$$\det(A^T) = \det(A)$$

Using the geometric definition of the determinant as the area spanned by the columns, could someone give a geometric interpretation of the property?

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    $\begingroup$ This may be relevant. $\endgroup$ Commented Dec 8, 2013 at 15:53

7 Answers 7


A geometric interpretation in four intuitive steps....

The Determinant is the Volume Change Factor

Think of the matrix as a geometric transformation, mapping points (column vectors) to points: $x \mapsto Mx$. The determinant $\mbox{det}(M)$ gives the factor by which volumes change under this mapping.

For example, in the question you define the determinant as the volume of the parallelepiped whose edges are given by the matrix columns. This is exactly what the unit cube maps to, so again, the determinant is the factor by which the volume changes.

A Matrix Maps a Sphere to an Ellipsoid

Being a linear transformation, a matrix maps a sphere to an ellipsoid. The singular value decomposition makes this especially clear.

If you consider the principal axes of the ellipsoid (and their preimage in the sphere), the singular value decomposition expresses the matrix as a product of (1) a rotation that aligns the principal axes with the coordinate axes, (2) scalings in the coordinate axis directions to obtain the ellipsoidal shape, and (3) another rotation into the final position.

The Transpose Inverts the Rotation but Keeps the Scaling

The transpose of the matrix is very closely related, since the transpose of a product is the reversed product of the transposes, and the transpose of a rotation is its inverse. In this case, we see that the transpose is given by the inverse of rotation (3), the same scaling (2), and finally the inverse of rotation (1).

(This is almost the same as the inverse of the matrix, except the inverse naturally uses the inverse of the original scaling (2).)

The Transpose has the Same Determinant

Anyway, the rotations don't change the volume -- only the scaling step (2) changes the volume. Since this step is exactly the same for $M$ and $M^\top$, the determinants are the same.

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    $\begingroup$ Matt, after reading your answer I NEED to ask - do you have any recommendations for linear algebra books that explain things geometrically, like in the way you have? $\endgroup$ Commented May 26, 2019 at 19:04
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    $\begingroup$ @Joshua: Sorry, no, linear algebra made very little sense to me the way it was taught. I had to re-figure it out on my own terms. If there is some book out there that develops the material in this way, always staying close to the geometric intuition, I'd like to see it too! $\endgroup$
    – Matt
    Commented May 27, 2019 at 14:30
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    $\begingroup$ Wow this makes me realise WHY the inverse of an orthogonal matrix is just it's transpose. If a matrix can be decomposed into a scaling and a rotation, the orthogonal matrix is just a rotation, it scales by 1. So for it a transpose is the inverse! Thanks for making me see the transpose in a different light. :( why is this not common knowledge! $\endgroup$
    – Aditya P
    Commented Oct 8, 2019 at 10:27
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    $\begingroup$ Orthogonal matrix could also be a reflection not a rotation. In that case the determinant is -1, not 1. But the inverse also has -1. $\endgroup$
    – Smithey
    Commented Nov 6, 2023 at 16:11
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    $\begingroup$ @joshuaronis Linear Algebra Through Geometry by Banchoff looks good. I also like Ted Schiffrin's lectures. Treil book is also very good, though geometry is not necessarily main focus. $\endgroup$
    – Smithey
    Commented Nov 6, 2023 at 16:13

This is more-or-less a reformulation of Matt's answer. He relies on the existence of the SVD-decomposition, I show that $\det(A)=\det(A^T)$ can be stated in a little different way.

Every square matrix can be represented as the product of an orthogonal matrix (representing an isometry) and an upper triangular matrix (QR decomposition)- where the determinant of an upper (or lower) triangular matrix is just the product of the elements along the diagonal (that stay in their place under transposition), so, by the Binet formula, $A=QR$ gives: $$\det(A^T)=\det(R^T Q^T)=\det(R)\det(Q^T)=\det(R)\det(Q^{-1}),$$ $$\det(A^T)=\frac{\det{R}}{\det{Q}}=\det(Q)\det(R)=\det(QR)=\det(A),$$ where we used that the transpose of an orthogonal matrix is its inverse, and the determinant of an orthogonal matrix belongs to $\{-1,1\}$ - since an orthogonal matrix represents an isometry.

You can also consider that $(*)$ the determinant of a matrix is preserved under Gauss-row-moves (replacing a row with the sum of that row with a linear combination of the others) and Gauss-column-moves, too, since the volume spanned by $(v_1,\ldots,v_n)$ is the same of the volume spanned by $(v_1+\alpha_2 v_2+\ldots,v_2,\ldots,v_n)$. By Gauss-row-moves you can put $A$ in upper triangular form $R$, then have $\det A=\prod R_{ii}.$ If you apply the same moves as column moves on $A^T$, you end with $R^T$ that is lower triangular and has the same determinant of $R$, obviously. So, in order to provide a "really geometric" proof that $\det(A)=\det(A^T)$, we only need to provide a "really geometric" interpretation of $(*)$. An intuition is that the volume of the parallelepiped originally spanned by the columns of $A$ is the same if we change, for istance, the base of our vector space by sending $(e_1,\ldots,e_n)$ into $(e_1,\ldots,e_{i-1},e_i+\alpha\, e_j,e_{i+1},\ldots,e_n)\,$ with $i\neq j$, since the geometric object is the same, and we are only changing its "description".

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    $\begingroup$ I agree with @Tim. Your answer has a part suggesting that Matt's answer claims something that is not true. To me, that part of your answer is not true. $\endgroup$
    – Tunococ
    Commented Jan 13, 2014 at 21:23
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    $\begingroup$ I think the SVD is arguably a better choice since it exposes the spectral norm. That helps a lot when trying to visualize the geometry. $\endgroup$ Commented Jan 13, 2014 at 21:24
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    $\begingroup$ The OP wants a geometric interpretation of a fact about determinants. You aren't required to prove anything, so I don't see a problem with using the SVD as a tool. Especially since it makes it extremely clear what happens when you transpose a matrix. $\endgroup$ Commented Jan 13, 2014 at 21:39
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    $\begingroup$ I find it extremely difficult also to think of Binet formula geometrically. $\endgroup$
    – Tunococ
    Commented Jan 13, 2014 at 22:49
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    $\begingroup$ Not if you think that the determinant is a ratio of signed volumes! $\endgroup$ Commented Jan 13, 2014 at 23:47

Since $\text{sign}(\sigma^{-1})=\text{sign}(\sigma)$ and $\phi:S_n\to S_n,\sigma\mapsto\sigma^{-1}$ is a bijection, we have $$\begin{align}\det(A)&=\sum_{\sigma\in S_n}\text{sign}(\sigma)\prod_{i=1}^na_{i\sigma(i)}\\&=\sum_{\sigma\in S_n}\text{sign}(\sigma^{-1})\prod_{i=1}^na_{\sigma^{-1}(i)i}\\&=\sum_{\sigma\in S_n}\text{sign}(\sigma)\prod_{i=1}^na_{\sigma(i)i}\\&=\det(A^t)\end{align}$$

Note: I'm not using the geometric definition, but I could only post this here since the question without the geometric requirement was flagged incorrectly as a duplicate of this problem.


I came up with the same question when I was watching a linear algebra tutorial by Professor Wildberger. And I think there's another interesting explanation in it. My personal math background is not very good, so I suggest watching the first ten lectures of the series. Here's the link.

The series first talked about wedge product. so if you have two vectors $v_1$ and $v_2$, the wedge product $v_1 \lor v_2$ represents a signed area or volume.

Another idea is to consider a matrix transformation to be a coordinates change. Take 2x2 matrix as example:

$$ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} * \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} = \begin{pmatrix} y_1 \\ y_2 \\ \end{pmatrix} $$

Let $e_1$ and $e_2$ to be the base of the plane. If we substitute $(1, 0)$ and $(0, 1)$ into $(x_1, x_2)$ separately, we will have $(a, c)$ and $(b, d)$ to be the new base after change of coordinates. If we assume the area of each grid in the old coordinates to be 1, namely $e_1 \lor e_2 = 1$, we will have $(a, c) \lor (b, d)$ be the area of each grid in the new coordinates. The result of $(a, c) \lor (b, d)$ happens to be $ad - bc$, which is the determinant. From a geometric point of view, it represents the ratio of area change.

Now let's take $x_1$ and $x_2$ themselves as bases, namely, let $x_1 = e_1$ and $x_2 = e_2$. We will then have $y_1 = a e_1 + b e_2$ and $y_2 = c e_1 + d e_2$. The wedge product of $y_1$ and $y_2$ also represents the same area change ratio, and the value is $(a, b) \lor (c, d) = ad - bc$ as well.

In the above calculation, we're using matrix columns in the first case, and matrix rows in the second case, and they are transpose of each other, thus the determinant of a matrix is the same as that of its transpose.

This is probably not a rigour way to think, but it's quite interesting.


Here's an algebraic proof, followed by a geometric interpretation of it, which could be summarized as follows.

Volumes can be computed with scalar products by cut-and-pasting, and because of the definition of the transpose, the volume spanned by some $Av_i$ and the volume spanned by the corresponding $A^tv_i$ can be decomposed using the same pieces.

Let $V$ be a vector space. For all vectors $x_1,\ldots, x_n$ of $V$ and $y_1,\ldots y_n$ linear functionals on $V$ define \begin{equation*} V(x,y)= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{i} y_i(x_{\sigma(i)}). \end{equation*}

It is easy to check that for fixed $y$'s this is a multilinear alternating form on the $x$'s. Conversely for fixed $x$'s it is an alternating multilinear form on the $y$'s. It follows that

\begin{align*} \det(A^t)V(x,y) &=V(x,A^ty)\\ &=\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{i} A^ty_i(x_{\sigma(i)})\\ &=\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{i} y_i(Ax_{\sigma(i)})\\ &=V(Ax,y) \\ &=\det(A)V(x,y) \end{align*} whence $\det(A)=\det(A^t)$ for properly chosen $x$ and $y$'s.

For the geometric interpretation we now replace the linear functionals $y$ by vectors and evaluation $y(x)$ by scalar product $(y,x)$. Furthermore we chose $x$ and $y$ to be equal to the canonical basis.

The above formula for $V(x,y)$ gives a decomposition of the volume spanned by the $Ax$ into a sum of volumes of 'hyperrectangles' whose sides are given by scalar products $(Ax_i,x_j)$. The exact same factors $(x_i,A^tx_j)$ and the same 'hyperrectangles' intervene when decomposing the volume spanned by the $A^t x$.

Hence the $Ax$ and the $A^tx$ span the same volume so that $A$ and $A^t$ have same determinant.

One can argue that it is difficult to see in a concrete example why the same factors appear. But then one should also say that it is difficult to see in a concrete example what the transpose of a transformation looks like, because of the indirect nature of its definition.


First, a non-technical summary: If a matrix $A$ transforms some (vector) space, the transpose (also known as dual) is telling you how "linear measurements on the space" are transformed. Its very definition is that to transform a measurement on something is to transform that something and then measure on the result (as witnessed by the formula $(A^Tv)^Tu=v^T(Au)$). Now, the volume in this dual "space of linear measurements" is given by evaluating various measurements on the tuples in the original space. So when one transforms the measurements and then computes the volume, it's by definition the same as measuring on the transformed inputs. Thus the result is scaled by the scaling of the volume of the inputs, i.e. the volume scalings in the two spaces are equal. Now let's make this mumbo-jumbo into math.


Given a way to evaluate volume in a vector space $V$ (i.e. a completely alternating multilinear form $\omega$ sending $v_1, ..., v_k$ to a number), the standard way to evaluate volume in the dual space $V^*$ is to pick any tuple $u_1, \ldots, u_n$ with non-zero volume $\omega(u_1, \ldots, u_n)$ and send $v_1^*, \ldots, v_n^*$ to

$$\omega^*(v_1^*, \ldots, v_n^*)=\frac{\sum_\sigma (-1)^{|\sigma|} \prod_i v^*_i(u_{\sigma(i)})}{\omega(u_1, \ldots, u_n)}.$$

Here the sum is over all permutations $\sigma$. (We can almost trivially check that as a function of the $u$s this is a ratio of two completely antisymmetric multilinear functions, hence a constant, showing that this is independent of the choice of $u$s. As a function of $v$s it is multilinear completely antisymmetric, so is indeed a volume, and if $v$s and $u$s are dual this gives the right answer, so this is indeed "the standard volume" in the dual.)

Now, for any automorphism $A:V\to V$, the dual map $A^*:V^*\to V^*$ has the defining property that $A^*(v)(u)=v(Au)$. Thus

$$\omega^*(A^*v_1^*, \ldots, A^*v_n^*)=\frac{\sum_\sigma (-1)^{|\sigma|} \prod_i A^*v^*_i(u_{\sigma(i)})}{\omega(u_1, \ldots, u_n)}=$$

$$=\frac{\sum_\sigma (-1)^{|\sigma|} \prod_i v^*_i(Au_{\sigma(i)})}{\omega(u_1, \ldots, u_n)}=$$ $$=\frac{\sum_\sigma (-1)^{|\sigma|} \prod_i v^*_i(Au_{\sigma(i)})}{\omega(Au_1, \ldots, Au_n)}\frac{\omega(Au_1, \ldots, Au_n)}{\omega(u_1, \ldots, u_n)}=$$

$$=\omega(v_1^*, \ldots, v_n^*)\frac{\omega(Au_1, \ldots, Au_n)}{\omega(u_1, \ldots, u_n)}.$$

In other words,

$$\frac{\omega^*(A^*v_1^*, \ldots, A^*v_n^*)}{\omega(v_1^*, \ldots, v_n^*)}=\frac{\omega(Au_1, \ldots, Au_n)}{\omega(u_1, \ldots, u_n)} -$$

the volume-stretching factors for $A$ and $A^*$ are the same. Of course, in coordinates, if $A$ is given by some matrix, $A^*$ is given by the transpose of that matrix, and the volume-stretching is the determinant. QED.


For any square matrix $A$, we have $PA=LDU$. Therefore, $A=P^{T}LDU$ (LU decomposition) where $P$ is permutation matrix, $D$ is diagonal matrix, $L$ is lower triangular matrix with diagonal is 1 or 0, $U$ is upper triangular matrix.

$|A^{T}|=|U^{T} D^{T} L^{T} P|=|U^{T}| |D^{T}| |L^{T}| |P|=|P||U^{T}||D^{T}||L^{T}|$

Because $P^{T}P=I$, then $|P^{T}||P|=1$. Therefore, we have $|P||U^{T}||D^{T}||L^{T}|=|P||U||D||L|=|P^{T}||U||D||L|=|A|$.


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