truncation for diagonal UniTensor

[hunghaoti]

When create a diagonal UniTensor with shape [3,3] and then call truncate(idx=0, dim=2) to truncate the bond dimension for first bond. Currently, it will return a diagonal UniTensor with shape [2,2]. Is it expected or a bug ?

The related unit test is and see line #L4562

Cytnx/tests/DenseUniTensor_test.cpp

Lines 4545 to 4566 in f62f0ce

	/*=====test info=====
	describe:test truncate with diangonal UniTensor
	====================*/
	TEST_F(DenseUniTensorTest, truncate_diag) {
	auto row_rank = 1u;
	std::vector<Bond> bonds = {Bond(3), Bond(3)};
	std::vector<std::string> labels = {"a", "b"};
	bool is_diag = true;
	auto seed = 0;
	auto ut_diag = UniTensor(bonds, labels, row_rank, Type.ComplexDouble, Device.cpu, is_diag);
	random::uniform_(ut_diag, 0.0, 5.0, seed);
	int dim = 2;
	auto clone = ut_diag.clone();
	auto ut_trunc = ut_diag.truncate("a", dim);
	EXPECT_TRUE(AreEqUniTensor(ut_diag, clone));
	auto shape = ut_trunc.shape();
	EXPECT_EQ(shape[0], dim);
	EXPECT_EQ(shape[1], dim); // expected ?
	for (int i = 0; i < shape[0]; i++) {
	EXPECT_EQ(ut_diag.at({i}), ut_trunc.at({i}));
	}
	}

[yingjerkao]

I think it is expected but misleading. Maybe we should throw a warning so that the user may not misuse it.

[ianmccul]

Diagonal matrices don't have to be square; there is no conceptual problem to having a $m \times n$ matrix with only diagonal elements and $m \neq n$.