0

Suppose we are given a matrix

$$A = \begin{pmatrix} x & y \\ -y & x \end{pmatrix} $$

where $x,y \in \mathbb{R}$ and $x^2+y^2=1$.

Then is, $\textrm{tr}(A)$ not equal to $0$?

If yes, then how is this possible?

Ken
  • 3,751
Dwaipayan Gupta
  • 1
  • If $A = \left(\matrix{x&y\-y&x}\right)$ then the trace of $A$ is $2x$, the sum of the elements along the diagonal. – Arthur Jun 10 '15 at 19:54
  • $\operatorname{Tr}'A)=2x$, by definition. You may note $A$ is a rotation matrix around the origin in the plane. – Bernard Jun 10 '15 at 19:54

2 Answers2

1

If $$ A =\pmatrix{x & y \\ -y & x} $$ with $x^2 + y^2 = 1$, then the trace of $A$ is $$ \text{trace}(A) = x + x = 2x. $$ This is zero exactly when $x=0$ and this would mean $y = \pm 1$.

graydad
  • 14,077
Thomas
  • 43,555
1

The trace of a matrix is the sum of the entries along its main diagonal (the one from the top-left to bottom-right of the matrix) For the matrix you wrote down

$$ \text{trace}\begin{pmatrix}x & y \\ -y & x\end{pmatrix} = x + x = 2x, $$

which is $0$ only if $x = 0$.

However, it is true that

$$ \text{trace}\begin{pmatrix}x & y \\ y & -x\end{pmatrix} = x - x = 0 $$

for every choice of $x$ and $y$ you described. In general, the trace of a nonzero matrix need not be nonzero.

Dan
  • 7,951