The stress-energy density of the dust in it's rest frame is
$$T_{ij} = \begin {bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end {bmatrix} $$
To change frames to one where the dust is moving, we perform a Lorentz boost on ##T_{ij}##
$$T'_{uv} = T_{ij} \, \Lambda^i{}_u \Lambda^j{}_v$$
Letting ##\beta = ||v||/c## and ##\gamma = 1/\sqrt{1-\beta^2}## for a Lorentz boost in the ##x^1## direction we can write:
$$\Lambda = \begin {bmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end {bmatrix} $$
So we get that the only nonzero components are
$$T'_{00} = \Lambda^0{}_0 \, \Lambda^0{}_0 \,T_{00} = \gamma^2 \, \rho \quad T'_{01} = T'_{10} = \Lambda^0{}_0 \, \Lambda^0{}_1 \, T_{00} = -\beta \gamma^2 \rho \quad T'_{11} = \Lambda^0{}_1 \, \Lambda^0{}_1 \, T_{00} = \beta^2 \gamma^2 \rho $$
i.e.
$$T'_{ij} = \begin {bmatrix} \gamma^2 \rho & -\beta \gamma^2 \rho & 0 & 0 \\ -\beta \gamma^2 \rho & \beta^2 \gamma^2 \rho & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end {bmatrix} $$
So no, all the spatial components of ##T'_{ij}## are not zero, in particular ##T'_{11}## which is what I think you mean by "spatial component"(?) is nonzero in the example where v points in the ##x^1## direction.