You have a matrix representation $\lambda$ when the matrices satisfy $T^\lambda(g_1)T^{\lambda}(g_2)=T^\lambda(g)$ when $g_1\circ g_2=g$ in the group, for all $g_1,g_2,g$.

If a set of basis vectors (usually an orthonormal set) $\{\vert\psi_k\rangle\}$ is such that \begin{align} T^\lambda(g_1)T^{\lambda}(g_2)\vert\psi_k\rangle=T^\lambda(g)\vert\psi_k\rangle\tag{1} \end{align} then the vector space spanned by these basis vector "carries" the representation $\lambda$. Basically this follows from the way the matrix elements of $T^\lambda(g)$ are constructed, i.e. \begin{align} \langle \psi_i\vert T^\lambda(g)\vert\psi_j\rangle := T^\lambda_{ij}(g)\, .\tag{2} \end{align}

Arbitrary normalizable linear combinations of these vectors are permissible in the vector space, and (in the physics parlance) the vectors "carry" the irreducible representation.

As - for instance - you can have a 3-dimensional representation of $su(2)$, any linear combination of the basis vectors is allowed provided that $T^1(g_1)T^1(g_2)=T^1(g)$ as $3\times 3$ matrices, and where I've used $\lambda=j=1$ since the SU(2) irreps with $j=1$ has dimension $3$.

In particular, the combination $\vert\phi\rangle = (1,1,1)^\top /\sqrt{3}$ is allowed, which is clearly not of the form you suggest. To find how such a vector transform under $SU(2)$, one would write \begin{align} \vert\phi\rangle&=\sum_k a_k\vert\psi_k\rangle\, ,\\ T^1(g)\vert\phi\rangle&= \sum_k a_k T^1(g)\vert\psi_k\rangle\, ,\\ &= \sum_{km} a_k \vert\psi_m\rangle T^1_{mk}(g) \, . \end{align}

You have a matrix representation $\lambda$ when the matrices satisfy $T^\lambda(g_1)T^{\lambda}(g_2)=T^\lambda(g)$ when $g_1\circ g_2=g$ in the group, for all $g_1,g_2,g$.

If a set of basis vectors (usually an orthonormal set) $\{\vert\psi_k\rangle\}$ is such that \begin{align} T^\lambda(g_1)T^{\lambda}(g_2)\vert\psi_k\rangle=T^\lambda(g)\vert\psi_k\rangle\tag{1} \end{align} then the vector space spanned by these basis vector "carries" the representation $\lambda$. Basically this follows from the way the matrix elements of $T^\lambda(g)$ are constructed, i.e. \begin{align} \langle \psi_i\vert T^\lambda(g)\vert\psi_j\rangle := T^\lambda_{ij}(g)\, .\tag{2} \end{align}

It is not always trivial to find the $T^{\lambda}_{ij}$. For instance, the spherical harmonics with $\ell=1$ span a 3-dimensional representation of $so(3)$. Using \begin{align} Y_{11}(\theta,\phi)\mapsto (1,0,0)^\top\, ,\qquad Y_{10}(\theta,\phi)\mapsto (0,1,0)^\top\, ,\qquad Y_{1,-1}(\theta,\phi)\mapsto (0,0,1)^\top \end{align} the usual $L_z\mapsto -i\frac{\partial}{\partial \phi}$ etc, you can first obtain a representation of the algebra $su(2)$ using the standard inner product \begin{align} (L_k)_{ij}=\int \sin\theta d\theta d\phi Y_{1i}^*(\theta,\phi) \hat L_k Y_{1j}(\theta,\phi) \end{align} and then using $R_k(\beta)=e^{-i\beta L_k}$ plus the Euler factorization $R_z(\alpha)R_y(\beta)R_z(\gamma)$ obtain the 3-dimensional representation of SO(3) with $Y_{1,m}(\theta,\phi)$ as basis vectors.

Arbitrary normalizable linear combinations of these vectors are permissible in the vector space, and (in the physics parlance) the vectors "carry" the irreducible representation.

As - for instance - you can have a 3-dimensional representation of $su(2)$, any linear combination of the basis vectors is allowed provided that $T^1(g_1)T^1(g_2)=T^1(g)$ as $3\times 3$ matrices, and where I've used $\lambda=j=1$ since the SU(2) irreps with $j=1$ has dimension $3$.

In particular, the combination $\vert\phi\rangle = (1,1,1)^\top /\sqrt{3}$ is allowed, which is clearly not of the form you suggest. To find how such a vector transform under $SU(2)$, one would write \begin{align} \vert\phi\rangle&=\sum_k a_k\vert\psi_k\rangle\, ,\\ T^1(g)\vert\phi\rangle&= \sum_k a_k T^1(g)\vert\psi_k\rangle\, ,\\ &= \sum_{km} a_k \vert\psi_m\rangle T^1_{mk}(g) \, . \end{align}

You have a matrix representation $\lambda$ when the matrices satisfy $T^\lambda(g_1)T^{\lambda}(g_2)=T^\lambda(g)$ when $g_1\circ g_2=g$ in the group, for all $g_1,g_2,g$.

If a set of basis vectors (usually an orthonormal set)  $\{\vert\psi_k\rangle\}$ is such that \begin{align} T^\lambda(g_1)T^{\lambda}(g_2)\vert\psi_k\rangle=T^\lambda(g)\vert\psi_k\rangle \end{align} then the vector space spanned by these basis vector "carries" the representation $\lambda$. Basically this follows from the way the matrix elements of $T^\lambda(g)$ are constructed, i.e. \begin{align} \langle \psi_i\vert T^\lambda(g)\vert\psi_j\rangle := T^\lambda_{ij}(g)\, . \end{align}

Arbitrary normalizable linear combinations of these vectors are permissible in the vector space, and (in the physics parlance) the vectors "carry" the irreducible representation.

As - for instance - you can have a 3-dimensional representation of $su(2)$, any linear combination of the basis vectors is allowed provided that $T^1(g_1)T^1(g_2)=T^1(g)$ as $3\times 3$ matrices, and where I've used $\lambda=j=1$ since the SU(2) irreps with $j=1$ has dimension $3$.

In particular, the combination $\vert\phi\rangle = (1,1,1)^\top /\sqrt{3}$ is allowed, which is clearly not of the form you suggest. To find how such a vector transform under $SU(2)$, one would write \begin{align} \vert\phi\rangle&=\sum_k a_k\vert\psi_k\rangle\, ,\\ T^1(g)\vert\phi\rangle&= \sum_k a_k T^1(g)\vert\psi_k\rangle\, ,\\ &= \sum_{km} a_k \vert\psi_m\rangle T^1_{mk}(g) \, . \end{align}

You have a matrix representation $\lambda$ when the matrices satisfy $T^\lambda(g_1)T^{\lambda}(g_2)=T^\lambda(g)$ when $g_1\circ g_2=g$ in the group, for all $g_1,g_2,g$.

If a set of basis vectors (usually an orthonormal set)  $\{\vert\psi_k\rangle\}$ is such that \begin{align} T^\lambda(g_1)T^{\lambda}(g_2)\vert\psi_k\rangle=T^\lambda(g)\vert\psi_k\rangle\tag{1} \end{align} then the vector space spanned by these basis vector "carries" the representation $\lambda$. Basically this follows from the way the matrix elements of $T^\lambda(g)$ are constructed, i.e. \begin{align} \langle \psi_i\vert T^\lambda(g)\vert\psi_j\rangle := T^\lambda_{ij}(g)\, .\tag{2} \end{align}

Arbitrary normalizable linear combinations of these vectors are permissible in the vector space, and (in the physics parlance) the vectors "carry" the irreducible representation.

As - for instance - you can have a 3-dimensional representation of $su(2)$, any linear combination of the basis vectors is allowed provided that $T^1(g_1)T^1(g_2)=T^1(g)$ as $3\times 3$ matrices, and where I've used $\lambda=j=1$ since the SU(2) irreps with $j=1$ has dimension $3$.

In particular, the combination $\vert\phi\rangle = (1,1,1)^\top /\sqrt{3}$ is allowed, which is clearly not of the form you suggest. To find how such a vector transform under $SU(2)$, one would write \begin{align} \vert\phi\rangle&=\sum_k a_k\vert\psi_k\rangle\, ,\\ T^1(g)\vert\phi\rangle&= \sum_k a_k T^1(g)\vert\psi_k\rangle\, ,\\ &= \sum_{km} a_k \vert\psi_m\rangle T^1_{mk}(g) \, . \end{align}