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I am confused about this homework question.

It says "Show that :

$\det(A) = \frac 12 \begin{vmatrix}\operatorname{tr}(A)&1\\\operatorname{tr}(A^2)& \operatorname{tr}(A)\end{vmatrix}$

for every $2\times 2$ matrix."

I am not sure how to "show" this. I'm guessing it means a proof of sorts.

But I am also confused as to why the $\det(A) = 1/2$ times a matrix. I'm pretty sure determinants are just a number, not a matrix.

Any tips would be great. Thanks in advance.

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    $\begingroup$ In mathematical writing, "show that..." is used synonymously with "prove that...". $\endgroup$
    – Srivatsan
    Nov 1, 2011 at 19:55
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    $\begingroup$ The right hand side is not a matrix. Is a determinant, and therefore a number. You have $\left| \begin{matrix}a&b\\ c&d \end{matrix} \right|=ad-bc$. $\endgroup$ Nov 1, 2011 at 20:06

5 Answers 5

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Here is a (requested) generalization for larger dimensions: If A is an n×n matrix, then one has the following expression for the determinant of A in terms of the determinant of a matrix whose entries are traces of powers of A:

$$\newcommand{\tr}{\operatorname{tr}} \newcommand{\det}{\operatorname{det}} \det(A) = \frac{1}{n!}\left|\begin{array}{cccccccc} \tr(A) & 1 & . & . & . & \dots & . & . \\ \tr(A^2) & \tr(A) & 2 & . & . & \dots & . & . \\ \tr(A^3) & \tr(A^2) & \tr(A) & 3 & . & \dots & . & . \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ \tr(A^{n-1}) & \tr(A^{n-2}) & \tr(A^{n-3}) & \tr(A^{n-4}) & \tr(A^{n-5}) & \dots & \tr(A) & n-1 \\ \tr(A^n) & \tr(A^{n-1}) & \tr(A^{n-2}) & \tr(A^{n-3}) & \tr(A^{n-4}) & \dots & \tr(A^2) & \tr(A) \end{array}\right|$$

In particular, we have for $3\times3$ that: $$\det(A) = \frac{1}{6}\left|\begin{array}{cccc} \tr(A) & 1 & 0 \\ \tr(A^2) & \tr(A) & 2 \\ \tr(A^3) & \tr(A^2) & \tr(A) \end{array}\right|$$

The matrix on the right is defined in general by: $$B_{ij} = \begin{cases} i & \text{ if } j = i + 1 \\ \tr(A^{i-j+1}) & \text{ if } j \leq i \\ 0 & \text{ if } j > i+1 \end{cases}$$

I suggest using this recursively to create even more complicated formulas (all in lower Hessenberg form!) to be entered into the big book of bad algorithms.

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    $\begingroup$ +1, I was also eager to know the generalization. Its really a pretty one. $\endgroup$
    – Tapu
    Nov 2, 2011 at 5:29
  • $\begingroup$ This is indeed very interesting. Could you please give a reference where it may have been proved ? Thanks $\endgroup$
    – user90041
    May 14, 2014 at 21:13
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Assuming you are familiar with Cayley–Hamilton theorem, this may be an alternative way:

Any $2\times2$ matrix satisfies $A^2-\text{tr}(A)A+\det(A)I_2=0 $. Taking trace of both sides, $\begin{align*}&\text{tr}(A^2)-\text{tr}(A)\text{tr}(A)+2\det(A)=0 \Rightarrow\\ &\det(A)=\frac12[\text{tr}(A)\text{tr}(A)-\text{tr}(A^2)]=\frac12\begin{vmatrix}\text{tr}(A)&1\\\text{tr}(A^2)&\text{tr}(A)\end{vmatrix}\end{align*}$

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  • $\begingroup$ Very nice. It is a nice trick I encountered several times to trace the Cayley Hamilton equation. $\endgroup$ Nov 1, 2011 at 20:08
  • $\begingroup$ This is one of the more ellegant ways to prove the result. @OP: I would highly recommend to have a look at this prove. Because it does not only solve the problem but also offers some insight into the (non-obvious) usefulness of the Cayley-Hamilton-theorem which is one of the most important theorems in basic linear algebra. $\endgroup$ Nov 1, 2011 at 20:26
  • $\begingroup$ Thanks! I just remember a quite similar question (taken from NBHM test, 2010): "Find out all (or at least one, I forgot) $3\times3$ non-diagonal real matrices $A$ such that $A^3=I$". In this case CHT make the problem easy to solve. $\endgroup$
    – Tapu
    Nov 1, 2011 at 20:46
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The "|" to the left and the right of your matrix probably indicate that you should take the determinant of this matrix, too. It is not completly uncommon to write $|A|$ instead of $det(A)$.

To prove the identity just take an abstract $2 \times 2$-Matrix and compute the left-hand and the right-hand side of the equation.

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You could use an eigenvalue formulation, since if $\lambda_1,\lambda_2$ are the eigenvalues of $A$ then $\det(A)=\lambda_1 \lambda_2$, $tr A=\lambda_1+\lambda_2$, $tr(A^2)=\lambda_1^2+\lambda_2^2$. Plug these in your equation and obvious equality holds.

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The RHS is determinant, i.e., $1/2[(tr A)^2-tr(A^2)]=1/2[(a_{11}+a_{22})^2-(a_{11}^2+2a_{12}a_{21}+a_{22}^2)]=\cdots$

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