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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...?

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

1 Answer 1

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Try to see how your argument deals with $$ A=\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}. $$

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

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?

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