Looking for elementary proof for $\rm~det(\textbf{P})$ = $\pm 1$ if $P$ is an orthogonal matrix. I prefer a proof without using determinant of transpose matrix.

My First Proof, with $\det(\textbf{P}^{t}) = \det(\textbf{P})$

If $P$ is orthogonal matrix, $\textbf{P}^{t}=\textbf{P}^{-1}$. So, $$\det(\textbf{P}^{t}\textbf{P})=\det(\textbf{I}) \implies \det(\textbf{P}^{t}\textbf{P}) = 1 \implies \det(\textbf{P}^{t}) \det(\textbf{P}) = 1$$ because $ \det(\textbf{P}^{t}) = \det(\textbf{P})$.
Therefore, $\det(\textbf{P}^{t})=\det(\textbf{P}) = \pm 1$.

My Second Proof, without $\det(\textbf{P}^{t}) = \det(\textbf{P})$

Let $\lambda$ be an eigenvalue for orthogonal matrix $\textbf{P}$. $$\textbf{P}\vec{v}=\lambda\vec{v},\quad\vec{v} \neq \vec{0}$$ $$\implies \rVert \textbf{P}\vec{v}\lVert = \lVert \lambda\vec{v} \lVert \implies \rVert \textbf{P}\vec{v} \lVert = \rvert \lambda \ \lvert \lVert \vec{v} \rVert = 1$$ because $\lVert \vec{v} \rVert = 1$

Therefore $\lvert \lambda \rvert = 1$ because $\textbf{P}$ is orthonormal matrix. $$\textbf{P} = \textbf{P}^{t}$$ Therefore $\textbf{P} = \textbf{U}^{t}\Lambda\textbf{U} $ $$\det(\textbf{U}^{t}\Lambda\textbf{U}) = \det(\textbf{U}^{t})\det(\Lambda)\det(\textbf{U})$$ $$\det(\textbf{U}^{t}\Lambda\textbf{U}) = \det(\textbf{U}^{t})\det(\textbf{U})\det(\Lambda)$$ $$\det(\textbf{U}^{t})\det(\textbf{U}) = 1 $$ (because $\textbf{U}$ is orthogonal matrix})

$$\det(\textbf{P}) = \det(\textbf{U}^{t}\Lambda\textbf{U}) = \det(\Lambda) = \prod_{i=1}^{n}\lambda_i = \pm 1 $$ $$\implies \det(\textbf{P}) = \pm 1$$

  • 3
    $\begingroup$ The claim isn't true; unitary matrices have $|\det{P}| = 1$. $\endgroup$ – user61527 Oct 11 '13 at 3:18
  • 2
    $\begingroup$ What you are getting is $(\det P)^2=1$, but no more than that. $\endgroup$ – Pedro Tamaroff Oct 11 '13 at 3:20
  • $\begingroup$ thx to point it out $\endgroup$ – bsdshell Oct 11 '13 at 3:21
  • $\begingroup$ Shouldn't it be $P^t P = I$ instead of "$\implies P^t = P$" at the top? $\endgroup$ – Pratyush Sarkar Oct 11 '13 at 3:25
  • $\begingroup$ And the proof is $\det(P^tP) = \det(I) \implies \det(P)^2 = 1$ which is what you already have. It is as elementary as it can get. $\endgroup$ – Pratyush Sarkar Oct 11 '13 at 3:30

Here is a geometric argument that can be used for real orthogonal matrices: If $T:\ V\to V$ is an arbitrary linear transformation of a finite dimensional euclidean vector space $V$ then the volumes of arbitrary measurable sets $A\subset V$, in particular of balls or cubes, are multiplied by $\bigl|\det(T)\bigr|$. That is to say, one has $${\rm vol}\bigl(T(A)\bigr)=\bigl|\det(T)\bigr|\ {\rm vol}(A)\ .$$ Now an orthogonal transformation $T$ transforms the unit ball $B:=\bigl\{x\in{\mathbb R}^n\bigm| |x|\leq1\bigr\}$ onto itself, and ${\rm vol}(B)>0$. It follows that $|\det(T)\bigr|=1$.

  • $\begingroup$ Thanks for the insight from Christian Blatter, can you explain more about what is "reasonable set A"? $\endgroup$ – bsdshell Oct 11 '13 at 22:07
  • $\begingroup$ @bsdshell : By reasonable we imply those sets for which the notion of volume is defined, which means those functions for which the indicator function can be integrated (using the Lebesgue integral to be as general as possible). $\endgroup$ – Patrick Da Silva Oct 13 '13 at 16:49
  • $\begingroup$ Is this a generalization to the higher dimensions of the idea that orthogonal transformations only reflects or rotates in 2D euclidean space, and that is why the volume is invariant under the transformation? $\endgroup$ – Aditya P Jul 23 '18 at 11:31
  • $\begingroup$ Quick question: Is the converse true? Are all matrices with det(X)=1, orthogonal? $\endgroup$ – Ali Dec 7 '19 at 11:08

The absolute value of the determinant of a real, square matrix is the volume of the parallelepiped whose sides are the columns of the matrix. For an orthogonal matrix, the columns are pairwise orthogonal unit vectors, so this parallelepiped is a cube with side length 1. So the volume is 1, and therefore the determinant is $\pm1$.


If $|\det(Q)|>1$, then for any positive integer $n$, $|\det(Q^n)|=|\det(Q)|^n$ will blow up. Since for any positive integer $n$, $Q^n$ is also an orthogonal matrix, whose determinant will not blow up due to the geometric meaning of determinant, we know that $|\det(Q^n)|$ will not blow up and $|\det(Q)|=1$.


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