# Existence of real symmetric orthogonal matrix

Q. There does not exist $$4\times4$$ real symmetric orthogonal matrix with all diagonal entries zero?

My approach

If a real symmetric matrix is orthogonal, then its columns (or rows) form an orthonormal basis of the vector space. In particular, the length of each column is $$1$$, and the dot product of any two different columns is $$0$$. Moreover, if the diagonal entries are all zero, then the matrix represents a linear transformation that sends the standard basis vectors to vectors that lie in a hyperplane, which is also called a subspace of codimension $$1$$.

Now suppose that there exists a $$4\times4$$ real symmetric orthogonal matrix with all diagonal entries zero. Let $$A$$ be such a matrix. Then the columns of $$A$$ form an orthonormal basis of $$\mathbb R^4$$, and since the diagonal entries are zero, each column must have a non-zero entry in one of the last three rows (otherwise, the column would be zero). Without loss of generality, we can assume that the first column of $$A$$ has a non-zero entry in the second row.

Let $$a$$, $$b$$, $$c$$, $$d$$ be the entries of the first column of $$A$$, with a non-zero entry in the second row. Since the first column is a unit vector, we have $$a^2 + b^2 + c^2 + d^2=1$$. Moreover, the dot product of the first column with the second, third, and fourth columns of $$A$$ must be $$0$$, which gives us the following equations:

$$ab + be + cf + df = 0$$

$$ac + ce + dg = 0$$

$$ad + de + dh = 0$$

Note that the first equation implies that $$b = 0$$, because a is non-zero. Therefore, the second and third equations simplify to:

$$ac = -ce - dg$$

$$ad = -de - dh$$

Squaring both sides of these equations and adding them up, we get:

$$a^2(c^2 + d^2) = (c^2 + d^2)(e^2 + g^2) + (d^2 + h^2)(e^2 + f^2)$$

Since $$a$$ is non-zero, we can divide both sides by $$a^2$$ and simplify:

$$c^2 + d^2 = (e^2 + g^2) + (d^2 + h^2)(e^2 + f^2)/(a^2)$$

Since the left-hand side is a constant, the right-hand side must be constant as well. However, $$e$$, $$f$$, $$g$$, $$h$$ are all entries of a unit vector in $$\mathbb R^3$$, so their squares add up to $$1$$. Therefore, the right-hand side is a sum of non-negative terms that is strictly less than $$1$$, which contradicts the fact that the left-hand side is $$1$$.

Hence, there does not exist a $$4\times4$$ real symmetric orthogonal matrix with all diagonal entries zero.

Is it okay...?

• have you tried looking at $4\times 4$ permutation matrices? Commented Apr 20, 2023 at 5:47

Try to see how your argument deals with $$A=\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}.$$
This is not true, an orthogonal matrix (like $$A$$ above) will send the standard basis to another orthonormal basis. In the particular example as above $$A$$ simply permutes $$e_1$$ with $$e_2$$ and $$e_3$$ with $$e_4$$.
Now suppose that there exists a $$4\times4$$ real symmetric orthogonal matrix with all diagonal entries zero. Let $$A$$ be such a matrix. Then the columns of $$A$$ form an orthonormal basis of $$\mathbb R^4$$, and since the diagonal entries are zero, each column must have a non-zero entry in one of the last three rows (otherwise, the column would be zero)
No. Nothing prevents a column like the second column of the $$A$$ in my example.
As for your equations, it is unclear what your variables are. If $$a,b,c,d$$ are the entries of the first column of $$A$$ then $$a=0$$, but you say it's nonzero. Are you considering them in a different order?