I am trying to understand the role of determinant on matrix-vector multiplication. I am reading Math Insight in order to get some intuition regarding the determinant's role.

Let's take 2x2 matrix A (as in the blog)

A = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}

According to the blog, the determinant's role can be seen by multiplying TWO vectors (that span a basis) by the matrix. And in this case the determinant shows how the square of the parallelogram formed by the original vectors and by the transformed vectors changes.

x=\begin{bmatrix} 1 \\ 0 \end{bmatrix}

y=\begin{bmatrix} 0 \\ 1 \end{bmatrix}

The parallelogram formed by x and y has a square of 1. And the parallelogram formed by Ax and Ay has a square of det(A) = 4.

But what the role of determinant when you have just one vector x? How the determinant influences the result of Ax?

  • $\begingroup$ The determinant tells you how volume (and orientation) is altered by the transformation. Say the determinant is $1$. Then the volume is unchanged by the transformation. If the determinant is $>1$ then the volume will increase, less than $1$ and it will decrease. If it is negative, the orientation will reverse. But say the determinant is $10$, then in some directions this may lengthen the vector, in others it may shrink it. If you chose vectors at random, you would expect them to lengthen. $\endgroup$ – Alex Clark Nov 25 '17 at 9:40

Quite honestly, I think the direct relation between determinant and matrix-vector multiplication that you are looking for does not exist. There will certainly be ways to relate the one to the other, and if anyone then it will be the smart people of this site who will go out of their way to find them, but I fear that most of these possibly indirect relations might not be satisfying for you.

As pointed out in the comments and the website you refer to, the determinant of a matrix affects volume. However for a single vector, this could mean all kinds of things. Take for example $$ A = \begin{pmatrix} 9 & 0 \\ 0 & \tfrac19 \end{pmatrix} $$ which has determinant $1$ as can be seen by how it maps the unit cube to the $9$ by $\frac19$ rectangle, which has area $1$ - but its effect on the two vectors \begin{align*} x_1&=\begin{pmatrix} 1\\0 \end{pmatrix} & x_2&=\begin{pmatrix} 0\\1\end{pmatrix} \end{align*} is quite different: One is stretched ninefold, the other is compressed ninefold.

The one thing you can say is that $\det(A)=0$ is equivalent to the existence of a vector $v$ with $Av=0$, and also equivalent to the fact that there is some vector $w$ such that $\forall v: Av \ne w$. This is because $\det(A)\ne0$ is equivalent to $A$ representing an invertible linear transformation.

To be more precise in my response, let me give you the following

Proposition. Let $\Bbbk$ be a field. For every nonzero $a\in\Bbbk$, every $n\in\Bbb N$, $n\ge 2$ and any two nonzero vectors $v,w\in\Bbbk^n$, there is a matrix $A\in\Bbbk^{n\times n}$ with $\det(A)=a$ and $Av=w$.

If you are interested, I can write a proof for this, but it might also be a good exercise. The prosaic phrasing of this proposition says that regardless of the determinant (excluding zero) you can have at least one vector that is mapped any way you like (except for sending it to zero).


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