# Quod Erat Demonstrandum

## 2010/11/18

### 唔打：矩陣簡單例子

Filed under: Pure Mathematics — johnmayhk @ 11:17 下午

## 5 則迴響 »

1. Extended question: Let A be the 7 by 7 matrix in your exercise.

(a) What are the eigenvalues of A.

(b) What are the eigenvalues of A^{2010}.

迴響 由 apook — 2010/11/19 @ 2:26 下午 | 回應

2. FYI(for other readers who have not heard of eigenvalues):

t is an eigenvalue of A iff det(A – tI) = 0.

迴響 由 apook — 2010/11/19 @ 2:28 下午 | 回應

3. 複平面上，單位圓內接正七邊形之頂點（其中一頂點為 z=1）。

迴響 由 johnmayhk — 2010/11/19 @ 5:44 下午 | 回應

4. I am more interested in how you get your answer? Did you compute det(A – tI)?

迴響 由 apook — 2010/11/22 @ 10:42 上午 | 回應

• I guess there are many ways to obtain the result

$\det(A - \lambda I) = 1 - \lambda^7$

For me, I just observe that every column in $A$ contains 2 non-zero entries (1 and $-\lambda$).

Just consider the definition of deteminant.

When taking the entry $-\lambda$ in $C_1$, then we must also take $-\lambda$‘s from other columns, obtaining $(-\lambda)^7$ as one of the non-zero terms in sum of $\det A$.

When taking the entry 1 in $C_1$, it is easy to see that it is a must to take 1’s from other columns, obataining the term $1^7$ as one of the non-zero terms.

And all other terms in the sum are zeros, hence result follows.

::::::::::

For not computing $\det(A - \lambda I)$, how about:

$A\overrightarrow{v} = \lambda \overrightarrow{v}$ (where $\overrightarrow{v}$ is an eigenvector corresponding to $\lambda$)

yield

$\overrightarrow{v} = A^7\overrightarrow{v} = \lambda^7v$

$(1 - \lambda^7)\overrightarrow{v} = \overrightarrow{0}$

thus

$1 - \lambda^7 = 0$

迴響 由 johnmayhk — 2010/11/22 @ 4:21 下午 | 回應