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Question: Let $V$ be a vector space of $2\times 2$ matrices over a field $F$. Let $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\in V$ and let $T:V\rightarrow V$ be a linear transformation defined by $T(X)=A(X)$. Compute $\det(T)$.

My thoughts: So, $\det(A)=ad-bc$ (not sure if this will help). Let's take a $2\times 2$ matrix $X=\begin{bmatrix} e & f \\ g & h \end{bmatrix}$, and compute $AX$. So, $AX=\begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}$. So, since $T(x)=AX$, then wouldn't the determinant of $T(x)$ just be $(cf+dh)(ae+bg)-(ce+dg)(af+bh)$?

Bernard
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User7238
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  • Fix a basis and find the matrix of the transformation – Varadharajan R Sep 09 '21 at 17:57
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    @User Note that the determinant of $T$ is not the same thing as the determinant of $T(X)$. – Ben Grossmann Sep 09 '21 at 18:28
  • For $n\gt 2$ you may find the Kronecker Product to be interesting. E.g. it gives the identity $\text{vec}\big(ABC\big) = \big(C^T \otimes A\big)\text{vec}\big( B\big)$, so $T$ is isomorphic to $\big(I\otimes A\big)\text{vec}\big( X\big)$, and Kronecker products have easy to understand properties for e.g. determinants. – user8675309 Sep 09 '21 at 22:59

2 Answers2

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Given a matrix $X \in V$, you've computed $T(X)$. Now to represent matrices in $V$ as vectors, you can just reshape the $2 \times 2$ matrices as vectors in $\mathbb{R}^4$. Then the matrix representation of $T$ is the $4\times 4$ matrix $M$ such that $$M \begin{bmatrix} e \\ f \\ g \\ h\end{bmatrix} = \begin{bmatrix} ae+bg \\ af+bh \\ ce+dg \\ cf+dh\end{bmatrix}$$ and compute $\det M$.

(Note that different "reshapings" of a $2 \times 2$ matrix into a vector will lead to slightly different matrices $M$, but the determinant will always be the same because the determinant does not change under changes of basis.)

angryavian
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  • Ah, okay, so I just need to find a suitable matrix $M$, and then take the determinant of that $4\times 4$ matrix. When trying to find $M$, is there a "nifty" way of trying to find $M$, or is it best to just play around with things that "feel" are good, and then refine from there? – User7238 Sep 09 '21 at 20:04
  • @User7238 The entry $ae+bg$ will tell you what the first row of $M$ is; the entry $af+bh$ will tell you the second row, etc. – angryavian Sep 09 '21 at 20:25
  • So, I get that $M$ is the matrix with $0$ everywhere, but each entry of the diagonal is $\frac{1}{e}(ae+bg)$, $\frac{1}{f}(af+bh)$, $\frac{1}{g}(ce+dg)$, and $\frac{1}{h}(cf+dh)$, respectively. Then, since $M$ is diagonal, $det(M)$ is just the product of the four entries I have above.... Does that look good? – User7238 Sep 11 '21 at 23:52
  • @User7238 No, $M$ is $\begin{bmatrix}a & 0 & b & 0 \ 0 & a & 0 & b \ c & 0 & d & 0 \ 0 & c & 0 & d\end{bmatrix}$. – angryavian Sep 12 '21 at 00:05
  • Ah, I see. And then we get the determinant is $a^2d^2-2abcd+b^2c^2$ – User7238 Sep 12 '21 at 00:23
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We can start by the case that $A$ is diagonalisable and let $\lambda$ and $\mu$ be its eigen values and respectively $x$ and $y$ its eigen vectors. Since $(x,y)$ is a basis of $F^2$ then you can easily prove that $\left(xx^T, xy^T, yx^T, yy^T\right)$ is a basis of $V$. In this basis the matrix of $T(A)$ is : \begin{align}\begin{bmatrix}\lambda & 0 & 0 & 0\\ 0 & \lambda & 0 & 0\\ 0 & 0 & \mu & 0\\ 0 & 0 & 0 & \mu\end{bmatrix}\end{align} So $\det(T(A)) = \lambda^2 \mu^2 = \det(A)^2$.

This will hold also for every matrix $A$ since every matrix is a limit of diagonalisable matrices.

This method also can be easily generalized to the $n\times n$ case :

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

Kroki
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  • "Every matrix is a limit of diagonalizable matricies".... really? I've never heard of this before. – User7238 Sep 09 '21 at 20:02
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    Yes. But the diagonalization has to be in the algebraic closure of $F$. And for a proof of that you can see : https://math.stackexchange.com/questions/459522/density-of-diagonalizable-matrices – Kroki Sep 09 '21 at 20:07
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    This answer is problematic e.g. for the field $\mathbb F_p$. Algebraic closure uses a lot of machinery and explaining what you mean by limits over a field with positive characteristic is also an issue. – user8675309 Sep 09 '21 at 22:54