What is an idempotent matrix?

2 answers:

7 0

$An\ idempotent\ matrix\ is\ a\ matrix\ n\times n\ (square\ matrix\ A)\ which:A^2=A\\\\Example:\\ A=\left[\begin{array}{ccc}2&-1\\2&-1\end{array}\right]\\A^2=\left[\begin{array}{ccc}2&-1\\2&-1\end{array}\right]\cdot\left[\begin{array}{ccc}2&-1\\2&-1\end{array}\right]=\left[\begin{array}{ccc}2\cdot2-1\cdot2&2\cdot(-1)-1\cdot(-1)\\2\cdot2-1\cdot2&2\cdot(-1)-1\cdot(-1)\end{array}\right]\\= \left[\begin{array}{ccc}4-2&-2+1\\4-2&-2+1\end{array}\right] = \left[\begin{array}{ccc}2&-1\\2&-1\end{array}\right] =A$

Fofino [41]3 years ago

7 0

A matrix E satisfying the equation E2 = E.

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If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.