _ {C} z \rangle = 0. \tag {39}
$$
We are now ready to prove Lemma 3.4.
Proof of Lemma 3.4. For the if part, we assume w.l.o.g. that  $G = \hat{Y}^{\top}\hat{Y}$  is doubly non-negative. This implies that the angle between each pair of optimal predictors is non-obtuse. Denote  $\mathcal{A} = \mathrm{cone}(\{\hat{y}_i\}_{i\in [k]})$  and  $\mathcal{A}^*$  as its dual cone. We have  $\mathcal{A}\subset \mathcal{A}^*$ .
Our goal is to show that, for every  $\| s \| = 1$ , we can always find  $s' \in \mathcal{A}^*$ , such that 1)  $\| s' \| = 1$ ; 2)  $\langle s', \hat{y}_i \rangle \geq |\langle s, \hat{y}_i \rangle|$ ,  $\forall i \in [k]$ . This implies that when restricting our discussion to the Pareto front, we can safely ignore the absolute value which appears in the proof of Theorem 3.1. As a consequence, the Pareto front belongs to the surface determined by  $Q = G^{-1}$ .
Consider  $s' = 2P_{\mathcal{A}^*}s - s$ . It is straightforward to see that  $\| s' \| = 1$  since it is the reflection of  $s$  w.r.t.  $\mathcal{A}^*$ . To show 2), we will prove:
$$
\langle s ^ {\prime}, x \rangle \geq | \langle s, x \rangle |, \quad \forall x \in \mathcal {A}. \tag {40}
$$
We break the discussion into two cases.
Case 1:  $\langle s,x\rangle \geq 0$ . Since  $x\in \mathcal{A}\subset \mathcal{A}^*$ , we have
$$
\langle s ^ {\prime}, x \rangle \geq \langle s, x \rangle = | \langle s, x \rangle | \tag {41}
$$
by Lemma B.6.
Case 2:  $\langle s,x\rangle < 0$ . Consider  $u = -s$  and denote  $\mathcal{C} \coloneqq \{a \mid \langle a, P_{\mathcal{A}^*} s \rangle \geq 0\}$ . It is straightforward to see that  $\mathcal{A} \subset \mathcal{C}$ .
We will show  $P_{\mathcal{C}}u = P_{\mathcal{A}^*}s - s$ . In fact, for any  $z \in \mathcal{C}$ , we have
$$
\begin{array}{l} \left\| z - u \right\| ^ {2} = \left\| z - \left(P _ {\mathcal {A} ^ {*}} s - s\right) + P _ {\mathcal {A} ^ {*}} s \right\| ^ {2} \\ = \| z - \left(P _ {\mathcal {A} ^ {*}} s - s\right) \| ^ {2} + \| P _ {\mathcal {A} ^ {*}} s \| ^ {2} + 2 \langle z - \left(P _ {\mathcal {A} ^ {*}} s - s\right), P _ {\mathcal {A} ^ {*}} s \rangle \\ \geq \left\| P _ {\mathcal {A} ^ {*}} s \right\| ^ {2} + 2 \langle z, P _ {\mathcal {A} ^ {*}} s \rangle \\ \geq \left\| P _ {\mathcal {A} ^ {*}} s \right\| ^ {2} \\ = \left\| P _ {\mathcal {A} ^ {*}} s - s - u \right\| ^ {2}. \\ \end{array}
$$
$$
s ^ {\prime} = 2 P _ {\mathcal {A} ^ {*}} s - s = 2 P _ {\mathcal {C}} u - u. \tag {42}
$$
With another application of Lemma B.6, we have
$$
\langle s ^ {\prime}, x \rangle \geq \langle u, x \rangle = \langle - s, x \rangle = | \langle s, x \rangle |. \tag {43}
$$
This finishes the proof of the if part.
For the only if part, we consider the point  $p_i$  which achieves maximum value along the  $i$ -axis. For a given  $i$ , it is straightforward to see that such  $p_i$  is unique, and therefore is a PO of the feasible region.
Now,  $p_i$  belongs to a surface determined by  $Q'$ , if and only if there exists a non-negative vector  $v_i$  such that  $Q'v_i = e_i$  (the normal vector at  $p_i$  aligns with the  $i$ -th axis). In other words,  $G'e_i \geq 0$  where  $G' = (Q')^{-1}$ . When C1 does not hold, there does not exist a  $G^* \in \mathcal{G}$ , such that  $G^*e_i \geq 0$  for all  $i \in [k]$ . This implies that these  $p_i$  must belong to different surfaces.
B.2 Proof of Lemma 3.5
Proof of Lemma 3.5. For the if part, we assume w.l.o.g. that  $Q_{\emptyset} = G_{\emptyset}^{-1}$  is doubly non-negative. This essentially implies that  $E_{\emptyset}$  is dominated by all other surfaces in  $\mathcal{E}$ , which further implies that  $I_{\emptyset}$  dominates all other surfaces in  $\mathcal{I}$ . Therefore, the Pareto front must belong to  $I_{\emptyset}$ .
For the only if part, we study the Pareto front of  $\mathcal{F}_{q-1}$  through the lens of  $\mathcal{F}_1$ . Specifically, a point  $z$  is a PO of  $\mathcal{F}_{q-1}$  if and only if  $\mathcal{F}_{q-1}$  and the non-negative orthant of  $z$  intersects only at  $z$ . Denote the corresponding point of  $z$  on  $\mathcal{F}_1$  as  $z' = t - z$ . An equivalent characterization is that  $\mathcal{F}_1$  and the non-positive orthant of  $z'$  intersects only at  $z'$ . We can therefore consider a special type of  $z'$  whose coordinates are zero except for  $i, j$ -th entry. If  $z'$  further lies on a surface  $Q$  with  $q_{ij} > 0$ , then the corresponding  $z$  will be a PO of  $\mathcal{F}_{q-1}$ .
When C2 does not hold, there exists a row of  $Q_{\emptyset}$  which contains both positive and negative entries. We assume w.l.o.g. the first row of  $Q_{\emptyset}$  satisfies this condition. By the above observation, we can find two PO of  $\mathcal{F}_{q-1}$  that correspond to  $Q_{\emptyset}$  and  $Q_{1}$ , respectively. This finishes the proof as desired.
B.3 Proof of Theorem 3.8
Proof of Theorem 3.8. We only prove for the case of  $q = 1$ , and the proof for  $q = k - 1$  can be done similarly. Suppose  $P$  (whose coordinate is  $v$ ) lies at the intersection of two surfaces defined by  $Q$  and  $Q'$ . We denote the intersection of  $Q$  and  $Q'$  as  $S$ , which is a non-linear manifold with dimension  $(k - 2)$ . Since  $P$  is a relative interior point of the PF, there exists some  $\varepsilon > 0$ , such that any point in  $S \cap B_{\varepsilon}(P)$  is a PO. We will show there exists some  $P' \in S \cap B_{\varepsilon}(P)$  (whose coordinate is  $v'$ ), such that the gradients at  $P'$  w.r.t. the two surfaces disagree, i.e.,
$$
Q \sqrt {v ^ {\prime}} \neq Q ^ {\prime} \sqrt {v ^ {\prime}}. \tag {44}
$$
To see why this is true, note that  $Q'$  can be expressed as  $Q' = DQD$ , where  $D$  is a diagonal matrix whose diagonal entries are either 1 or -1. As a consequence, the set
$$
\{v \mid Q v = Q ^ {\prime} v \} \tag {45}
$$
is a subspace of  $\mathbb{R}^k$  whose dimension is at most  $(k - 2)$ , so it cannot fully contain a non-linear manifold whose dimension is  $(k - 2)$ . This finishes the proof as desired.
B.4 Proof of Theorem 3.8 without assumptions  $(k = 3)$
Here we provide a proof for Theorem 3.8 under  $k = 3$ , without relying on Assumptions 3.6 and 3.7.
Proof. We prove the necessity of C1 and C2 separately.
C1 is necessary. Suppose C1 is not true. We assume w.l.o.g. that  $\langle \hat{y}_1, \hat{y}_2 \rangle < 0$ ,  $\langle \hat{y}_1, \hat{y}_3 \rangle > 0$ ,  $\langle \hat{y}_2, \hat{y}_3 \rangle > 0$ . Now we can write
$$
Q = \left[ \begin{array}{c c c} q _ {1 1} & q _ {1 2} & - q _ {1 3} \\ q _ {2 1} & q _ {2 2} & - q _ {2 3} \\ - q _ {3 1} & - q _ {3 2} & q _ {3 3} \end{array} \right], \tag {46}
$$
where  $q_{ij} > 0$  for all pairs of  $(i,j)$ . We also have
$$
q _ {1 1} q _ {2 3} > q _ {1 2} q _ {1 3}, q _ {2 2} q _ {1 3} > q _ {1 2} q _ {2 3}, q _ {3 3} q _ {1 2} > q _ {1 3} q _ {2 3}. \tag {47}
$$
The boundary of the feasible region is formed by  $E_{\emptyset}, E_1, E_2$ . Our goal is to find a point  $I \in E_{\emptyset} \cap E_1$  such that 1)  $I$  is a PO of the feasible region; 2) the gradients at  $I$  w.r.t.  $E_{\emptyset}$  and  $E_1$  disagree. We write the coordinate of  $I$  as  $(x, y = q_{13}^2 / s, z = q_{12}^2 / s)^{\top}$ , where
$$
q _ {1 1} x + \frac {q _ {2 2} q _ {1 3} ^ {2} + q _ {3 3} q _ {1 2} ^ {2} - 2 q _ {1 2} q _ {2 3} q _ {3 1}}{s} = 1, \tag {48}
$$
and  $s < q_{11}q_{23}^2 +q_{22}q_{13}^2 +q_{33}q_{12}^2 -2q_{12}q_{23}q_{31}$